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## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x+12)(x- 1)2
= x4 + 16x3 + 37x2 -126x + 72  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+12)(x- 1)2 = x4 + 16x3 + 37x2 -126x + 72

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-13,1372), (-10,-968), (-7,-320), (-4,400), (-1,220), (2,112), (5,2992) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = -12.
There is a double root at x = 1. Not just the function but also its first derivative are zero at this point.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x+12)(x- 1)2 = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 48x2 + 74x -126
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 48x2 + 74x -126 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-9.7, -3.2, 1.0]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 96x2 + 37x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -7.14 and -0.86. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x- 12)(x- 1)2
= x4 -8x3 -59x2 + 138x -72  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x- 12)(x- 1)2 = x4 -8x3 -59x2 + 138x -72

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,1216), (-4,-800), (-1,-260), (2,-80), (5,-1232), (8,-2744), (11,-1700), (14,6760), (17,29440) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = 12.
There is a double root at x = 1. Not just the function but also its first derivative are zero at this point.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x- 12)(x- 1)2 = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 -24x2 -118x + 138
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 -24x2 -118x + 138 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-3.8, 1.0, 8.9]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 -48x2 -59x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -1.72 and 5.72. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x+12)(x+13)(x- 1)
= x4 + 30x3 + 275x2 + 630x -936  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+12)(x+13)(x- 1) = x4 + 30x3 + 275x2 + 630x -936

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-14,240), (-11,120), (-8,360), (-5,-336), (-2,-1320), (1,0), (4,8160) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = -12.
There is a single, unique root at x = -13.
There is a single, unique root at x = 1.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x+12)(x+13)(x- 1) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 90x2 + 550x + 630
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 90x2 + 550x + 630 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-12.5, -8.4, -1.4]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 180x2 + 275x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -10.73 and -4.27. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x- 12)(x+13)(x- 1)
= x4 + 6x3 -157x2 -786x + 936  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x- 12)(x+13)(x- 1) = x4 + 6x3 -157x2 -786x + 936

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-14,3120), (-11,-2760), (-8,-1800), (-5,816), (-2,1848), (1,0), (4,-4080), (7,-7800), (10,-6624), (13,5928), (16,38280) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = 12.
There is a single, unique root at x = -13.
There is a single, unique root at x = 1.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x- 12)(x+13)(x- 1) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 18x2 -314x -786
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 18x2 -314x -786 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-10.2, -2.3, 8.2]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 36x2 -157x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -6.83 and 3.83. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x+12)(x+13)(x- 4)
= x4 + 27x3 + 182x2 -288x -3744  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+12)(x+13)(x- 4) = x4 + 27x3 + 182x2 -288x -3744

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-14,288), (-11,150), (-8,480), (-5,-504), (-2,-2640), (1,-3822), (4,0), (7,14820) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = -12.
There is a single, unique root at x = -13.
There is a single, unique root at x = 4.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x+12)(x+13)(x- 4) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 81x2 + 364x -288
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 81x2 + 364x -288 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-12.5, -8.4, 0.7]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 162x2 + 182x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -10.65 and -2.85. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x- 12)(x+13)(x- 4)
= x4 + 3x3 -178x2 -336x + 3744  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x- 12)(x+13)(x- 4) = x4 + 3x3 -178x2 -336x + 3744

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-14,3744), (-11,-3450), (-8,-2400), (-5,1224), (-2,3696), (1,3234), (4,0), (7,-3900), (10,-4416), (13,4446), (16,30624) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = 12.
There is a single, unique root at x = -13.
There is a single, unique root at x = 4.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x- 12)(x+13)(x- 4) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 9x2 -356x -336
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 9x2 -356x -336 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-10.1, -0.9, 8.9]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 18x2 -178x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -6.25 and 4.75. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x+12)(x+13)(x+5)
= x4 + 36x3 + 461x2 + 2466x + 4680  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+12)(x+13)(x+5) = x4 + 36x3 + 461x2 + 2466x + 4680

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-14,144), (-11,60), (-8,120), (-5,0), (-2,1320) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = -12.
There is a single, unique root at x = -13.
There is a single, unique root at x = -5.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x+12)(x+13)(x+5) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 108x2 + 922x + 2466
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 108x2 + 922x + 2466 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-12.5, -9.0, -5.4]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 216x2 + 461x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -11.04 and -6.96. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x- 12)(x+13)(x+5)
= x4 + 12x3 -115x2 -1686x -4680  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x- 12)(x+13)(x+5) = x4 + 12x3 -115x2 -1686x -4680

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-14,1872), (-11,-1380), (-8,-600), (-5,0), (-2,-1848), (1,-6468), (4,-12240), (7,-15600), (10,-11040), (13,8892), (16,53592) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = 12.
There is a single, unique root at x = -13.
There is a single, unique root at x = -5.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x- 12)(x+13)(x+5) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 36x2 -230x -1686
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 36x2 -230x -1686 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-10.6, -5.4, 7.2]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 72x2 -115x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -8.31 and 2.31. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x+12)(x- 1)(x- 4)
= x4 + 13x3 -14x2 -288x + 288  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+12)(x- 1)(x- 4) = x4 + 13x3 -14x2 -288x + 288

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-13,1666), (-10,-1232), (-7,-440), (-4,640), (-1,550), (2,-224), (5,748), (8,7840) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = -12.
There is a single, unique root at x = 1.
There is a single, unique root at x = 4.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x+12)(x- 1)(x- 4) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 39x2 -28x -288
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 39x2 -28x -288 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-9.7, -2.7, 2.8]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 78x2 -14x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -6.84 and 0.34. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x- 12)(x- 1)(x- 4)
= x4 -11x3 -38x2 + 336x -288  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x- 12)(x- 1)(x- 4) = x4 -11x3 -38x2 + 336x -288

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,1672), (-4,-1280), (-1,-650), (2,160), (5,-308), (8,-1568), (11,-1190), (14,5200), (17,23920) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = 12.
There is a single, unique root at x = 1.
There is a single, unique root at x = 4.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x- 12)(x- 1)(x- 4) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 -33x2 -76x + 336
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 -33x2 -76x + 336 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-3.5, 2.6, 9.4]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 -66x2 -38x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -0.98 and 6.48. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x+12)(x- 15)(x- 1)
= x4 + 2x3 -201x2 -882x + 1080  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+12)(x- 15)(x- 1) = x4 + 2x3 -201x2 -882x + 1080

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-13,2744), (-10,-2200), (-7,-880), (-4,1520), (-1,1760), (2,-1456), (5,-7480), (8,-13720), (11,-15640), (14,-6760), (17,21344), (20,79040) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = -12.
There is a single, unique root at x = 15.
There is a single, unique root at x = 1.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x+12)(x- 15)(x- 1) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 6x2 -402x -882
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 6x2 -402x -882 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-9.5, -2.2, 10.4]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 12x2 -201x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -6.31 and 5.31. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x- 12)(x- 15)(x- 1)
= x4 -22x3 + 39x2 + 1062x -1080  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x- 12)(x- 15)(x- 1) = x4 -22x3 + 39x2 + 1062x -1080

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,3344), (-4,-3040), (-1,-2080), (2,1040), (5,3080), (8,2744), (11,680), (14,-520), (17,3680), (20,19760) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = 12.
There is a single, unique root at x = 15.
There is a single, unique root at x = 1.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x- 12)(x- 15)(x- 1) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 -66x2 + 78x + 1062
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 -66x2 + 78x + 1062 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-3.2, 6.1, 13.7]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 -132x2 + 39x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 0.63 and 10.37. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x+12)(x- 15)(x- 4)
= x4 -1x3 -210x2 -288x + 4320  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+12)(x- 15)(x- 4) = x4 -1x3 -210x2 -288x + 4320

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-13,3332), (-10,-2800), (-7,-1210), (-4,2432), (-1,4400), (2,2912), (5,-1870), (8,-7840), (11,-10948), (14,-5200), (17,17342), (20,66560) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = -12.
There is a single, unique root at x = 15.
There is a single, unique root at x = 4.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x+12)(x- 15)(x- 4) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 -3x2 -420x -288
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 -3x2 -420x -288 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-9.5, -0.6, 11.0]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 -6x2 -210x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -5.67 and 6.17. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x- 12)(x- 15)(x- 4)
= x4 -25x3 + 102x2 + 1008x -4320  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x- 12)(x- 15)(x- 4) = x4 -25x3 + 102x2 + 1008x -4320

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,4598), (-4,-4864), (-1,-5200), (2,-2080), (5,770), (8,1568), (11,476), (14,-400), (17,2990), (20,16640) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = 12.
There is a single, unique root at x = 15.
There is a single, unique root at x = 4.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x- 12)(x- 15)(x- 4) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 -75x2 + 204x + 1008
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 -75x2 + 204x + 1008 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-2.4, 7.6, 13.7]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 -150x2 + 102x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 1.55 and 10.95. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x+12)(x- 15)(x+5)
= x4 + 8x3 -183x2 -2070x -5400  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+12)(x- 15)(x+5) = x4 + 8x3 -183x2 -2070x -5400

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-13,1568), (-10,-1000), (-7,-220), (-4,-304), (-1,-3520), (2,-10192), (5,-18700), (8,-25480), (11,-25024), (14,-9880), (17,29348), (20,104000) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = -12.
There is a single, unique root at x = 15.
There is a single, unique root at x = -5.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x+12)(x- 15)(x+5) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 24x2 -366x -2070
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 24x2 -366x -2070 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-9.9, -5.4, 9.5]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 48x2 -183x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -7.87 and 3.87. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x- 12)(x- 15)(x+5)
= x4 -16x3 -87x2 + 1170x + 5400  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x- 12)(x- 15)(x+5) = x4 -16x3 -87x2 + 1170x + 5400

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,836), (-4,608), (-1,4160), (2,7280), (5,7700), (8,5096), (11,1088), (14,-760), (17,5060), (20,26000) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = 12.
There is a single, unique root at x = 15.
There is a single, unique root at x = -5.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x- 12)(x- 15)(x+5) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 -48x2 -174x + 1170
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 -48x2 -174x + 1170 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-5.5, 3.9, 13.7]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 -96x2 -87x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -1.52 and 9.52. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x+12)(x- 1)(x+5)
= x4 + 22x3 + 139x2 + 198x -360  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+12)(x- 1)(x+5) = x4 + 22x3 + 139x2 + 198x -360

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-13,784), (-10,-440), (-7,-80), (-4,-80), (-1,-440), (2,784), (5,7480) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = -12.
There is a single, unique root at x = 1.
There is a single, unique root at x = -5.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x+12)(x- 1)(x+5) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 66x2 + 278x + 198
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 66x2 + 278x + 198 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-10.1, -5.5, -0.8]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 132x2 + 139x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -8.16 and -2.84. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x- 12)(x- 1)(x+5)
= x4 -2x3 -101x2 -258x + 360  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x- 12)(x- 1)(x+5) = x4 -2x3 -101x2 -258x + 360

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,304), (-4,160), (-1,520), (2,-560), (5,-3080), (8,-5096), (11,-2720), (14,9880), (17,40480) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = 12.
There is a single, unique root at x = 1.
There is a single, unique root at x = -5.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x- 12)(x- 1)(x+5) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 -6x2 -202x -258
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 -6x2 -202x -258 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-5.5, -1.3, 8.5]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 -12x2 -101x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -3.63 and 4.63. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)2(x+12)(x- 1)
= x4 + 23x3 + 156x2 + 252x -432  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)2(x+12)(x- 1) = x4 + 23x3 + 156x2 + 252x -432

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-13,686), (-10,-352), (-7,-40), (-4,-160), (-1,-550), (2,896), (5,8228) Let us inspect the roots of the given polynomial function.
There is a double root at x = -6. Not just the function but also its first derivative are zero at this point.
There is a single, unique root at x = -12.
There is a single, unique root at x = 1.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)2(x+12)(x- 1) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 69x2 + 312x + 252
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 69x2 + 312x + 252 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-10.2, -6.0, -1.0]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 138x2 + 156x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -8.41 and -3.09. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)2(x- 12)(x- 1)
= x4 -1x3 -108x2 -324x + 432  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)2(x- 12)(x- 1) = x4 -1x3 -108x2 -324x + 432

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,152), (-4,320), (-1,650), (2,-640), (5,-3388), (8,-5488), (11,-2890), (14,10400), (17,42320) Let us inspect the roots of the given polynomial function.
There is a double root at x = -6. Not just the function but also its first derivative are zero at this point.
There is a single, unique root at x = 12.
There is a single, unique root at x = 1.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)2(x- 12)(x- 1) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 -3x2 -216x -324
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 -3x2 -216x -324 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-6.0, -1.6, 8.4]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 -6x2 -108x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -4.0 and 4.5. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)2(x+12)(x- 4)
= x4 + 20x3 + 84x2 -288x -1728  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)2(x+12)(x- 4) = x4 + 20x3 + 84x2 -288x -1728

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-13,833), (-10,-448), (-7,-55), (-4,-256), (-1,-1375), (2,-1792), (5,2057), (8,15680) Let us inspect the roots of the given polynomial function.
There is a double root at x = -6. Not just the function but also its first derivative are zero at this point.
There is a single, unique root at x = -12.
There is a single, unique root at x = 4.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)2(x+12)(x- 4) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 60x2 + 168x -288
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 60x2 + 168x -288 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-10.1, -6.0, 1.2]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 120x2 + 84x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -8.32 and -1.68. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)2(x- 12)(x- 4)
= x4 -4x3 -108x2 + 1728  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)2(x- 12)(x- 4) = x4 -4x3 -108x2 + 1728

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,209), (-4,512), (-1,1625), (2,1280), (5,-847), (8,-3136), (11,-2023), (14,8000), (17,34385) Let us inspect the roots of the given polynomial function.
There is a double root at x = -6. Not just the function but also its first derivative are zero at this point.
There is a single, unique root at x = 12.
There is a single, unique root at x = 4.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)2(x- 12)(x- 4) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 -12x2 -216x + 0
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 -12x2 -216x + 0 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-6.0, 0.0, 9.0]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 -24x2 -108x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -3.36 and 5.36. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)2(x+12)(x+5)
= x4 + 29x3 + 300x2 + 1332x + 2160  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)2(x+12)(x+5) = x4 + 29x3 + 300x2 + 1332x + 2160

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-13,392), (-10,-160), (-7,-10), (-4,32), (-1,1100) Let us inspect the roots of the given polynomial function.
There is a double root at x = -6. Not just the function but also its first derivative are zero at this point.
There is a single, unique root at x = -12.
There is a single, unique root at x = -5.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)2(x+12)(x+5) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 87x2 + 600x + 1332
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 87x2 + 600x + 1332 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-10.4, -6.0, -5.3]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 174x2 + 300x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -8.85 and -5.65. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)2(x- 12)(x+5)
= x4 + 5x3 -108x2 -972x -2160  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)2(x- 12)(x+5) = x4 + 5x3 -108x2 -972x -2160

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,38), (-4,-64), (-1,-1300), (2,-4480), (5,-8470), (8,-10192), (11,-4624), (14,15200), (17,58190) Let us inspect the roots of the given polynomial function.
There is a double root at x = -6. Not just the function but also its first derivative are zero at this point.
There is a single, unique root at x = 12.
There is a single, unique root at x = -5.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)2(x- 12)(x+5) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 15x2 -216x -972
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 15x2 -216x -972 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-6.0, -5.3, 7.6]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 30x2 -108x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -5.67 and 3.17. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x+12)(x- 8)(x- 1)
= x4 + 9x3 -82x2 -504x + 576  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+12)(x- 8)(x- 1) = x4 + 9x3 -82x2 -504x + 576

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-13,2058), (-10,-1584), (-7,-600), (-4,960), (-1,990), (2,-672), (5,-2244), (8,0), (11,11730) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = -12.
There is a single, unique root at x = 8.
There is a single, unique root at x = 1.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x+12)(x- 8)(x- 1) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 27x2 -164x -504
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 27x2 -164x -504 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-9.6, -2.4, 5.4]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 54x2 -82x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -6.58 and 2.08. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x- 12)(x- 8)(x- 1)
= x4 -15x3 -10x2 + 600x -576  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x- 12)(x- 8)(x- 1) = x4 -15x3 -10x2 + 600x -576

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,2280), (-4,-1920), (-1,-1170), (2,480), (5,924), (8,0), (11,-510), (14,3120), (17,16560) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = 12.
There is a single, unique root at x = 8.
There is a single, unique root at x = 1.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x- 12)(x- 8)(x- 1) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 -45x2 -20x + 600
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 -45x2 -20x + 600 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-3.3, 4.3, 10.4]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 -90x2 -10x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -0.22 and 7.72. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x+12)(x- 8)(x- 4)
= x4 + 6x3 -112x2 -288x + 2304  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+12)(x- 8)(x- 4) = x4 + 6x3 -112x2 -288x + 2304

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-13,2499), (-10,-2016), (-7,-825), (-4,1536), (-1,2475), (2,1344), (5,-561), (8,0), (11,8211) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = -12.
There is a single, unique root at x = 8.
There is a single, unique root at x = 4.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x+12)(x- 8)(x- 4) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 18x2 -224x -288
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 18x2 -224x -288 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-9.5, -1.2, 6.3]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 36x2 -112x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -6.07 and 3.07. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x- 12)(x- 8)(x- 4)
= x4 -18x3 + 32x2 + 672x -2304  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x- 12)(x- 8)(x- 4) = x4 -18x3 + 32x2 + 672x -2304

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,3135), (-4,-3072), (-1,-2925), (2,-960), (5,231), (8,0), (11,-357), (14,2400), (17,13455) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = 12.
There is a single, unique root at x = 8.
There is a single, unique root at x = 4.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x- 12)(x- 8)(x- 4) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 -54x2 + 64x + 672
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 -54x2 + 64x + 672 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-2.7, 5.9, 10.5]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 -108x2 + 32x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) 0.64 and 8.36. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x+12)(x- 8)(x+5)
= x4 + 15x3 -22x2 -936x -2880  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+12)(x- 8)(x+5) = x4 + 15x3 -22x2 -936x -2880

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-13,1176), (-10,-720), (-7,-150), (-4,-192), (-1,-1980), (2,-4704), (5,-5610), (8,0), (11,18768) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = -12.
There is a single, unique root at x = 8.
There is a single, unique root at x = -5.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x+12)(x- 8)(x+5) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 45x2 -44x -936
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 45x2 -44x -936 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-10.0, -5.4, 4.3]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 90x2 -22x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -7.96 and 0.46. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x- 12)(x- 8)(x+5)
= x4 -9x3 -94x2 + 456x + 2880  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x- 12)(x- 8)(x+5) = x4 -9x3 -94x2 + 456x + 2880

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,570), (-4,384), (-1,2340), (2,3360), (5,2310), (8,0), (11,-816), (14,4560), (17,22770) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = 12.
There is a single, unique root at x = 8.
There is a single, unique root at x = -5.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x- 12)(x- 8)(x+5) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 -27x2 -188x + 456
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 -27x2 -188x + 456 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-5.5, 2.1, 10.3]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 -54x2 -94x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -2.3 and 6.8. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x+4)(x- 1)2
= x4 + 8x3 + 5x2 -38x + 24  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+4)(x- 1)2 = x4 + 8x3 + 5x2 -38x + 24

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,192), (-4,0), (-1,60), (2,48), (5,1584) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = -4.
There is a double root at x = 1. Not just the function but also its first derivative are zero at this point.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x+4)(x- 1)2 = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 24x2 + 10x -38
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 24x2 + 10x -38 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-5.1, -1.8, 1.0]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 48x2 + 5x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -3.78 and -0.22. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x- 4)(x- 1)2
= x4 -27x2 + 50x -24  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x- 4)(x- 1)2 = x4 -27x2 + 50x -24

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,704), (-4,-400), (-1,-100), (2,-16), (5,176), (8,2744) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = 4.
There is a double root at x = 1. Not just the function but also its first derivative are zero at this point.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x- 4)(x- 1)2 = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 -54x + 50
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 -54x + 50 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-4.0, 1.0, 3.1]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 -27x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -2.12 and 2.12. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x+4)(x+13)(x- 1)
= x4 + 22x3 + 131x2 + 158x -312  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+4)(x+13)(x- 1) = x4 + 22x3 + 131x2 + 158x -312

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-14,1200), (-11,-840), (-8,-360), (-5,48), (-2,-264), (1,0), (4,4080) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = -4.
There is a single, unique root at x = -13.
There is a single, unique root at x = 1.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x+4)(x+13)(x- 1) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 66x2 + 262x + 158
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 66x2 + 262x + 158 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-10.7, -5.0, -0.7]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 132x2 + 131x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -8.4 and -2.6. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x- 4)(x+13)(x- 1)
= x4 + 14x3 -13x2 -314x + 312  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x- 4)(x+13)(x- 1) = x4 + 14x3 -13x2 -314x + 312

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-14,2160), (-11,-1800), (-8,-1080), (-5,432), (-2,792), (1,0), (4,0), (7,4680) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = 4.
There is a single, unique root at x = -13.
There is a single, unique root at x = 1.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x- 4)(x+13)(x- 1) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 42x2 -26x -314
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 42x2 -26x -314 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-10.3, -2.7, 2.7]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 84x2 -13x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -7.3 and 0.3. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x+4)(x+13)(x- 4)
= x4 + 19x3 + 62x2 -304x -1248  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+4)(x+13)(x- 4) = x4 + 19x3 + 62x2 -304x -1248

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-14,1440), (-11,-1050), (-8,-480), (-5,72), (-2,-528), (1,-1470), (4,0), (7,8580) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = -4.
There is a single, unique root at x = -13.
There is a single, unique root at x = 4.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x+4)(x+13)(x- 4) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 57x2 + 124x -304
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 57x2 + 124x -304 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-10.6, -4.9, 1.5]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 114x2 + 62x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -8.25 and -1.25. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x- 4)2(x+13)
= x4 + 11x3 -58x2 -320x + 1248  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x- 4)2(x+13) = x4 + 11x3 -58x2 -320x + 1248

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-14,2592), (-11,-2250), (-8,-1440), (-5,648), (-2,1584), (1,882), (4,0), (7,2340) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a double root at x = 4. Not just the function but also its first derivative are zero at this point.
There is a single, unique root at x = -13.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x- 4)2(x+13) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 33x2 -116x -320
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 33x2 -116x -320 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-10.3, -1.9, 4.0]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 66x2 -58x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -6.9 and 1.4. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x+4)(x+13)(x+5)
= x4 + 28x3 + 269x2 + 1082x + 1560  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+4)(x+13)(x+5) = x4 + 28x3 + 269x2 + 1082x + 1560

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-14,720), (-11,-420), (-8,-120), (-5,0), (-2,264), (1,2940) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = -4.
There is a single, unique root at x = -13.
There is a single, unique root at x = -5.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x+4)(x+13)(x+5) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 84x2 + 538x + 1082
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 84x2 + 538x + 1082 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-11.0, -5.5, -4.4]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 168x2 + 269x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -9.04 and -4.96. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x- 4)(x+13)(x+5)
= x4 + 20x3 + 77x2 -302x -1560  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x- 4)(x+13)(x+5) = x4 + 20x3 + 77x2 -302x -1560

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-14,1296), (-11,-900), (-8,-360), (-5,0), (-2,-792), (1,-1764), (4,0), (7,9360) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = 4.
There is a single, unique root at x = -13.
There is a single, unique root at x = -5.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x- 4)(x+13)(x+5) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 60x2 + 154x -302
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 60x2 + 154x -302 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-10.7, -5.4, 1.3]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 120x2 + 77x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -8.49 and -1.51. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x+4)(x- 1)(x- 4)
= x4 + 5x3 -22x2 -80x + 96  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+4)(x- 1)(x- 4) = x4 + 5x3 -22x2 -80x + 96

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,264), (-4,0), (-1,150), (2,-96), (5,396), (8,4704) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = -4.
There is a single, unique root at x = 1.
There is a single, unique root at x = 4.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x+4)(x- 1)(x- 4) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 15x2 -44x -80
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 15x2 -44x -80 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-5.1, -1.3, 2.8]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 30x2 -22x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -3.54 and 1.04. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x- 4)2(x- 1)
= x4 -3x3 -30x2 + 128x -96  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x- 4)2(x- 1) = x4 -3x3 -30x2 + 128x -96

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,968), (-4,-640), (-1,-250), (2,32), (5,44), (8,1568) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a double root at x = 4. Not just the function but also its first derivative are zero at this point.
There is a single, unique root at x = 1.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x- 4)2(x- 1) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 -9x2 -60x + 128
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 -9x2 -60x + 128 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-3.8, 2.1, 4.0]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 -18x2 -30x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -1.61 and 3.11. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x+4)(x- 15)(x- 1)
= x4 -6x3 -121x2 -234x + 360  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+4)(x- 15)(x- 1) = x4 -6x3 -121x2 -234x + 360

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,528), (-4,0), (-1,480), (2,-624), (5,-3960), (8,-8232), (11,-10200), (14,-4680), (17,15456), (20,59280) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = -4.
There is a single, unique root at x = 15.
There is a single, unique root at x = 1.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x+4)(x- 15)(x- 1) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 -18x2 -242x -234
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 -18x2 -242x -234 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-5.1, -1.0, 10.7]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 -36x2 -121x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -3.23 and 6.23. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x- 4)(x- 15)(x- 1)
= x4 -14x3 -41x2 + 414x -360  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x- 4)(x- 15)(x- 1) = x4 -14x3 -41x2 + 414x -360

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,1936), (-4,-1520), (-1,-800), (2,208), (5,-440), (8,-2744), (11,-4760), (14,-2600), (17,9568), (20,39520) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = 4.
There is a single, unique root at x = 15.
There is a single, unique root at x = 1.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x- 4)(x- 15)(x- 1) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 -42x2 -82x + 414
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 -42x2 -82x + 414 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-3.5, 2.6, 11.5]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 -84x2 -41x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -0.87 and 7.87. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x+4)(x- 15)(x- 4)
= x4 -9x3 -106x2 + 144x + 1440  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+4)(x- 15)(x- 4) = x4 -9x3 -106x2 + 144x + 1440

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,726), (-4,0), (-1,1200), (2,1248), (5,-990), (8,-4704), (11,-7140), (14,-3600), (17,12558), (20,49920) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = -4.
There is a single, unique root at x = 15.
There is a single, unique root at x = 4.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x+4)(x- 15)(x- 4) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 -27x2 -212x + 144
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 -27x2 -212x + 144 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-5.0, 0.7, 11.2]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 -54x2 -106x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -2.52 and 7.02. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x- 4)2(x- 15)
= x4 -17x3 -2x2 + 576x -1440  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x- 4)2(x- 15) = x4 -17x3 -2x2 + 576x -1440

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,2662), (-4,-2432), (-1,-2000), (2,-416), (5,-110), (8,-1568), (11,-3332), (14,-2000), (17,7774), (20,33280) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a double root at x = 4. Not just the function but also its first derivative are zero at this point.
There is a single, unique root at x = 15.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x- 4)2(x- 15) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 -51x2 -4x + 576
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 -51x2 -4x + 576 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-3.0, 4.0, 11.9]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 -102x2 -2x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -0.04 and 8.54. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x+4)(x- 15)(x+5)
= x4 -151x2 -990x -1800  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+4)(x- 15)(x+5) = x4 -151x2 -990x -1800

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,132), (-4,0), (-1,-960), (2,-4368), (5,-9900), (8,-15288), (11,-16320), (14,-6840), (17,21252), (20,78000) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = -4.
There is a single, unique root at x = 15.
There is a single, unique root at x = -5.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x+4)(x- 15)(x+5) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 -302x -990
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 -302x -990 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-5.5, -4.4, 10.1]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 -151x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -5.02 and 5.02. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x- 4)(x- 15)(x+5)
= x4 -8x3 -119x2 + 90x + 1800  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x- 4)(x- 15)(x+5) = x4 -8x3 -119x2 + 90x + 1800

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,484), (-4,304), (-1,1600), (2,1456), (5,-1100), (8,-5096), (11,-7616), (14,-3800), (17,13156), (20,52000) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = 4.
There is a single, unique root at x = 15.
There is a single, unique root at x = -5.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x- 4)(x- 15)(x+5) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 -24x2 -238x + 90
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 -24x2 -238x + 90 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-5.5, 0.4, 11.2]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 -48x2 -119x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -2.88 and 6.88. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x+4)(x- 1)(x+5)
= x4 + 14x3 + 59x2 + 46x -120  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+4)(x- 1)(x+5) = x4 + 14x3 + 59x2 + 46x -120

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,48), (-4,0), (-1,-120), (2,336), (5,3960) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = -4.
There is a single, unique root at x = 1.
There is a single, unique root at x = -5.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x+4)(x- 1)(x+5) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 42x2 + 118x + 46
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 42x2 + 118x + 46 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-5.5, -4.4, -0.4]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 84x2 + 59x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -5.05 and -1.95. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x- 4)(x- 1)(x+5)
= x4 + 6x3 -21x2 -106x + 120  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x- 4)(x- 1)(x+5) = x4 + 6x3 -21x2 -106x + 120

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,176), (-4,80), (-1,200), (2,-112), (5,440), (8,5096) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = 4.
There is a single, unique root at x = 1.
There is a single, unique root at x = -5.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x- 4)(x- 1)(x+5) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 18x2 -42x -106
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 18x2 -42x -106 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-5.5, -1.7, 2.8]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 36x2 -21x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -3.9 and 0.9. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)2(x+4)(x- 1)
= x4 + 15x3 + 68x2 + 60x -144  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)2(x+4)(x- 1) = x4 + 15x3 + 68x2 + 60x -144

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,24), (-4,0), (-1,-150), (2,384), (5,4356) Let us inspect the roots of the given polynomial function.
There is a double root at x = -6. Not just the function but also its first derivative are zero at this point.
There is a single, unique root at x = -4.
There is a single, unique root at x = 1.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)2(x+4)(x- 1) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 45x2 + 136x + 60
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 45x2 + 136x + 60 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-6.0, -4.7, -0.5]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 90x2 + 68x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -5.4 and -2.1. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)2(x- 4)(x- 1)
= x4 + 7x3 -20x2 -132x + 144  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)2(x- 4)(x- 1) = x4 + 7x3 -20x2 -132x + 144

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,88), (-4,160), (-1,250), (2,-128), (5,484), (8,5488) Let us inspect the roots of the given polynomial function.
There is a double root at x = -6. Not just the function but also its first derivative are zero at this point.
There is a single, unique root at x = 4.
There is a single, unique root at x = 1.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)2(x- 4)(x- 1) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 21x2 -40x -132
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 21x2 -40x -132 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-6.0, -2.0, 2.8]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 42x2 -20x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -4.28 and 0.78. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)2(x+4)(x- 4)
= x4 + 12x3 + 20x2 -192x -576  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)2(x+4)(x- 4) = x4 + 12x3 + 20x2 -192x -576

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,33), (-4,0), (-1,-375), (2,-768), (5,1089), (8,9408) Let us inspect the roots of the given polynomial function.
There is a double root at x = -6. Not just the function but also its first derivative are zero at this point.
There is a single, unique root at x = -4.
There is a single, unique root at x = 4.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)2(x+4)(x- 4) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 36x2 + 40x -192
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 36x2 + 40x -192 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-6.0, -4.7, 1.8]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 72x2 + 20x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -5.38 and -0.62. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)2(x- 4)2
= x4 + 4x3 -44x2 -96x + 576  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)2(x- 4)2 = x4 + 4x3 -44x2 -96x + 576

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,121), (-4,256), (-1,625), (2,256), (5,121), (8,3136) Let us inspect the roots of the given polynomial function.
There is a double root at x = -6. Not just the function but also its first derivative are zero at this point.
There is a double root at x = 4. Not just the function but also its first derivative are zero at this point.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)2(x- 4)2 = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 12x2 -88x -96
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 12x2 -88x -96 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-6.0, -1.0, 4.0]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 24x2 -44x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -3.89 and 1.89. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)2(x+4)(x+5)
= x4 + 21x3 + 164x2 + 564x + 720  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)2(x+4)(x+5) = x4 + 21x3 + 164x2 + 564x + 720

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,6), (-4,0), (-1,300) Let us inspect the roots of the given polynomial function.
There is a double root at x = -6. Not just the function but also its first derivative are zero at this point.
There is a single, unique root at x = -4.
There is a single, unique root at x = -5.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)2(x+4)(x+5) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 63x2 + 328x + 564
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 63x2 + 328x + 564 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-6.0, -5.3, -4.3]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 126x2 + 164x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -5.73 and -4.77. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)2(x- 4)(x+5)
= x4 + 13x3 + 28x2 -204x -720  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)2(x- 4)(x+5) = x4 + 13x3 + 28x2 -204x -720

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,22), (-4,-32), (-1,-500), (2,-896), (5,1210), (8,10192) Let us inspect the roots of the given polynomial function.
There is a double root at x = -6. Not just the function but also its first derivative are zero at this point.
There is a single, unique root at x = 4.
There is a single, unique root at x = -5.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)2(x- 4)(x+5) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 39x2 + 56x -204
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 39x2 + 56x -204 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-6.0, -5.3, 1.6]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 78x2 + 28x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -5.68 and -0.82. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x+4)(x- 8)(x- 1)
= x4 + 1x3 -58x2 -136x + 192  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+4)(x- 8)(x- 1) = x4 + 1x3 -58x2 -136x + 192

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,360), (-4,0), (-1,270), (2,-288), (5,-1188), (8,0), (11,7650) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = -4.
There is a single, unique root at x = 8.
There is a single, unique root at x = 1.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x+4)(x- 8)(x- 1) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 3x2 -116x -136
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 3x2 -116x -136 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-5.1, -1.1, 5.6]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 6x2 -58x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -3.37 and 2.87. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x- 4)(x- 8)(x- 1)
= x4 -7x3 -34x2 + 232x -192  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x- 4)(x- 8)(x- 1) = x4 -7x3 -34x2 + 232x -192

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,1320), (-4,-960), (-1,-450), (2,96), (5,-132), (8,0), (11,3570) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = 4.
There is a single, unique root at x = 8.
There is a single, unique root at x = 1.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x- 4)(x- 8)(x- 1) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 -21x2 -68x + 232
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 -21x2 -68x + 232 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-3.6, 2.5, 6.5]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 -42x2 -34x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -1.2 and 4.7. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x+4)(x- 8)(x- 4)
= x4 -2x3 -64x2 + 32x + 768  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+4)(x- 8)(x- 4) = x4 -2x3 -64x2 + 32x + 768

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,495), (-4,0), (-1,675), (2,576), (5,-297), (8,0), (11,5355) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = -4.
There is a single, unique root at x = 8.
There is a single, unique root at x = 4.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x+4)(x- 8)(x- 4) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 -6x2 -128x + 32
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 -6x2 -128x + 32 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-5.0, 0.3, 6.4]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 -12x2 -64x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -2.8 and 3.8. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x- 4)2(x- 8)
= x4 -10x3 -16x2 + 352x -768  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x- 4)2(x- 8) = x4 -10x3 -16x2 + 352x -768

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,1815), (-4,-1536), (-1,-1125), (2,-192), (5,-33), (8,0), (11,2499) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a double root at x = 4. Not just the function but also its first derivative are zero at this point.
There is a single, unique root at x = 8.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x- 4)2(x- 8) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 -30x2 -32x + 352
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 -30x2 -32x + 352 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-3.2, 4.0, 6.8]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 -60x2 -16x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -0.49 and 5.49. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x+4)(x- 8)(x+5)
= x4 + 7x3 -46x2 -472x -960  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x+4)(x- 8)(x+5) = x4 + 7x3 -46x2 -472x -960

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,90), (-4,0), (-1,-540), (2,-2016), (5,-2970), (8,0), (11,12240) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = -4.
There is a single, unique root at x = 8.
There is a single, unique root at x = -5.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x+4)(x- 8)(x+5) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 + 21x2 -92x -472
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 + 21x2 -92x -472 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-5.5, -4.4, 4.8]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 + 42x2 -46x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -5.03 and 1.53. At these points, the concavity changes.

## Example:

Let's analyze the following polynomial function.

f(x) = (x+6)(x- 4)(x- 8)(x+5)
= x4 -1x3 -70x2 -8x + 960  (obtained on multiplying the terms)

Let us analyze the graph of this function which is a quartic polynomial. A quartic polynomial is a fourth degree polynomial.
• These have zero to four real roots
• They may have one, two or three extrema (maxima/minima - turning points)
• Zero, one or two inflection (inflexion) points.
Let's do some curve sketching take a look at the graph of y = f(x) = (x+6)(x- 4)(x- 8)(x+5) = x4 -1x3 -70x2 -8x + 960

We can plot the curve by computing some sample points other than the zeroes already known.
Some sample points are: A few computed points on the curve, apart from the zero(s) which are known:
(-7,330), (-4,192), (-1,900), (2,672), (5,-330), (8,0), (11,5712) Let us inspect the roots of the given polynomial function.
There is a single, unique root at x = -6.
There is a single, unique root at x = 4.
There is a single, unique root at x = 8.
There is a single, unique root at x = -5.
These roots are the solutions of the quartic equation f(x) = 0

These values of x are the roots of the quadratic equation (x+6)(x- 4)(x- 8)(x+5) = 0

Let us analyze the turning points in this curve.
The derivative of the given function = f'(x) = 4x3 -3x2 -140x -8
This is a cubic function.
Turning points are those where the derivative of the function (obtained on differentiating the function) is zero.
These may be obtained by solving the cubic equation 4x3 -3x2 -140x -8 = 0.
Some of these are local maximas and some are local minimas. The roots of the above cubic equation are the ones where the turning points are located.
The turning points of this curve are approximately at x = [-5.5, 0.0, 6.4]. At these points, the curve has either a local maxima or minima. These are the extrema - the peaks and troughs in the graph plot. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero..

Now, let us take a look at the inflexion points.
These are the points where the curve changes concavity.
The second derivative f'''(x) = 12x3 -6x2 -70x
This is a quadratic function. The value(s) of x for which this quadratic function is zero, are the inflexion points.
This curve has two inflexion points at (approx) -3.17 and 3.67. At these points, the concavity changes.