Graph of Cubic Functions/Cubic Equations for zeros and roots (-11,10,-14)

Let us consider the cubic function f(x) = (x+11)(x- 10)(x+14) = x3 + 15x2 -96x -1540. 
We will inspect the graph, the zeroes, the turning and inflection points in the cubic curve curve y = f(x).

Cubic Polynomials and Equations

A cubic polynomial is a polynomial of degree 3.
where a is nonzero.
An equation involving a cubic polynomial is called a cubic equation and is of the form f(x) = 0. There is also a closed-form solution known as the cubic formula which exists for the solutions of an arbitrary cubic equation. A cubic polynomial is represented by a function of the form. And f(x) = 0 is a cubic equation. The points at which this curve cuts the X-axis are the roots of the equation.

Graph of y = f(x) = (x+11)(x- 10)(x+14) = x3 + 15x2 -96x -1540
Cubic Polynomial Curve Plot on Graph

A few computed points on the curve, apart from the zero(s) which are known:
(-15,-100), (-12,44), (-9,-190), (-6,-640), (-3,-1144), (0,-1540), (3,-1666), (6,-1360), (9,-460), (12,1196), (15,3770)

Characteristics of the Graph Plot, Curve Sketching of Cubic Curves

If the given cubic function is: f(x) = ax3 + bx2 + cx + d 
The derivative of this function is: f'(x) = 3ax2 + 2bx + c
The function given to us us f(x) = (x+11)(x- 10)(x+14) = x3 + 15x2 -96x -1540
And the derivative for this is f'(x) = 3x2 + 30x + 1

Consider the cubic equation f(x) = (x+11)(x- 10)(x+14) = x3 + 15x2 -96x -1540 = 0
The roots of this cubic equation are at: 
(x - (-11)) = 0 => x = -11, 
OR (x - (10)) = 0 => x = 10, 
OR (x - (-14)) = 0 => x =-14
This cubic equation has real and unique roots at -11, 10, -14.
The plotted curve cuts the x-axis at these values of x: i.e, these are the zeroes of the given cubic polynomial. 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.

The turning or stationary points is where f'(x) = 0 => 3x2 + 30x + 1 = 0 => x = -12.55, x = 2.55  
These are also called the "critical" points where the derivative is zero. 
Coming to other geometrical features of this curve: What we see here is the graph of a nonlinear function. The y-intercept of this curve is at y=-1540.
And the second derivative of this curve becomes zero at x = -5.0. At this point the curve changes concavity. A cubic curve has point symmetry around the point of inflection or inflexion. These are just some of the important features and aspects to keep in mind while trying to visualize and analyze a plot of an algebraic function. A graphical and visual inspection helps in several ways.

The zeroes of a polynomial, if they are known, and the coefficients of that polynomial are two different sets of numbers that have interesting relations.  If we know the zeroes, then we can write down algebraic expressions for the coefficients. Going the other way is much harder and cannot be done in general.
A cubic function has a bit more variety in its shape than the quadratic polynomials which are always parabolas. We can get a lot of information from the factorization of a cubic function. We get a fairly generic cubic shape when we have three distinct linear factors

Compute the Area between the given curve and the X-Axis

Here are some examples of computing the area under curves or between a given curve and the X-Axis
We'll use integral calculus and definite integrals here. Skip this part if you haven't yet started integral calculus.
F(x) = Integral of f(x) = ∫ ( x3 + 15x2 -96x -1540) dx = x4/3 + 15x3/2 -96x2 -1540x + C
Where C is the constant of integration.
Values of F at each of the roots -14,-11,10:
F(-14) = (-14)4/3 + (15 * (-14)3)/2 + -96*(-14)2 + -1540*(-14) + C = -5030.67 + C
Similarly, we can compute: F(-11) = 221.83 + C and F(10) = -14166.67 + C
Where C is the constant of integration.

Area between the curve and the X-axis = |Area enclosed for x in interval (-14,-11)| + |Area enclosed for x in interval (-11,10)|
= |Area enclosed for x in interval (-14,-11)| + |Area enclosed for x in interval (-11,10)|
= |Definite integral of f(x) between limits -14 and -11| + |Definite integral of f(x) between limits -11 and 10|
= |F(-11) - F(-14)| + |F(10) - F(-11)|
= |221.83 + C - (-5030.67 + C)| + |-14166.67 + C - (221.83 + C)|
Terms involving the constant of integration cancel out.
Area = |5252.5| + |-14388.5|=> Area = 19641.0 Square Units

Check the plot of another cubic curve here with roots at -8, 10, -14

Here's another cubic curve here with roots at -11, 10, -8
Many of these concepts are a part of the Grade 9,10,11,12 (High School) Mathematics syllabus of the UK GCSE/GCE curriculum, Common Core Standards in the US, ICSE/CBSE/SSC/NTSE syllabus in Indian high schools. You may check out our free and printable worksheets for Common Core and GCSE.

Prashant Bhattacharji,
Mar 21, 2017, 1:35 AM