We have to plot this 3rd-grade polynomial.
We start by finding the points at which the axis are intersected.
First, we start with the vertical axis. The y-intercept is equal to the value of f(x) when x = 0.
Then, its value is equal to the independent term:
[tex]\begin{gathered} f(0)=0^3+2(0)^2-64(0)-128 \\ f(0)=0+0-0-128 \\ f(0)=-128 \end{gathered}[/tex]
Then, the point (0,-128) is part of the function.
We can now try to find the x-intercepts of roots of the function.
There is some similarity between the terms that we can use to factorize the polynomial, as there is no simple formula to calculate the roots of a third grade polynomial:
[tex]\begin{gathered} f(x)=x^3+2x^2-64x-128 \\ f(x)=x^2\cdot x+x^2\cdot2+(-64)\cdot x+(-64)\cdot2 \\ f(x)=x^2(x+2)-64(x+2) \\ f(x)=(x^2-64)(x+2) \\ f(x)=(x+8)(x-8)(x+2) \end{gathered}[/tex]
Now we now we have roots at x = -2, x = -8 and x = 8.
If we plot this points we get:
From the y-intercept at y = -128, we can deduce the sign of each interval of the polynomial.
We can then graph it as:
