Question

Analyze each polynomial function.

$f\left(x\right)=x^3+x^2-12x$

Short answer

The graph of the function is :

Step-by-step solution

Given Information

The given function is$x+y$

$f\left(x\right)=x^3+x^2-12x$

The polynomial function

$f\left(x\right)=x^3+x^2-12x=x(x-3)(x+4)$

The polynomial function is of degree $3.$The graph of $f$behaves like$y=x^3.$

Find the x intercept

The $y$-intercept is $f\left(0\right)=0.$

$x(x-3)(x+4)=0$$x=0$

or

$$\begin{gathered} x-3=0 \\ x=3 \end{gathered}$$

or

$$\begin{gathered} x+4=0 \\ x=-4 \end{gathered}$$

$x$-intercepts are $0,3$and $-4.$

The zero $0$is a zero of multiplicity $1,$so the graph of $f$crosses the $x$-axis at $x=0.$

The zero $3$is a zero of multiplicity $1,$so the graph of $f$crosses the $x-$axis at $x=3.$

The zero $-4$is a zero of multiplicity $1,$so the graph of $f$crosses the $x-$axis at $x=-4.$

Since the degree of the function is $3,$the graph of the function has$3-1=2$ turning points.

Find a line with a slope

Near $0:$

$$\begin{gathered} f\left(x\right)=x(x-3)(x+4) \\ =x(0-3)(0+4) \\ =-12x \end{gathered}$$

a line with slope $-12.$

Near $3:$

$$\begin{gathered} f\left(x\right)=x(x-3)(x+4) \\ =3(x-3)(3+4) \\ =2x-63 \end{gathered}$$

a line with a slope $2.$

Near $-4:$

$$\begin{gathered} f\left(x\right)=x(x-3)(x+4) \\ =-4(-4-3)(x+4) \\ =28x+112 \end{gathered}$$

a line with a slope$28.$

Interval of increase and decrease.

The given function$f\left(x\right)=x^3+x^2-12x=x(x-3)(x+4)$

increases in the interval $\left(3,\infty \right)\cup \left(-4,0\right)$

And decrease in the interval $\left(-\infty ,-4\right)\cup \left(0,3\right)$.

Step 6. The graph 

The graph of the given function is as shown below: