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

Step 1. Given Information

The given function is

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

Step 2. Simplify the polynomial function.

The function

$$\begin{gathered} f\left(x\right)=x^3+x^2-12x \\ =x(x-3)(x+4) \end{gathered}$$

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

The function is defined for all values of $x$.

So, the domain of the given function is$D_f=\mathrm{ℝ}$.

Step 3.  Find the x intercept.

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

So,$$\begin{gathered} \Rightarrow x(x-3)(x+4)=0 \\ \Rightarrow x=0\mathrm{or}x=3\mathrm{or}x=-4 \end{gathered}$$

x-intercepts are $0,3\mathrm{and}-4$

The zero o is a zero of multiplicity 1 so the graph of the function crosses the x-axis at x=0

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

The zero -4 is a zero of multiplicity 1, so the graph of the function 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.

Step 4. 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 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 slope $28$.

Step 5. 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: