Question

Analyze each polynomial function.

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

Short answer

The graph of the function is defined as,

Step-by-step solution

Step 1. Given Information.

Consider the given function.

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

The polynomial function.

Factor the function.

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

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

Step 3. Find the x-intercept

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

And the x-intercept is,

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

Or,

$\begin{array}{rcl}x-3& =& 0\\ x& =& 3\end{array}$

Or,

$\begin{array}{rcl}x+4& =& 0\\ x& =& -4\end{array}$

The 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,

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

A line with slope -12.

Near 3:

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

A line with a slope is 2.

Near -4:

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

A line with slope is 28.

Interval of increase and decrease.

The given function is $f\left(x\right)=x^3+x^2-12x=x(x-3)(x+4)$and increases in the interval $\left(3,\infty \right)\cup \left(-4,0\right)$and decreases in the interval$\left(-\infty ,-4\right)\cup \left(0,3\right)$.

Step 6. Graph of the polynomial.

The graph of the given function is as shown below,