91Ó°ÊÓ

Consider the given function.

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

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$y=x^3$.

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.

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.

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

The graph of the given function is as shown below,