91Ó°ÊÓ

The given function is$f\left(x\right)=-2(x+2){(x-2)}^3$.

Solve the given function.

$\begin{array}{rcl}f\left(x\right)& =& -2(x+2){(x-2)}^3\\ & =& (-2x-4)(x^3-6x^2+12x-8)\\ & =& -2x^4+12x^3-24x^2+16x-4x^3+24x^2-48x+32\\ & =& -2x^4+8x^3-32x+32\end{array}$

The polynomial function $f$is of degree $4$.

Therefore, the graph of the given function behaves like localid="1646309512347" $y=-2x^4$.

Substitute $x=0$to find the $y$-intercepts.

localid="1646306534094" $\begin{array}{rcl}f\left(x\right)& =& -2(x+2){(x-2)}^3\\ f\left(0\right)& =& -2(0+2){(0-2)}^3\\ & =& -2\left(2\right){(-2)}^3\\ & =& 32\end{array}$

Thus, the $y$-intercept is at the point $(0,32)$.

Now, substitute $f\left(x\right)=0$to find the $x$-intercepts.

$\begin{array}{rcl}f\left(x\right)& =& -2(x+2){(x-2)}^3\\ 0& =& -2(x+2){(x-2)}^3\\ & ⇒& x+2=0\\ & & or\\ {(x-2)}^3& =& 0\end{array}$

$\begin{array}{rcl}& ⇒& x=-2\\ & & or\\ x& =& 2\end{array}$

Thus, the $x$-intercepts are at the points $(-2,0)$and$(2,0)$.

• If the multiplicity is even, then the graph touches the $x$-axis.
• If the multiplicity is odd, then the graph crosses the $x$-axis.

From the given function $f\left(x\right)=-2(x+2){(x-2)}^3$, conclude that the zeros of the given function are $-2$and $2$.

The zero $-2$has multiplicity $1$, so the graph of $f$crosses the $x$-axis at $x=-2$.

The zero $2$has multiplicity $3$, so the graph of $f$crosses the $x$-axis at $x=2$.

The graph of the given function is shown below:

From the graph of $f$, observe that $f$has one turning point.

Using MAXIMUM, the turning point is at $(-1,54)$.

The graph is shown below:

Since the function is a polynomial function.

So, the domain is the set of all real numbers.

Thus, domain $=(-∞,∞)$.

From the graph, the range is the set of all possible values for which the function is defined.

Thus, range $(-∞,54]$.

From the graph of $f$, observe the following:

$f$is increasing on the interval $(-∞,-1)$.

$f$is decreasing on the intervals $(-1,2)$and $(2,∞)$.