Chapter 2.3, Problem 40E

Solution

Find the degree and zeros of the given function.

Explain the multiplicity of zeros and show whether the graph pass through the $x-$ axis at the corresponding intercept.

Sketch the graph.

The degree of the function is $4$ .

The zeros of the function $f\left(x\right)=-x^3\left(x-2\right)$ are $0$ and $2.$

Given: $f\left(x\right)=-x^3\left(x-2\right)$

Expand the term to find the degree of the function:

  $f\left(x\right)=-x^4+2x^3$

The largest exponent is the degree of the polynomial, so the degree is $4.$

Find the zeros:

To find the zeros, solve the related equation $f\left(x\right)=0$ by factoring.

  $-x^3\left(x-2\right)=0$

Apply the zero-factor property:

  $x=0\left(\text{multiplicity}3\right)\text{or }x-2=0$

  $x=0\left(\text{multiplicity}3\right)\text{or }x=2$

Hence, the zeros of function $f\left(x\right)=-x^3\left(x-2\right)$ are $0$ and $2.$

Explain the multiplicity of zeros and show whether the graph pass through the $x-$axis at the corresponding intercept:

The graph crosses the $x-$ axis at $x=0$ because the multiplicity $3$ is odd. Also, the graph crosses the $x-$ axis at $x=2$ because the multiplicity $1$ is odd.

Notice that values of $f\left(x\right)=-x^3\left(x-2\right)$ are negative for $x<0,$ positive for $0<x<2$ , and negative for $x>2$ .

Sketch the graph:

The graph of the function $f\left(x\right)=-x^3\left(x-2\right)$ is: