Chapter 2.3, Problem 39E

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 $3$ .

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

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

Find the degree of the function:

Rewrite ${\left(x-3\right)}^2$ as $\left(x-3\right)\left(x-3\right)$

  $x\left(\left(x-3\right)\left(x-3\right)\right)$

Expand $\left(x-3\right)\left(x-3\right)$ using the FOIL Method:

  $x\left(x\cdot x+x\cdot -3-3x-3\cdot -3\right)$

Simplify and combine like terms:

  $x\left(x^2-6x+9\right)$

Expand the term:

  $x^3-6x^2+9x$

The largest exponent is the degree of the polynomial, so the degree is $3.$

Find the zeros:

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

  $x{\left(x-3\right)}^2=0$

Apply the zero-factor property:

  $x=0\text{ or }x-3=0$

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

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

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 $1$ is odd. The graph does not cross the $x-$ axis at $x=3$ because the multiplicity $2$ is even.

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

Sketch the graph:

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