Chapter 4.1, Problem 59AYU

In Problems 57-68, for each polynomial function:

(a) List each real zero and its multiplicity.

(b) Determine whether the graph crosses or touches the $x\text{-axis}$ at each $x\text{-intercept}$

(c) Determine the maximum number of turning points on the graph.

(d) Determine the end behavior; that is, find the power function that the graph of $f$ resembles for large values of $\left|x\right|$.

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

To determine

To find:

a. Each real zero and its multiplicity.

Answer to Problem 59AYU

a. $\text{The real zeros }=0,0,\sqrt{2},\sqrt{2}$.

$\text{The multiplicity of the polynomial }=0\left(\text{Multiplicity }=2\right)$.

$\text{The multiplicity of the polynomial }=\sqrt{2}\left(\text{Multiplicity }=\text{ 2}\right)$.

Explanation of Solution

Given:

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

To find the real zeros

$f\left(x\right)=0$

$-2x^2=0,(x^2-2)=0$

$x^2=0, x^2=2$

$x=0,0,\sqrt{2 },\sqrt{2}$

a. The real zeros = 0, 0, $\sqrt{2}$, $\sqrt{2}$.

The multiplicity of the polynomial = 0 (Multiplicity = 2).

The multiplicity of the polynomial = $\sqrt{2}$ (Multiplicity = 2).

To determine

To find:

b. To determine whether the graph crosses or touches the $x\text{-axes}$ at each $x\text{-intercept}$.

Answer to Problem 59AYU

b. $\text{Zero }0\text{has even multiplicity}\text{. Therefore the graph touches the }x\text{-axis}$.

$\text{Zero }\sqrt{2}\text{has even multiplicity}\text{. Therefore the graph touches the }x\text{-axis}$.

Explanation of Solution

Given:

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

To find the real zeros

$f\left(x\right)=0$

$-2x^2=0,(x^2-2)=0$

$x^2=0, x^2=2$

$x=0,0,\sqrt{2 },\sqrt{2}$

b. Zero 0 has even multiplicity. Therefore the graph touches the $x\text{-axis}$.

Zero $\sqrt{2}$ has even multiplicity. Therefore the graph touches the $x\text{-axis}$.

To determine

To find:

c. To determine the maximum number of turning points on the graph.

Answer to Problem 59AYU

c. $\text{The maximum number of turning points}=\left(n-1\right)=3$.

Explanation of Solution

Given:

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

To find the real zeros

$f\left(x\right)=0$

$-2x^2=0,(x^2-2)=0$

$x^2=0, x^2=2$

$x=0,0,\sqrt{2 },\sqrt{2}$

c. The degree of $f\left(x\right)=n=4$.

The maximum number of turning points $=\left(n-1\right)=3$.

To determine

To find:

d. To determine the end behavior (power function that the graph of $f$ resembles for large values of $x$).

Answer to Problem 59AYU

d. $\text{For large values of }x\text{in the positive direction or negative direction value of }f\text{approaches negative infinity }\underset{x\to \infty }{\lim }f\left(x\right)=-\infty$.

Unbounded in the negative direction.

Explanation of Solution

Given:

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

To find the real zeros

$f\left(x\right)=0$

$-2x^2=0,(x^2-2)=0$

$x^2=0, x^2=2$

$x=0,0,\sqrt{2 },\sqrt{2}$

d. Polynomial $f\left(x\right)=a_n{x}^n$.

Here $a_n=-2<0$.

For large values of $x$ in the positive direction or negative direction value of $f$ approaches negative infinity $\underset{x\to \infty }{\lim }f\left(x\right)=-\infty$.

Unbounded in the negative direction.