Chapter 1.5, Problem 32E

a)

To determine

To find: a graph of the function and to find the zeros of the function.

Answer to Problem 32E

Zeros are $x=-\frac{3}{\sqrt{2}}$ and $x=\frac{3}{\sqrt{2}}$

Explanation of Solution

Given:

The given function is $f\left(x\right)=\frac{2x^2-9}{3-x}$ .

Calculation:

The equation is $f\left(x\right)=\frac{2x^2-9}{3-x}$

Now, draw the graph by using zoom in feature.

Enter the function in $Y=\frac{2x^2-9}{3-x}$

The display is as shown below.

Now adjust the window by entering the values into $x\max ,x\min ,xscl,y\max ,y\min ,andyscl$

The display is as shown below.

Draw the graph by using zoom in feature. To enter into Zoom in feature, we need to clock the zoom button then click $2^{nd}$ button.

The display is as shown below.

Now, click enter in button to obtain the graph.

The graph is as shown below.

From the above graph it is clear that the function has two zero at $x=-\frac{3}{\sqrt{2}}$ and $x=\frac{3}{\sqrt{2}}$ .

b)

To determine

To verify: the results from part (a) algebraically.

Answer to Problem 32E

Explanation of Solution

Calculation:

To find the zeros of the function $f\left(x\right)=\frac{2x^2-9}{3-x}$

Equate it to zero

  $\begin{array}{c}\frac{2x^2-9}{3-x}=0\\ 2x^2-9=0\\ 2x^2=9\\ x^2=\frac{9}{2}\end{array}$

Taking square root on both sides as below

  $x=\pm \frac{3}{\sqrt{2}}$

So the function $f\left(x\right)=\frac{2x^2-9}{3-x}$ has zero at $x=\frac{3}{\sqrt{2}}$ and $x=-\frac{3}{\sqrt{2}}$ .

Thus, the zero of the function is verified and it is correct.