Chapter 4.3, Problem 42AYU

To determine

To analyze: the graph of given function

Answer to Problem 42AYU

  

Explanation of Solution

Given:

  $f\left(x\right)=2x^2+\frac{16}{x}$

Calculation:

Given rational function is $R\left(x\right)=2x^2+\frac{16}{x}$

First write $R\left(x\right)$ in the lowest term.

  $\begin{array}{l}R\left(x\right)=2x^2+\frac{16}{x}\\ =\frac{2x^2+16}{x}\\ =\frac{2\left(x+2\right)\left(x^2-2x+4\right)}{x}\end{array}$

Step 1:

The domain of $R\left(x\right)is\left\{x|x\cancel{=}0\right\}$ . There is no $y$ intercept.

Step 2:

Since $R\left(x\right)$ is in the lowest term, the real zeros of the numerator satisfies the equation

  $\begin{array}{l}2x^3+16=0\hfill \\ 2x^3=-16\hfill \\ x^3=-8\hfill \\ x=-2\hfill \\ Hencethexinterceptis-2.\hfill \end{array}$

  $\begin{array}{l}Near-2:\\ R\left(x\right)=\frac{2\left(x+2\right)\left(x^2-2x+4\right)}{x}\\ \approx \frac{2\left(x+2\right)\left(4+4+4\right)}{-2}\\ =\frac{2\left(x+2\right)\left(12\right)}{-2}\\ -12\left(x+2\right)\end{array}$

Step 3:

Since $R\left(x\right)$ is in the lowest term. the real zeros of the denominator are real solutions of the equation $x=0,thatis\text{ 0}$ . The graph of $R\left(x\right)$ has vertical asymptote the lines $x=0$ .Graph the asymptotes using a dashed line.

Step 4:

  $\begin{array}{l}Asx\to -\infty orasx\to \infty ,\\ \frac{16}{x}\to 0\end{array}$

Therefore there is no horizontal or oblique asymptote.

Step 5:

The zero of the numerator $-2$ and the zero of denominator $0$ divides the $xaxis$ into three intervals:

  $\left(-\infty ,-2\right) \left(-2,0\right)\left(0,\infty \right)$

Now construct the table:

Interval	$\left(-\infty ,-2\right)$	$\left(-2,0\right)$	$\left(0,\infty \right)$
Number chosen	$-4$	$-1$	$1$
Value of R	$28$	$-14$	$18$
Location of graph	$abovexaxis$	$\text{below }xaxis$	$abovexaxis$
Point of graph	$\left(-4,28\right)$	$\left(-1,-14\right)$	$\left(1,18\right)$

Plot the points from above table.

Step 6:

Complete graph of given ration function is shown below:

  

Conclusion:

Thus, the graph of $f$ is obtained which illustrates the information obtained from Steps 1 through 6.