Question

In Problems 7-44, follow Steps 1 through 7 on page 228 to analyze the graph of each function.

$f\left(x\right)=2x^2+\frac{16}{x}$

Short answer

The graph is

Step-by-step solution

Given information

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

Factoring the numerator and denominator

Factoring, we get

$$\begin{gathered} f\left(x\right)=2x^2+\frac{16}{x} \\ f\left(x\right)=\frac{2x^3+16}{x} \end{gathered}$$

Calculate the domain

Setting denominator to zero, we get

$x=0$

The domain is all real numbers except$x=0$

Calculate the x intercept 

Setting the numerator to zero, we get

$$\begin{gathered} 2x^3+16=0 \\ {x}^3=-8 \\ x=-2 \end{gathered}$$

There is no$y$intercept as$f\left(0\right)=undefined$

Calculating the vertical asymptote

The zeroes of denominator are $x=0$so the vertical asymptote is$x=0$

Calculating the horizontal asymptote

The degree of numerator is greater than denominator so there is an oblique asymptote

$f\left(x\right)=2x^2+\frac{16}{x}$

So $y=2x^2$is the oblique asymptote

$$\begin{gathered} f\left(x\right)=\frac{2(x+2)\left(x^2-2x+4\right)}{x} \\ =2x^2 \\ 2(x+2)\left(x^2-2x+4\right)=2x^2\left(x\right) \\ 2x^3+16=2x^3 \\ 0=-16 \end{gathered}$$

So it doesn't intersect$f$

Sketching the graph

The graph is