Question

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

$R\left(x\right)=\frac{8x^2+26x+15}{2x^2-x-15}$

Short answer

The graph is

Step-by-step solution

Given information

Given function$R\left(x\right)=\frac{8x^2+26x+15}{2x^2-x-15}$

Factoring the numerator and denominator

Factoring, we get

$$\begin{gathered} R\left(x\right)=\frac{8x^2+26x+15}{2x^2-x-15} \\ R\left(x\right)=\frac{8x^2+30x+4x+15}{2x^2-6x+5x-15} \\ R\left(x\right)=\frac{2x(4x+15)+1(4x+15)}{2x(x-3)+5(x-3)} \\ R\left(x\right)=\frac{(2x+1)(4x+15)}{(2x+5)(x-3)} \end{gathered}$$

Calculating the domain

Setting the denominator to zero, we get

$$\begin{gathered} (2x+5)(x-3)=0 \\ x=-\frac{5}{2},3 \end{gathered}$$

The domain is all real numbers except$-\frac{5}{2},3$

Calculating the x intercept

Setting numerator to zero, we get

$$\begin{gathered} (2x+1)(4x+15)=0 \\ x=-\frac{1}{2},\frac{-4}{15} \end{gathered}$$

The$y$intercept is$R\left(0\right)=-1$i.e.$(0,-1)$

Calculating the vertical asymptote

The zeroes of denominator is$x=-\frac{5}{2},3$so the vertical asymptote is$x=-\frac{5}{2},3$

Calculating the horizontal asymptote

The degree of numerator is equal to denominator so the horizontal asymptote is

$$\begin{gathered} y=\frac{a}{b} \\ y=\frac{1}{1} \\ y=1 \end{gathered}$$

$$\begin{gathered} R\left(x\right)=\frac{8x^2+26x+15}{2x^2-x-15} \\ =4 \\ 8x^2+26x+15=4\left(2x^2-x-15\right) \\ 8x^2+26x+15=8x^2-8x-60 \\ x=-\frac{15}{7} \end{gathered}$$

The intersection point is$(-\frac{15}{7},4)$

Sketching the graph

The graph is