Chapter 8, Problem 11CR

To determine

To graph: The rational function $R\left(x\right)=\frac{2x^2-7x-4}{x^2+2x-15}$

Explanation of Solution

Given information:

The rational function $R\left(x\right)=\frac{2x^2-7x-4}{x^2+2x-15}$

Graph:

Step-1: The factors of $2x^2-7x-4$ are $2x^2-7x-4=2x-8x+x-4=2x(x-4)+1(x-4)=\left(x-4\right)\left(x+\frac{1}{2}\right)$ and factors of $x^2+2x-15=x^2+5x-3x-15$ are $\left(x+5\right)\left(x-3\right)$

The function is $R\left(x\right)=\frac{2x^2-7x-4}{x^2+2x-15}$ becomes undefined, when $x=-5,3$. The domain of function $R\left(x\right)$ is set of all real numbers except $-5,3$

Therefore, domain of the rational function $R\left(x\right)=\frac{2x^2-7x-4}{x^2+2x-15}$ is $\left(-\infty ,-5\right)\cup \left(-5,3\right)\cup \left(3,\infty \right)$.

Step-2: $x-$ intercept is a point on the graph, where $y=0$

Therefore, $R\left(x\right)=\frac{\left(x-4\right)\left(x+\frac{1}{2}\right)}{\left(x+5\right)\left(x-3\right)}=0\Rightarrow \left(x-4\right)\left(x+\frac{1}{2}\right)=0$

That is, $x=4$ and $x=-\frac{1}{2}$ are $x-$ intercepts of $R\left(x\right)$

$y-$ intercept is a point on the graph, where $x=0$

Therefore, $R\left(0\right)=\frac{2{\left(0\right)}^2-7\left(0\right)-4}{{\left(0\right)}^2+2\left(0\right)-15}=\frac{-4}{-15}=\frac{4}{15}$

Thus, the function $R\left(x\right)=\frac{2x^2-7x-4}{x^2+2x-15}$ has $x-$ intercepts $\left(4,0\right)$ and, $\left(-\frac{1}{2},0\right)$ and $y-$ intercept $\left(0,\frac{4}{15}\right)$. The multiplicity of $\left(4,0\right)$ and $\left(-\frac{1}{2},0\right)$ intercept is odd, graph of $R\left(x\right)$ will cross $x-$ axis at point $\left(4,0\right)$ and $\left(-\frac{1}{2},0\right)$

Step-4: As $x=-5,3$ are zeroes of denominator and not of numerator, therefore $x=-5,3$ are vertical asymptotes of $R\left(x\right)$

The multiplicity of $-5$ and $3$ is odd; therefore, the graph of $R\left(x\right)$ approaches to $\infty$ and $-\infty$ on either side of vertical asymptotes.

Step-5: The degree of numerator is $2$ and degree of denominator is $2$. As degree of the denominator equals to degree of numerator, the horizontal asymptote of $G\left(x\right)$ is the ratio of leading coefficients of numerator and denominator

Therefore, the horizontal asymptote of $R\left(x\right)$ is $y=2$

To check whether graph of $R\left(x\right)$ intersects horizontal asymptote $y=1$, solve equation $R\left(x\right)=2$

$\Rightarrow 2=\frac{2x^2-7x-4}{x^2+2x-15}$

$\Rightarrow 2x^2+4x-30=2x^2-7x-4$

$\Rightarrow 11x=26$

$\Rightarrow x=\frac{26}{11}$

That means, graph of $R\left(x\right)$ intersects horizontal asymptote $y=2$ at $x=\frac{26}{11}$

Step-6: The real zeroes of the numerator are $x=-\frac{1}{2},4$; the real zero of the denominator are $x=-5,3$

The intervals using the numbers $x=-2,-1,2$ are $\left(-\infty ,-5\right),\left(-5,-\frac{1}{2}\right),\left(-\frac{1}{2},3\right),\left(3,4\right),\left(4,\infty \right)$.

$\begin{array}{cccccc}\text{Intervals}& \left(-\infty ,-5\right)& \left(-5,-\frac{1}{2}\right)& \left(-\frac{1}{2},3\right)& \left(3,4\right)& \left(4,\infty \right)\\ \text{Test points}& -6& -1& 0& \frac{7}{2}& 5\\ \text{Value of R}\left(x\right)& \frac{110}{9}& -\frac{5}{16}& \frac{4}{15}& -\frac{16}{17}& \frac{11}{20}\\ \text{Location of graph}& \text{Above }x-\text{axis}& \text{Below }x-\text{axis}& \text{Above }x-\text{axis}& \text{Below }x-\text{axis}& \text{Above }x-\text{axis}\\ \text{Point on graph}& \left(-6,\frac{110}{9}\right)& \left(-1,-\frac{5}{16}\right)& \left(0,\frac{4}{15}\right)& \left(\frac{7}{2},-\frac{16}{17}\right)& \left(5,\frac{11}{20}\right)\end{array}$

Step-7: The graph of $R\left(x\right)$ is below $x-$ axis for $3<x<4$, the graph of $R\left(x\right)$ approaches to $-\infty$ as $G\left(x\right)$ approaches to vertical asymptotes $x=-5,3$ from left side.

The graph of function $R\left(x\right)$ is above $x-\text{axis}$ for $-\infty <x<-5$, also the graph intersects the horizontal asymptote $y=2$ at $x=\frac{26}{11}$, so graph of $R\left(x\right)$ approaches to $y=2$ from above as $x\to -\infty$.

Graph of $R\left(x\right)$ crosses $y=2$ from above $x\to \infty$

The graph of $R\left(x\right)$ is below $x-$ axis for $-5<x<-\frac{1}{2}$, the graph of $R\left(x\right)$ approaches to $-\infty$ as $R\left(x\right)$ approaches to vertical asymptote $x=-5$ from right side.

The graph of $R\left(x\right)$ is above $x-$ axis for $-\frac{1}{2}<x<3$, the graph of $R\left(x\right)$ approaches to $\infty$ as $R\left(x\right)$ approaches to vertical asymptote $x=3$ from left side

The graph of $R\left(x\right)$ is above $x-$ axis for $4<x<\infty$, the graph of $R\left(x\right)$ approaches to $\infty$ as $R\left(x\right)$ approaches to horizontal asymptote $y=2$

Therefore, the graph of rational function $R\left(x\right)=\frac{2x^2-7x-4}{x^2+2x-15}$ is

Interpretation:

The graph represents rational function $R\left(x\right)=\frac{2x^2-7x-4}{x^2+2x-15}$