College Algebra, Chapter 4, Chapter Review, Section Review, Problem 70

Graph the rational function $\displaystyle r(x) = \frac{2x^2 - 6x - 7}{x - 4}$. Show clearly all $x$ and $y$ intercepts and asymptotes.

We first find the intercepts, so

$x$ intercept: The $x$ intercepts are the zeros of the numerator,


$
\begin{equation}
\begin{aligned}

2x^2 - 6x - 7 =& 0
\\
\\
x^2 - 3x - \frac{7}{2} =& 0
\qquad \text{Divide } 2

\end{aligned}
\end{equation}
$


By using Quadratic Formula,

$\displaystyle x = \frac{3 \pm \sqrt{23}}{2}$

$y$ intercept: To find the $y$ intercept, we substitute $x = 0$.

$\displaystyle r(0) = \frac{2(0)^2 - 6(0) - 7}{0 - 4} = \frac{-7}{-4} = \frac{7}{4}$

Vertical Asymptotes: The vertical asymptotes occur where the denominator is , that is where the function is undefined. Hence, the line $x = 4$ is the vertical asymptote.

Next, to know the behavior near vertical asymptote, we set $4.1$ and $3.9$ for the values of $y$ to the right and left of $4$ respectively so..

$\displaystyle y = \frac{2(4.1)^2 - 6 (4.1) - 7}{4.1 - 4}$ whose sign is $\displaystyle \frac{(+)}{(+)}$ (positive)

Hence, $y \to \infty$ as $x \to 4^+$. Then by substituting $x = 3.9$ to the function, we will find out that $y \to - \infty$ as $x \to 4^-$.

Since, the degree of the numerator is greater than the degree of the denominator, then the function has no horizontal asymptote but rather slant asymptote. So by applying long division







Thus, the line $y = 2x + 2$ is the slant asymptote of the function.

Therefore, the graph will look like..