Chapter 9, Problem 17PT

To determine

the equation of any vertical asymptote and the values of $x$ for any holes in the graph of the function

Answer to Problem 17PT

The discontinuity at $x=-5$ and the vertical asymptote is $x=7$

Explanation of Solution

Given information: The function is $f\left(x\right)=\frac{x+5}{x^2-2x-35}$

Calculation:

Factor the denominator of the equation

  $\begin{array}{l}f\left(x\right)=\frac{x+5}{x^2-2x-35}\\ f\left(x\right)=\frac{x+5}{\left(x-7\right)\left(x+5\right)}\\ f\left(x\right)=\frac{1}{\left(x-7\right)}\end{array}$

Discontinuity at $x=-5$

Since $x=-5$ would yield a zero in the denominator and this factor canceled, there is discontinuity at $x=-5$

Vertical asymptote is $x=7$

  $\begin{array}{l}x-7\\ x=7\end{array}$