Question

attempt 1: 10 attempts remaining. the function $f(x) = \frac{x^2 - 8}{x - 5}$ has a vertical asymptote. at what $x$ value does this occur? answer: submit answer

Explanation:

Step1: Recall vertical asymptote rule

A vertical asymptote of a rational function \( f(x)=\frac{g(x)}{h(x)} \) occurs where \( h(x) = 0 \) and \( g(x)
eq0 \) at that \( x \)-value.
For \( f(x)=\frac{x^{2}-8}{x - 5} \), the denominator is \( h(x)=x - 5 \) and the numerator is \( g(x)=x^{2}-8 \).

Step2: Solve denominator for zero

Set \( h(x)=0 \), so \( x - 5=0 \).
Solving for \( x \), we get \( x = 5 \).

Step3: Check numerator at \( x = 5 \)

Substitute \( x = 5 \) into the numerator \( g(x)=x^{2}-8 \).
\( g(5)=5^{2}-8=25 - 8 = 17
eq0 \).
So, the vertical asymptote occurs at \( x = 5 \).

Answer:

\( 5 \)