Question

question 18 · 1 point
consider the graph of the function ( f(x)=\frac{x^{2}+x - 12}{x^{2}-6x + 8} ).
what are the vertical asymptotes? list the ( x )-values separated by commas.
provide your answer below:

Explanation:

Step1: Factor numerator and denominator

Numerator: \(x^{2}+x - 12=(x + 4)(x-3)\)
Denominator: \(x^{2}-6x + 8=(x - 2)(x - 4)\)
So \(f(x)=\frac{(x + 4)(x-3)}{(x - 2)(x - 4)}\)

Step2: Find values that make denominator zero

Set \(x^{2}-6x + 8 = 0\), i.e., \((x - 2)(x - 4)=0\)
Solve \(x-2=0\) gives \(x = 2\), solve \(x - 4=0\) gives \(x=4\)
Also, check if these values make numerator zero. When \(x = 2\), numerator \((2 + 4)(2-3)=-6
eq0\). When \(x = 4\), numerator \((4 + 4)(4-3)=8
eq0\)

Answer:

\(2,4\)