Question

question 4 (multiple choice worth 5 points) (oa 03r mc) find the vertical asymptote(s) of $f(x) = \frac{2x^2 + 3x + 6}{x^2 - 1}$. $x = -1, 1$ $x = 1, 2$ $x = -1, 2$ $x = -2, 2$

Explanation:

Step1: Recall vertical asymptote rule

For a rational function \( f(x)=\frac{N(x)}{D(x)} \), vertical asymptotes occur where \( D(x) = 0 \) and \( N(x)
eq0 \) at those points.

Step2: Factor denominator

Denominator \( D(x)=x^{2}-1 \). Factor using difference of squares: \( x^{2}-1=(x - 1)(x + 1) \).

Step3: Solve \( D(x)=0 \)

Set \( (x - 1)(x + 1)=0 \). Solutions are \( x = 1 \) and \( x=-1 \).

Step4: Check numerator at these x - values

Numerator \( N(x)=2x^{2}+3x + 6 \).

• At \( x = 1 \): \( N(1)=2(1)^{2}+3(1)+6=2 + 3+6 = 11

eq0 \).

• At \( x=-1 \): \( N(-1)=2(-1)^{2}+3(-1)+6=2-3 + 6 = 5

eq0 \).
So vertical asymptotes are at \( x=-1 \) and \( x = 1 \).

Answer:

A. \( x=-1,1 \) (assuming the first option is labeled as A with \( x=-1,1 \))