Question

hw 9 rational functions
due wednesday by 11:59pm points 100 submitting an external tool
consider the graph of the function $f(x)=\frac{x^{2}-1}{x^{2}+3x - 4}$.
which describes the vertical asymptote and the removable discontinuity?
select the correct answer below:
the vertical asymptote is $x=-4$ and the removable discontinuity is at $x=-1$.
the vertical asymptote is $x=-4$ and the removable discontinuity is at $x = 1$.
the vertical asymptote is $x = 1$ and the removable discontinuity is at $x=-4$.
the vertical asymptote is $x = 1$ and the removable discontinuity is at $x = 4$.
the vertical asymptote is $x = 4$ and the removable discontinuity is at $x=-1$.
the vertical asymptote is $x = 4$ and the removable discontinuity is at $x = 1$.

Explanation:

Step1: Factor the numerator and denominator

The numerator $x^{2}-1=(x + 1)(x - 1)$. The denominator $x^{2}+3x - 4=(x + 4)(x - 1)$. So $f(x)=\frac{(x + 1)(x - 1)}{(x + 4)(x - 1)}$.

Step2: Find the vertical asymptote

Vertical asymptotes occur where the denominator is zero and the numerator is non - zero. Set the denominator $(x + 4)(x - 1)=0$. Solving gives $x=-4$ or $x = 1$. But since we can cancel out the factor $(x - 1)$ in the numerator and denominator, $x=-4$ is the vertical asymptote.

Step3: Find the removable discontinuity

A removable discontinuity occurs when a factor can be canceled out in the numerator and denominator. After canceling out $(x - 1)$ (where $x
eq1$), there is a removable discontinuity at $x = 1$.

Answer:

The vertical asymptote is $x=-4$ and the removable discontinuity is at $x = 1$. So the correct answer is: The vertical asymptote is $x=-4$ and the removable discontinuity is at $x = 1$.