Question

write the equation of the vertical asymptote(s) of the rational function. $y = \frac{x + 2}{x^{2}-3x - 4}$
$x=-2$
$x = 4$ and $x = 3$
$x = 1$ and $x=-4$
$x=-1$ and $x = 4$

Explanation:

Step1: Find the denominator's zeros

Set $x^{2}-3x - 4=0$.

Step2: Factor the quadratic equation

$(x - 4)(x+1)=0$.

Step3: Solve for x

$x - 4=0$ gives $x = 4$; $x + 1=0$ gives $x=-1$. The vertical asymptotes of a rational function $y=\frac{f(x)}{g(x)}$ occur at the values of $x$ that make $g(x)=0$ (assuming no common factors between $f(x)$ and $g(x)$ that can be cancelled out). Here, for $y=\frac{x + 2}{x^{2}-3x - 4}$, the vertical - asymptotes are at the zeros of the denominator.

Answer:

$x=-1$ and $x = 4$