Question

find all vertical asymptotes of the following function.
f(x)=\frac{x^{2}-4}{2x^{2}-10x}

Explanation:

Step1: Factor the numerator and denominator

The numerator $x^{2}-4=(x + 2)(x - 2)$ and the denominator $2x^{2}-10x=2x(x - 5)$. So $f(x)=\frac{(x + 2)(x - 2)}{2x(x - 5)}$.

Step2: Set the denominator equal to zero

Vertical asymptotes occur where the denominator of a rational - function is zero and the numerator is non - zero. Set $2x(x - 5)=0$.
Using the zero - product property, if $ab = 0$, then $a = 0$ or $b = 0$. So $2x=0$ gives $x = 0$ and $x-5=0$ gives $x = 5$.

Step3: Check the numerator at these values

When $x = 0$, the numerator $(0 + 2)(0 - 2)=-4
eq0$. When $x = 5$, the numerator $(5 + 2)(5 - 2)=21
eq0$.

Answer:

$x = 0$ and $x = 5$