question find all vertical asymptotes of the following function. f(x)=(x^2 - 4)/(2x^2 - 10x)

Explanation:

Step1: Factor the function

Factor 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 $2x(x - 5)=0$.

Step3: Solve for x

Using the zero - product property, if $2x(x - 5)=0$, then $2x=0$ or $x - 5=0$. Solving $2x=0$ gives $x = 0$, and solving $x - 5=0$ gives $x = 5$. When $x = 0$, the numerator $(0 + 2)(0 - 2)=-4\neq0$. When $x = 5$, the numerator $(5 + 2)(5 - 2)=21\neq0$.

Answer:

$x = 0,x = 5$