Question

for the function $f(x)=\frac{(x^{2}-4)}{x^{2}}$, at what x - value is the vertical asymptote located?

Explanation:

Step1: Recall the definition of vertical asymptote

Vertical asymptote occurs where the denominator of a rational function is zero (after simplifying the function).

Step2: Set the denominator equal to zero

For the function \(f(x)=\frac{x^{2}-4}{x^{2}}\), the denominator is \(x^{2}\). Set \(x^{2} = 0\).

Step3: Solve the equation

Solving \(x^{2}=0\) gives \(x = 0\).

Answer:

\(0\)