Question

what are the vertical and horizontal asymptotes for the function $f(x)=\frac{3x^2}{x^2 - 4}$?
\bigcirc\\ horizontal asymptote: $y = -2$, $y = 2$
\quad\\ \\ vertical asymptote: $x = 3$
\bigcirc\\ horizontal asymptote: $y = -4$, $y = 1$
\quad\\ \\ vertical asymptote: $x = 3$
\bigcirc\\ horizontal asymptote: $y = 3$
\quad\\ \\ vertical asymptote: $x = -4$, $x = 1$
\bigcirc\\ horizontal asymptote: $y = 3$
\quad\\ \\ vertical asymptote: $x = -2$, $x = 2$

Explanation:

Step1: Find vertical asymptotes

Set denominator \(x^{2}-4 = 0\).
Factor: \((x + 2)(x - 2)=0\).
Solve: \(x=-2\) or \(x = 2\).

Step2: Find horizontal asymptotes

For \(f(x)=\frac{3x^{2}}{x^{2}-4}\), since degree of numerator \(n = 2\) and degree of denominator \(m=2\).
When \(n=m\), horizontal asymptote \(y=\frac{a_{n}}{b_{m}}\), where \(a_{n}=3\) (leading coefficient of numerator) and \(b_{m}=1\) (leading coefficient of denominator). So \(y = 3\).

Answer:

horizontal asymptote: \(y = 3\), vertical asymptote: \(x=-2,x = 2\) (the fourth option)