Question

what are the vertical and horizontal asymptotes for the function $f(x)=\frac{3x^2}{x^2 - 4}$?

• horizontal asymptote: $y = -2$, $y = 2$

vertical asymptote: $x = 3$

• horizontal asymptote: $y = -4$, $y = 1$

vertical asymptote: $x = 3$

• horizontal asymptote: $y = 3$

vertical asymptote: $x = -4$, $x = 1$

• horizontal asymptote: $y = 3$

vertical asymptote: $x = -2$, $x = 2$

Explanation:

Step1: Find Vertical Asymptotes

Vertical asymptotes occur where the denominator is zero (and numerator is not zero). For \( f(x)=\frac{3x^2}{x^2 - 4} \), set denominator \( x^2 - 4 = 0 \). Solve \( x^2=4 \), so \( x = \pm 2 \) (i.e., \( x = -2, x = 2 \)).

Step2: Find Horizontal Asymptotes

For rational functions \( \frac{ax^n}{bx^m} \), if \( n = m \), horizontal asymptote is \( y=\frac{a}{b} \). Here, \( n = 2 \), \( m = 2 \), \( a = 3 \), \( b = 1 \), so \( y=\frac{3}{1}=3 \).

Answer:

horizontal asymptote: \( y = 3 \)
vertical asymptote: \( x = -2, x = 2 \) (corresponding to the last option: "horizontal asymptote: \( y = 3 \); vertical asymptote: \( x = -2, x = 2 \)")