Question

for what values of a and m does f(x) have a horizontal asymptote at y = 2 and a vertical asymptote at x = 1?
f(x) = \frac{2x^m}{x + a}
options:

• a = -1, m = 1
• a = 1, m = 0
• a = 1, m = 1
• a = -1, m = 0

Explanation:

Step1: Analyze Vertical Asymptote

A vertical asymptote of a rational function \( f(x)=\frac{N(x)}{D(x)} \) occurs where \( D(x) = 0 \) (and \( N(x)
eq0 \) at that point). For \( f(x)=\frac{2x^m}{x + a} \), the denominator is \( x + a \). We need a vertical asymptote at \( x = 1 \), so set \( x + a=0 \) when \( x = 1 \). Substituting \( x = 1 \), we get \( 1 + a=0\), so \( a=- 1 \).

Step2: Analyze Horizontal Asymptote

For horizontal asymptotes of rational functions:

• If the degree of the numerator (\( n \)) equals the degree of the denominator (\( d \)), the horizontal asymptote is \( y=\frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}} \).

The denominator \( x + a \) (with \( a=-1 \)) has degree 1. The numerator is \( 2x^m \), so for the degrees to be equal, \( m = 1 \). Then the horizontal asymptote is \( y=\frac{2}{1}=2 \), which matches the given horizontal asymptote \( y = 2 \).

Answer:

\( a=-1, m = 1 \)