Question

1 multiple choice 20 points which function has a horizontal asymptote at ( y = 2 ) ( y=\frac{x^{2}-2}{x - 4} ) ( y=\frac{x - 2}{x + 2} ) ( y=\frac{2x - 7}{x - 1} ) ( y=\frac{x - 2}{x^{2}-4x - 5} )

Explanation:

Step1: Recall horizontal asymptote rules

For a rational function \(y=\frac{f(x)}{g(x)}\) where \(f(x)=a_nx^n+\cdots\) and \(g(x)=b_mx^m+\cdots\). If \(n = m\), the horizontal asymptote is \(y=\frac{a_n}{b_m}\).

Step2: Analyze \(y=\frac{x^{2}-2}{x - 4}\)

Here \(n = 2\) (degree of numerator) and \(m=1\) (degree of denominator). Since \(n>m\), there is no horizontal asymptote (has an oblique asymptote).

Step3: Analyze \(y=\frac{x - 2}{x + 2}\)

Here \(n = 1\), \(m = 1\). \(a_n=1\), \(b_m=1\). Horizontal asymptote \(y=\frac{1}{1}=1\).

Step4: Analyze \(y=\frac{2x-7}{x - 1}\)

Here \(n = 1\), \(m = 1\). \(a_n = 2\), \(b_m=1\). Horizontal asymptote \(y=\frac{2}{1}=2\).

Step5: Analyze \(y=\frac{x - 2}{x^{2}-4x - 5}\)

Here \(n = 1\), \(m = 2\). Since \(n

Answer:

\(y=\frac{2x-7}{x - 1}\)