Question

10 which of the following has a horizontal asymptote at y = 0?
a. $f(x)=\frac{2x}{3 - x}$
b. $f(x)=\frac{1}{x - 4}$
c. $f(x)=\frac{x^{2}-5x + 6}{x - 3}$
d. $f(x)=\frac{x^{2}-x}{2x^{2}+4}$

Explanation:

Step1: Recall horizontal asymptote rules

For a rational function \( f(x)=\frac{a_nx^n+\cdots+a_0}{b_mx^m+\cdots + b_0}\), if \(n < m\), the horizontal asymptote is \(y = 0\); if \(n=m\), the horizontal asymptote is \(y=\frac{a_n}{b_m}\); if \(n>m\), there is no horizontal asymptote.

Step2: Analyze option A

For \(f(x)=\frac{2x}{3 - x}=\frac{2x}{-x + 3}\), \(n = 1\), \(m=1\). Then \(y=\frac{2}{-1}=- 2\).

Step3: Analyze option B

For \(f(x)=\frac{1}{x - 4}\), \(n = 0\), \(m = 1\). Since \(nightarrow\pm\infty}\frac{1}{x - 4}=0\).

Step4: Analyze option C

For \(f(x)=\frac{x^{2}-5x + 6}{x - 3}=\frac{(x - 2)(x - 3)}{x - 3}=x-2(x
eq3)\), \(n = 1\), \(m = 0\). There is no horizontal asymptote.

Step5: Analyze option D

For \(f(x)=\frac{x^{2}-x}{2x^{2}+4}\), \(n = 2\), \(m = 2\). Then \(y=\frac{1}{2}\).

Answer:

B. \(f(x)=\frac{1}{x - 4}\)