Question

question
which functions have graphs with a horizontal asymptote? select all that apply.
select all that apply:
$f(x)=\frac{2x}{x - 5}$
$f(x)=\frac{x^{2}-1}{x + 3}$
$f(x)=\frac{x^{3}+5}{x^{5}+2}$
$f(x)=\frac{9x^{3}}{x - 6}$
$f(x)=\frac{x - 1}{x^{2}}$

Explanation:

Step1: Recall horizontal - asymptote rules

For a rational function $f(x)=\frac{a_nx^n + a_{n - 1}x^{n - 1}+\cdots+a_0}{b_mx^m + b_{m - 1}x^{m - 1}+\cdots+b_0}$, if $n\lt m$, 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 $f(x)=\frac{2x}{x - 5}$

Here $n = 1$ and $m = 1$. Then $y=\frac{2}{1}=2$ is the horizontal asymptote since $n=m$.

Step3: Analyze $f(x)=\frac{x^2-1}{x + 3}$

Here $n = 2$ and $m = 1$. Since $n>m$, there is no horizontal asymptote.

Step4: Analyze $f(x)=\frac{x^3+5}{x^5+2}$

Here $n = 3$ and $m = 5$. Since $n\lt m$, the horizontal asymptote is $y = 0$.

Step5: Analyze $f(x)=\frac{9x^3}{x - 6}$

Here $n = 3$ and $m = 1$. Since $n>m$, there is no horizontal asymptote.

Step6: Analyze $f(x)=\frac{x - 1}{x^2}$

Here $n = 1$ and $m = 2$. Since $n\lt m$, the horizontal asymptote is $y = 0$.

Answer:

$f(x)=\frac{2x}{x - 5}$, $f(x)=\frac{x^3+5}{x^5+2}$, $f(x)=\frac{x - 1}{x^2}$