which function does not have a horizontal asymptote?\no f(x) = (x^3 + 2x)/(x^2 - 3)\no y = (x - 6)/(x^2…

Explanation:

Step1: Recall horizontal - asymptote rules

For a rational function $y = \frac{f(x)}{g(x)}=\frac{a_nx^n+\cdots+a_0}{b_mx^m+\cdots + b_0}$, 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 $f(x)=\frac{x^3 + 2x}{x^2-3}$

Here, $n = 3$ (degree of numerator) and $m = 2$ (degree of denominator). Since $n>m$, there is no horizontal asymptote.

Step3: Analyze $y=\frac{x - 6}{x^2-4}$

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

Step4: Analyze $f(x)=\frac{x - 1}{x + 5}$

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

Step5: Analyze $y=\frac{x^2-3x}{-3x^2 + 5}$

Here, $n = 2$ and $m = 2$. Then $y=\frac{1}{-3}=-\frac{1}{3}$ is the horizontal asymptote.

Answer:

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