which functions have a horizontal asymptote? check all that apply.\n f(x)=\frac{x^{3}-2x + 3}{x^{2}-5}\n…

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

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

Step3: Analyze $v(x)=\frac{x^2-1}{x^2 - 5}$

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

Step4: Analyze $g(x)=\frac{1 - x}{x^2+2}$

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

Step5: Analyze $w(x)=\frac{x + 3x^4}{x^2}$

Here $n = 4$ and $m = 2$. Since $n>m$, there is no horizontal asymptote.

Step6: Analyze $h(x)=\frac{x^3}{x^2-5x^4}$

Here $n = 3$ and $m = 4$. Since $n<m$, the horizontal asymptote is $y = 0$.

Answer:

$v(x)=\frac{x^2-1}{x^2 - 5}$, $g(x)=\frac{1 - x}{x^2+2}$, $h(x)=\frac{x^3}{x^2-5x^4}$