identify the rules that validate the following horizontal asymptotes. i know the ha for g(x)=(x^2 - 5x…

Explanation:

Step1: Recall horizontal - asymptote rules

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

Step2: Analyze $g(x)=\frac{x^{2}-5x - 6}{3x^{2}+7x - 6}$

The degree of the numerator $n = 2$ and the degree of the denominator $m = 2$. Since $n=m$, the horizontal asymptote is $y=\frac{1}{3}$ (the ratio of the leading - coefficients, where the leading coefficient of the numerator is $1$ and the leading coefficient of the denominator is $3$). So the rule is $n = d$.

Step3: Analyze $g(x)=\frac{x^{2}-5x - 6}{3x^{3}+7x^{2}-6x}$

The degree of the numerator $n = 2$ and the degree of the denominator $m = 3$. Since $n<m$, the horizontal asymptote is $y = 0$. So the rule is $n<d$.

Step4: Analyze $g(x)=\frac{x^{3}-5x^{2}-6x}{3x^{2}+7x - 6}$

The degree of the numerator $n = 3$ and the degree of the denominator $m = 2$. Since $n>m$, there is no horizontal asymptote. So the rule is $n>d$.

Answer:

For $g(x)=\frac{x^{2}-5x - 6}{3x^{2}+7x - 6}$, the rule is $n = d$. For $g(x)=\frac{x^{2}-5x - 6}{3x^{3}+7x^{2}-6x}$, the rule is $n<d$. For $g(x)=\frac{x^{3}-5x^{2}-6x}{3x^{2}+7x - 6}$, the rule is $n>d$.