Question

a rational function is given below. $r(x)=\frac{x^{2}-9}{x^{2}+2x - 10}$. what are the asymptotes of the graph of $r(x)$? drag and drop the correct equation into each box to make both statements true. the horizontal asymptote is . the vertical asymptotes are and . y = - 3 y = 0 y = 1 y = 3 x = - 5 x = - 3 x = - 2 x = 2 x = 3 x = 5

Explanation:

Step1: Find horizontal asymptote

For $R(x)=\frac{x^{2}+2x - 10}{x^{2}-9}$, since the degrees of the numerator and denominator are the same (both degree 2), we divide the leading - coefficients. The leading coefficient of the numerator is 1 and of the denominator is 1. So, $\lim_{x
ightarrow\pm\infty}R(x)=\lim_{x
ightarrow\pm\infty}\frac{x^{2}+2x - 10}{x^{2}-9}=\lim_{x
ightarrow\pm\infty}\frac{1+\frac{2}{x}-\frac{10}{x^{2}}}{1-\frac{9}{x^{2}}}=1$. So the horizontal asymptote is $y = 1$.

Step2: Find vertical asymptotes

Set the denominator equal to zero. We have $x^{2}-9=(x + 3)(x - 3)=0$. Solving $(x + 3)(x - 3)=0$ gives $x=-3$ and $x = 3$. So the vertical asymptotes are $x=-3$ and $x = 3$.

Answer:

Horizontal asymptote: $y = 1$
Vertical asymptotes: $x=- 3$, $x = 3$