Question

write the equation of all horizontal asymptotes of the function $f(x)=\frac{sqrt{9x^{2}+25}}{3x + 2}$. answer attempt 1 out of 3 one horizontal asymptote

Explanation:

Step1: Analyze limit as x approaches positive infinity

We find $\lim_{x
ightarrow+\infty}\frac{\sqrt{9x^{2}+25}}{3x + 2}$. Divide numerator and denominator by $x$ (since $x>0$ as $x
ightarrow+\infty$). We get $\lim_{x
ightarrow+\infty}\frac{\sqrt{9+\frac{25}{x^{2}}}}{3+\frac{2}{x}}$. As $x
ightarrow+\infty$, $\frac{25}{x^{2}}
ightarrow0$ and $\frac{2}{x}
ightarrow0$. So $\lim_{x
ightarrow+\infty}\frac{\sqrt{9+\frac{25}{x^{2}}}}{3+\frac{2}{x}}=\frac{\sqrt{9 + 0}}{3+0}=1$.

Step2: Analyze limit as x approaches negative infinity

We find $\lim_{x
ightarrow-\infty}\frac{\sqrt{9x^{2}+25}}{3x + 2}$. Divide numerator and denominator by $|x|=-x$ (since $x<0$ as $x
ightarrow-\infty$). The numerator becomes $\frac{\sqrt{9x^{2}+25}}{-x}=\sqrt{9+\frac{25}{x^{2}}}$ and the denominator becomes $\frac{3x + 2}{-x}=-3-\frac{2}{x}$. As $x
ightarrow-\infty$, $\frac{25}{x^{2}}
ightarrow0$ and $\frac{2}{x}
ightarrow0$. So $\lim_{x
ightarrow-\infty}\frac{\sqrt{9x^{2}+25}}{3x + 2}=\frac{\sqrt{9+0}}{-3 + 0}=-1$.

Answer:

$y = 1,y=-1$