Question

write the equation of all horizontal asymptotes of the function $f(x)=\frac{sqrt{9x^{6}-25x^{4}}}{5x^{3}+2}$. answer attempt 1 out of 2 one horizontal asymptote

Explanation:

Step1: Analyze for \(x\to+\infty\)

Divide numerator and denominator by \(x^{3}\). For \(x>0\), \(\sqrt{9x^{6}-25x^{4}}=x^{3}\sqrt{9 - \frac{25}{x^{2}}}\). Then \(f(x)=\frac{\sqrt{9x^{6}-25x^{4}}}{5x^{3}+2}=\frac{x^{3}\sqrt{9-\frac{25}{x^{2}}}}{x^{3}(5 + \frac{2}{x^{3}})}=\frac{\sqrt{9-\frac{25}{x^{2}}}}{5+\frac{2}{x^{3}}}\). As \(x\to+\infty\), \(\lim_{x\to+\infty}f(x)=\lim_{x\to+\infty}\frac{\sqrt{9-\frac{25}{x^{2}}}}{5+\frac{2}{x^{3}}}\). Since \(\lim_{x\to+\infty}\frac{1}{x^{n}} = 0\) for \(n>0\), we have \(\lim_{x\to+\infty}f(x)=\frac{\sqrt{9 - 0}}{5+0}=\frac{3}{5}\).

Step2: Analyze for \(x\to-\infty\)

For \(x<0\), \(\sqrt{9x^{6}-25x^{4}}=-x^{3}\sqrt{9 - \frac{25}{x^{2}}}\) (because \(\sqrt{x^{2}}=-x\) when \(x < 0\) and we are taking the square - root of a sixth - degree and fourth - degree polynomial). Then \(f(x)=\frac{\sqrt{9x^{6}-25x^{4}}}{5x^{3}+2}=\frac{-x^{3}\sqrt{9-\frac{25}{x^{2}}}}{x^{3}(5+\frac{2}{x^{3}})}=\frac{-\sqrt{9-\frac{25}{x^{2}}}}{5+\frac{2}{x^{3}}}\). As \(x\to-\infty\), \(\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}\frac{-\sqrt{9-\frac{25}{x^{2}}}}{5+\frac{2}{x^{3}}}=-\frac{3}{5}\).

Answer:

\(y = \frac{3}{5},y=-\frac{3}{5}\)