Answer
$g\left( x \right)=5\text{ or }y=5$ is the equation of the horizontal asymptote.
Work Step by Step
Write $g\left( x \right)$ in the form $\frac{p\left( x \right)}{q\left( x \right)}$.
$\begin{align}
& g\left( x \right)=\frac{15{{x}^{2}}}{3{{x}^{2}}+1} \\
& =\frac{p\left( x \right)}{q\left( x \right)}
\end{align}$
To find the horizontal asymptotes, divide the numerator and denominator by the highest degree of x, which is 2.
$\begin{align}
& g\left( x \right)=\frac{15{{x}^{2}}}{3{{x}^{2}}+1} \\
& =\frac{\frac{15{{x}^{2}}}{{{x}^{2}}}}{\frac{3{{x}^{2}}}{{{x}^{2}}}+\frac{1}{{{x}^{2}}}}
\end{align}$
Now, when x tends to infinity, $\frac{1}{x}$ tends to zero.
$\begin{align}
& g\left( x \right)=\frac{15{{x}^{2}}}{3{{x}^{2}}+1} \\
& =\frac{\frac{15{{x}^{2}}}{{{x}^{2}}}}{\frac{3{{x}^{2}}}{{{x}^{2}}}+\frac{1}{{{x}^{2}}}} \\
& =\frac{15}{3} \\
& =5
\end{align}$
Therefore, $g\left( x \right)=5\text{ or }y=5$ is the equation of the horizontal asymptote.
