Answer
The line $ y=\frac{1}{4}$ is the horizontal asymptote of the given function.
Work Step by Step
To find the horizontal asymptotes, we calculate the following limit
\begin{align*}
\lim _{x \rightarrow \pm \infty}\frac{2x^2-3x}{8 x^2+8}&= \lim _{x \rightarrow \pm\infty}\frac{2 -3x^{-1}}{8 +8x^{-1}}\\
&=\frac{2}{8}=\frac{1}{4} .
\end{align*}
Hence, the line $ y=\frac{1}{4}$ is the horizontal asymptote of the given function.
