Answer
$f\left( x \right)=-\frac{2}{3}\text{ or }y=-\frac{2}{3}$ will be the horizontal asymptote.
Work Step by Step
Write $f\left( x \right)$ in the form of $\frac{p\left( x \right)}{q\left( x \right)}$.
$\begin{align}
& f\left( x \right)=\frac{-3x+7}{5x-2} \\
& =\frac{p\left( x \right)}{q\left( x \right)}
\end{align}$
To find the horizontal asymptotes, divide the numerator and the denominator by the highest degree of x, which is 1.
$\begin{align}
& f\left( x \right)=\frac{-3x+7}{5x-2} \\
& =\frac{\frac{-3x+7}{x}}{\frac{5x}{x}-\frac{2}{x}}
\end{align}$
Now, when x tends to infinity, $\frac{1}{x}$ tends to zero.
$\begin{align}
& f\left( x \right)=\frac{-3x+7}{5x-2} \\
& =\frac{\frac{-3x+7}{x}}{\frac{5x}{x}-\frac{2}{x}} \\
& =\frac{-3+\frac{7}{x}}{5-\frac{2}{x}} \\
& =\frac{-3}{5}
\end{align}$
Therefore, $f\left( x \right)=-\frac{3}{5}\text{ or }y=-\frac{3}{5}$ will be the horizontal asymptote.
