## Precalculus (6th Edition) Blitzer

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