## Calculus with Applications (10th Edition)

$$y=\frac{8}{5-3x}$$ We find that: Asymptote: $y=0$ and $x=\frac{5}{3}$ $x$-intercept: non $y$ -intercept: $\frac{8}{5}$
$$y=\frac{8}{5-3x}$$ The function is undefined for $x=\frac{5}{3}$, so the line $x=\frac{5}{3}$ is a vertical asymptotic.. To find a horizontal asymptotic, let $x$ get larger and larger, so that $$y=\lim _{x \rightarrow \infty} \frac{8}{5-3x} =0$$ This means that the line $y=0$ is a horizontal asymptotic. When $x=0$ the y-intercept is $\frac{8}{5}$ So, Asymptotic: $y=0$ and $x=\frac{5}{3}$ $x$-intercept: non $y$ -intercept: $\frac{8}{5}$