## Calculus with Applications (10th Edition)

$$y=\frac{x-4}{x+1}$$ we find that : Asymptotes: $x=-1$ and $y=1$ $x$-intercept: 4, the value when $y= 0$ $y$-intercept: -4, the value when $x= 0$
$$y=\frac{x-4}{x+1}$$ The value $x=-1$ makes the denominator 0, but not the numerator, so the line $x=-1$ is a vertical asymptotic. To find a horizontal asymptote, let $x$ get larger and we find that $$y= \lim\limits_{x \to \infty}{\frac{x+1}{x-4}}=1$$ This means that the line $y=1$ is a horizontal asymptote. Asymptotes: $x=-1$ and $y=1$ $x$-intercept: 4, the value when $y= 0$ $y$-intercept: -4, the value when $x= 0$