# How do you graph f(x)=-3/(x-1)-1 using holes, vertical and horizontal asymptotes, x and y intercepts?

Feb 14, 2018

The green line is the vertical asymptote, the blue line is the horizontal asymptote.

#### Explanation:

Our first step is to make this function completely rational by combining the two fractions:

$f \left(x\right) = - \frac{3}{x - 1} - 1$

$f \left(x\right) = - \frac{3}{x - 1} - \frac{x - 1}{x - 1} = \frac{- 3 - x + 1}{x - 1}$

$f \left(x\right) = \frac{- 2 - x}{x - 1}$

There are no holes because the factors do not cancel on the top and bottom.

There are vertical asymptotes anywhere the denominator equals zero so:

$x - 1 = 0 \to x = 1$ Thus a vertical asymptote at x=1.

Since the degree is the same on the top and bottom, the horizontal asymptote is determined by dividing the leading coefficients in the numerator and denominator. So $y = - \frac{1}{1} = - 1$ is a horizontal asymptote.

Find the x-intercept by setting the numerator equal to zero, so $- 2 - x = 0 \to x = 2$

Finally find the y-intercept by setting $x = 0$, so:
$f \left(0\right) = \frac{- 2 - \left(0\right)}{0 - 1} = - \frac{2}{-} 1 = 2$.

After plotting these points and graphing these asymptotes just make sure the curve crosses through these points while also slowly approaching the asymptotes as the graph approaches infinity.