# How do you graph f(x)=2/(x-3)+1 using holes, vertical and horizontal asymptotes, x and y intercepts?

Aug 1, 2018

Below

#### Explanation:

$f \left(x\right) = 1 + \frac{2}{x - 3}$

For vertical asymptote, look at the denominator. It cannot equal to $0$ as the graph will be undefined at that point. Hence, you let denominator equal to $0$ to find at what point the graph cannot equal to $0$.

$x - 3 = 0$
$x = - 3$

For horizontal asymptote, imagine what happens to the graph when $x \to \infty$. As $x \to \infty$, $\frac{2}{x - 3} \to 0$ so $f \left(x\right) = 1 + 0 = 1$ ie $y = 1$

For intercepts,
When $y = 0$, $x = 1$
When $x = 0$, $y = \frac{1}{3}$

Below is what the graph looks like

graph{1+2/(x-3) [-10, 10, -5, 5]}