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

1 Answer
Aug 1, 2018

Below

Explanation:

f(x)=1+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 ->oo. As x ->oo, 2/(x-3) ->0 so f(x)=1+0=1 ie y=1

For intercepts,
When y=0, x=1
When x=0, y=1/3

Below is what the graph looks like

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