Question

How do you graph `f(x)=(5-2x)/(x-2)` using holes, vertical and horizontal asymptotes, x and y intercepts?

Answer 1

Below

Explanation:

Rearrange.
`y=(5-2x)/(x-2)=(-2(x-2)+1)/(x-2)=-2+1/(x-2)`

This means that `x=2` and `y=-2` are horizontal and vertical asymptotes.

Subbing `x=0` to find the y-intercept, we have `-2+1/(0-2)=-5/2` so `(0,-5/2)`

Subbing `y=0` to find the x-intercept, we have `0=(5-2x)/(x-2)`.
`x=5/2` so `(5/2,0)`

Plot your intercepts and draw your asymptotes and you should have a general idea of how your graph looks.
graph{(5-2x)/(x-2) [-10, 10, -5, 5]}