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

1 Answer
Jul 7, 2018

Answer:

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]}