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

How do you graph $f\left(x\right)=\frac{5-2x}{x-2}$ using holes, vertical and horizontal asymptotes, x and y intercepts?
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Kaya Garza
Rearrange.
$y=\frac{5-2x}{x-2}=\frac{-2\left(x-2\right)+1}{x-2}=-2+\frac{1}{x-2}$

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

Subbing x=0 to find the y-intercept, we have $-2+\frac{1}{0-2}=-\frac{5}{2}$ so $\left(0,-\frac{5}{2}\right)$

Subbing y=0 to find the x-intercept, we have $0=\frac{5-2x}{x-2}$
$x=\frac{5}{2}$ so $\left(\frac{5}{2},0\right)$

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]}