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

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Answer 1 · 2018-07-07T00:06:40

Below

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]}

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