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

furajat4h
2022-09-24
Answered

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

Answered 2022-09-25
Author has **8** answers

Rearrange.

$y=\frac{5-2x}{x-2}=\frac{-2(x-2)+1}{x-2}=-2+\frac{1}{x-2}$

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 $(0,-\frac{5}{2})$

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

$x=\frac{5}{2}$ so $(\frac{5}{2},0)$

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]}

$y=\frac{5-2x}{x-2}=\frac{-2(x-2)+1}{x-2}=-2+\frac{1}{x-2}$

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 $(0,-\frac{5}{2})$

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

$x=\frac{5}{2}$ so $(\frac{5}{2},0)$

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]}

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