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

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#### Explanation

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#### Explanation:

I want someone to double check my answer

1
Lucy Share
Jul 7, 2018

Below

#### Explanation:

Rearrange.
$y = \frac{5 - 2 x}{x - 2} = \frac{- 2 \left(x - 2\right) + 1}{x - 2} = - 2 + \frac{1}{x - 2}$

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 $0 = \frac{5 - 2 x}{x - 2}$.
$x = \frac{5}{2}$ 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]}

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