How do you graph f(x)=2/(x−3)+1 using holes, vertical and horizontal asymptotes, x and y intercepts?

opinaj 2022-09-29 Answered
How do you graph f ( x ) = 2 x - 3 + 1 using holes, vertical and horizontal asymptotes, x and y intercepts?
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Answers (1)

Emilia Boyle
Answered 2022-09-30 Author has 10 answers
For vertical asymptote, look at the denominator. It cannot equal to 0 as the graph will be undefined at that point. Hence, you let denominator equal to 0 to find at what point the graph cannot equal to 0.

x−3=0
x=−3

For horizontal asymptote, imagine what happens to the graph when x . As x , 2 x - 3 0 so f(x)=1+0=1 ie y=1

For intercepts,
When y=0, x=1
When x=0, y = 1 3

Below is what the graph looks like

graph{1+2/(x-3) [-10, 10, -5, 5]}
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