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

1 Answer
Lucy
Jul 11, 2018

Below

Explanation:

#f(x)=(2x-1)/(x+3)#
#f(x)=(2(x+3)-7)/(x+3)#
#f(x)=2-7/(x+3)#

Therefore, the horizontal asymptote is #y=2# and the vertical asymptote is #x=-3# What asymptotes mean is that the end points of your lines are APPROACHING #y=2# and #x=-3#

Now, to find the intercepts
When #y=0#, #x=1/2#
When #x=0#, #y=-1/3#

After plotting the intercepts and the asymptotes, you should get something like this
graph{(2x-1)/(x+3) [-10, 10, -5, 5]}

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