How do you graph #f(x)=(x^2+3x+2)/(-3x-12)# using holes, vertical and horizontal asymptotes, x and y intercepts?

1 Answer
Lucy
Jul 9, 2018

Below

Explanation:

#f(x)=(x^2+3x+2)/(-3x-12)#

Using long division, we can rearrange the polynomial
#f(x)=((-3x-12)(-1/3x-1/3)-2)/(-3x-12)#

#f(x)=-1/3x-1/3-2/(-3x-12)#

This means that your oblique asymptote is #y=-1/3x-1/3# which can be found by figuring out what happens when x approaches 0 and your horizontal asymptote is #x=-4# after solving #-3x-12=0# since your denominator cannot equal to 0.

To figure out your x and y intercepts, we let #y=0# and #x=0# respectively.

When #x=0#, #y=-1/6# so your y-intercept is #(0,-1/6)#
When #y=0#, #x=-2# and #x=-1# so your x-intercepts are #(-2,0)# and #(-1,0)#

graph{(x^2+3x+2)/(-3x-12) [-10, 10, -5, 5]}
Above is the graph

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