How do you graph #f(x)=(x^2+4x+3)/(-3x-6)# using holes, vertical and horizontal asymptotes, x and y intercepts?

1 Answer
Lucy
Jul 25, 2018

Below

Explanation:

#f(x)=(x^2+4x+3)/(-3x-6)#
#f(x)=-1/3((x^2+4x+3)/(x+2))#
#f(x)=-1/3((x+2)(x+2)-1)/(x+2)#
#f(x)=-1/3(x+2-1/(x+2))#
#f(x)=-1/3(x+2)+1/(3(x+2)#

For oblique asymptote, it is #y=-1/3(x+2)# which can be found by letting #x -> oo#.
For vertical asymptote, the denominator cannot equal to 0. so #x+2!=0# so at #x=-2#, there is an asymptote.

For intercepts,
When #x=0#, #y=-1/2#
When #y=0#, #x^2+4x+3=0#
#(x+3)(x+1)=0# so #x=-1# and #x=-3#

Plotting your x and y intercepts as well as your asymptotes, you should have a general idea as to how your graph should appear. The ends of your graph should be approaching your asymptotes but they should NEVER TOUCH the asymptotes.

Below is the graph
graph{(x^2+4x+3)/(-3x-6) [-10, 10, -5, 5]}

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