How do you identify the oblique asymptote of #f(x) = (2x^2+3x+8)/(x+3)#?

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Answer 1 · 2016-03-02T14:44:50

Oblique Asymptote is #y=2x-3#
Vertical Asymptote is #x=-3#

from the given:

#f(x)=(2x^2+3x+8)/(x+3)#

perform long division so that the result is

#(2x^2+3x+8)/(x+3)=2x-3+17/(x+3)#

Notice the part of the quotient

#2x-3#

equate this to #y# like as follows

#y=2x-3# this is the line which is the Oblique Asymptote

And the divisor #x+3# be equated to zero and that is the Vertical asymptote

#x+3=0# or #x=-3#

You can see the lines #x=-3# and #y=2x-3# and the graph of
#f(x)=(2x^2+3x+8)/(x+3)#
graph{(y-(2x^2+3x+8)/(x+3))(y-2x+3)=0[-60,60,-30,30]}

God bless...I hope the explanation is useful..