How do you identify all asymptotes or holes for #h(x)=(3x^2+2x-5)/(x-2)#?

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Answer 1 · 2018-05-31T00:33:11

Below

#h(x) = (3x^2+2x-5)/(x-2)#

#h(x) = ((3x+5)(x-1))/(x-2)#

For vertical asymptote, #x-2=0# since your denominator cannot equal to 0.

#x-2=0#
#x=2# is your vertical asymptote

Now we need to find the horizontal or oblique asymptote.

If you divide your polynomial, you will end up getting:

#h(x) = ((x-2)(3x+8)+11)/(x-2)#

#h(x) = 3x+8+11/(x-2)#

Therefore, your oblique asymptote is #y=3x+8#

To find your x-intercepts, let #y=0#

#0=(3x^2+2x-5)/(x-2)#

#0=(3x+5)(x-1)#

#x=1, -5/3#
ie #(1,0)# #(-5/3,0)#

To find your y-intercepts, let #x=0#

#y=(3x^2+2x-5)/(x-2)#

#y=(0+0-5)/(0-2)#

#y=-5/2#

ie #(0,-5/2)#

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