How do you find all the asymptotes for #(3x^2 + 2x - 3)/(x-1 )#?

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Answer 1 · 2015-11-09T13:41:44

graph{(3x^2+2x-3)/(x-1) [-35.6, 41.33, -13.23, 25.24]}

#x=1# is a vertical asymptote.
#y=3x+5# is an oblique asymptote.

First simplify the expression by performing long division.

#frac{3x^2+2x-3}{x-1}-=3x+5+frac{2}{x-1}#

As #x->+-oo#, #frac{3x^2+2x-3}{x-1}->3x+5#

#y=3x+5# is an oblique asymptote.

As #x->1^+#, #frac{3x^2+2x-3}{x-1}->+oo#
As #x->1^-#, #frac{3x^2+2x-3}{x-1}->-oo#

#x=1# is a vertical asymptote.