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.

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