SOLUTION: Give the locations of horizontal and vertical asymptotes and the x-coordinate of holes. y = {{{ (2x^3 + 2)/(x^2 - 9) }}}

Algebra ->  Rational-functions -> SOLUTION: Give the locations of horizontal and vertical asymptotes and the x-coordinate of holes. y = {{{ (2x^3 + 2)/(x^2 - 9) }}}       Log On


   

Question 860506: Give the locations of horizontal and vertical asymptotes and the x-coordinate of holes.
y = +%282x%5E3+%2B+2%29%2F%28x%5E2+-+9%29+
Answer by josgarithmetic(39625) About Me  (Show Source):
You can put this solution on YOUR website!
No holes, but two vertical asymptotes.

Doing division you get y=2x+some remainder, and the remainder becomes increasingly small as x gets extreme to either left or right. The slant asymptote is y=2x. I do not show this polynomial division, so ask for it if you need to see it.

The denominator is %28x-3%29%28x%2B3%29; so vertical asymptotes are at x=3 and x=-3.