Posted by **Justin** on Friday, March 4, 2011 at 12:21am.

F(X) = 2x^3-5x^2-19x+1 / x^2-9....I need the vertical asymptote, horizontal asymptote and the slant asymptote...please help!

- Trigonometry -
**agrin04**, Friday, March 4, 2011 at 2:43am
Vertical asymptote: denominator = 0, so: x^2 - 9 = 0

Horizontal: lim x->(infinity) f(x)

Since the degree of nominator is higher than the denominator, then the horizontal asymptote does not exist.

Slant: use long division method to find the quotient. That quotient is the slant asymptote

In this case: divide 2x^3-5x^2-19x+1 with x^2-9

- Trigonometry -
**Anonymous**, Sunday, March 1, 2015 at 7:27pm
k(x)= x^4-2x +7/x^2-9

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