Posted by Justin on .
F(X) = 2x^3-5x^2-19x+1 / x^2-9....I need the vertical asymptote, horizontal asymptote and the slant asymptote...please help!
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
k(x)= x^4-2x +7/x^2-9