Question 62456
Given y = (5x^2 - 8x + 1)/ (2x^2 - 5x - 12)

To find the vertical asymptotes , equate the denominator to zero.

==> 2x^2 - 5x - 12 = 0
==> 2x^2 - 8x + 3x - 12 = 0
==> 2x(x-4) + 3(x-4) = 0
==> (x-4)(2x + 3) = 0
==> x - 4 = 0  or 2x + 3 = 0
==> x = 4  or x = - 3/2

x = 4 and x = -3/2 are the two vertical asymptotes.

To find the horizontal asymptotes, 

Since the degrees of the numerator and the denominator are the same (each being 2), then this rational has a non-zero (non-x-axis) horizontal asymptote, not a slant asymptote, and the horizontal asymptote is found by dividing the leading terms:
= 5/2

Thus y = 5/2 is the horizontal asymptote and x = 4 and x = -3/2 are the vertical asymptotes.


Good Luck!!!