Question 62456: Find the vertical and horizontal asymptotes of Y=(5x^2-8x+1)/(2x^2-5x-12)
thank you for your time Found 2 solutions by uma, stanbon:Answer by uma(370) (Show Source):
You can put this solution on YOUR website! 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.
You can put this solution on YOUR website! Find the vertical and horizontal asymptotes of Y=(5x^2-8x+1)/(2x^2-5x-12)
-------------------
Factor where you can to get:
y=[5x^2-8x+1]/[(x-4)(2x+3)]
------------------
Vertical asymptotes where the denominator is zero at x=4 and x=-3/2
Horizontal asymtote at y=5/2 because coefficients of highest common
power in numerator and denominator are 5 and 2.
-------
Cheers,
Stan H.
(
