How do you find the Vertical, Horizontal, and Oblique Asymptote given #y = (x^2 + 2x - 3)/( x^2 - 5x - 6) #?

1 Answer
Jan 23, 2017

Answer:

vertical asymptotes at x = -1 and x = 6
horizontal asymptote at y = 1

Explanation:

The denominator of y cannot be zero as this would make y undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve : #x^2-5x-6=0rArr(x-6)(x+1)=0#

#rArrx=-1" and "x=6" are the asymptotes"#

Horizontal asymptotes occur as

#lim_(xto+-oo),ytoc" ( a constant)"#

divide terms on numerator/denominator by the highest power of x, that is #x^2#

#y=(x^2/x^2+(2x)/x^2-3/x^2)/(x^2/x^2-(5x)/x^2-6/x^2)=(1+2/x-3/x^2)/(1-5/x-6/x^2)#

as #xto+-oo,yto(1+0-0)/(1-0-0)#

#rArry=1" is the asymptote"#

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 2 ) Hence there are no oblique asymptotes.
graph{(x^2+2x-3)/(x^2-5x-6) [-10, 10, -5, 5]}