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

Question

Lincoln Anderson · Accepted Answer

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

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]}

Liam Brennan · Answer

undefined To find the vertical asymptotes, set the denominator equal to zero and solve for x. Vertical asymptotes occur where the function is undefined.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0. If the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients. If the degree of the numerator is greater, there is no horizontal asymptote.

To find the oblique asymptote, divide the numerator by the denominator using polynomial long division. The oblique asymptote is the quotient obtained from the division. If the degree of the numerator is less than the degree of the denominator, there is no oblique asymptote.