How do you find the asymptotes for #y = (x^2 + 2x - 3)/( x^2 - 5x - 6)#?

Question

Anna Allen · 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# #x=-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)# #y=1" is the asymptote"# graph{(x^2+2x-3)/(x^2-5x-6) [-20, 20, -10, 10]}

Elias Ackerman · Answer

undefined To find the asymptotes of $ y = \frac{x^2 + 2x - 3}{x^2 - 5x - 6} $, you first need to determine where the function approaches infinity or negative infinity. Asymptotes occur where the function approaches these values as $ x $ approaches certain values.

1. Vertical asymptotes: Vertical asymptotes occur where the denominator of the rational function equals zero but the numerator doesn't. To find vertical asymptotes, set the denominator equal to zero and solve for $ x $. In this case, the denominator is $ x^2 - 5x - 6 $. So, we solve $ x^2 - 5x - 6 = 0 $ to find the values of $ x $ where vertical asymptotes may occur.

The roots of $ x^2 - 5x - 6 = 0 $ are $ x = -1 $ and $ x = 6 $.

Therefore, the vertical asymptotes are $ x = -1 $ and $ x = 6 $.

2. Horizontal asymptotes: Horizontal asymptotes occur when the degree of the numerator is equal to the degree of the denominator. In this case, both the numerator and the denominator have the same degree, which is 2. Therefore, we divide the leading coefficient of the numerator by the leading coefficient of the denominator to find the horizontal asymptote.

The horizontal asymptote is $ y = \frac{1}{1} = 1 $.

So, the vertical asymptotes are $ x = -1 $ and $ x = 6 $, and the horizontal asymptote is $ y = 1 $.