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

1 Answer
Aug 8, 2018

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#

#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]}