How do you find the asymptotes for y = (x^2 + 2x - 3)/( x^2 - 5x - 6)y=x2+2x−3x2−5x−6?
1 Answer
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)=0solve x2−5x−6=0⇒(x−6)(x+1)=0
x=-1" and "x=6" are the asymptotes"x=−1 and x=6 are the asymptotes
"Horizontal asymptotes occur as"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]}
