How do you find vertical, horizontal and oblique asymptotes for (x^2+x-6)/(x^2-x-6)x2+x−6x2−x−6?
1 Answer
Explanation:
"let "f(x)=(x^2+x-6)/(x^2-x-6)let f(x)=x2+x−6x2−x−6 The denominator of f(x) cannot be zero as this would make f(x) 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-x-6)=0rArr(x-3)(x+2)=0solve (x2−x−6)=0⇒(x−3)(x+2)=0
x=-2" and "x=3" are the asymptotes"x=−2 and x=3 are the asymptotes
"Horizontal asymptotes occur as"Horizontal asymptotes occur as
lim_(xto+-oo),f(x)toc" ( a constant)"
"divide terms on numerator/denominator by the highest"
"power of "x" that is "x^2
f(x)=(x^2/x^2+x/x^2-6/x^2)/(x^2/x^2-x/x^2-6/x^2)=(1+1/x-6/x^2)/(1-1/x-6/x^2)
"as "xto+-oo,f(x)to(1+0-0)/(1-0-0)
y=1" is the asymptote"
"Oblique asymptotes occur when the degree of the "
"numerator is greater than the degree of the denominator."
"This is not the case here hence there are no oblique"
"asymptotes"
graph{(x^2+x-6)/(x^2-x-6) [-10, 10, -5, 5]}
