How do you find the vertical asymptotes and holes of f(x)=1/(x^2+5x+6)f(x)=1x2+5x+6?
1 Answer
Explanation:
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+5x+6=0rArr(x+3)(x+2)=0solve x2+5x+6=0⇒(x+3)(x+2)=0
x=-3" and "x=-2" are the asymptotes"x=−3 and x=−2 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)=(1/x^2)/(x^2+x^2+(5x)/x^2+6/x^2)=(1/x^2)/(1+5/x+6/x^2)
"as "xto+-oo,f(x)to0/(1+0+0)
y=0" is the asymptote"
"holes occur when a common factor is removed from"
"the numerator/denominator. This is not the case here"
"hence there are no holes"
graph{1/(x^2+5x+6) [-10, 10, -5, 5]}
