Question

How do you find the vertical asymptotes and holes of `f(x)=1/(x^2+5x+6)`?

Answer 1

`"vertical asymptotes at "x=-3" and "x=-2`
`"horizontal asymptote at "y=0`

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)=0`

`x=-3" and "x=-2" are the asymptotes"`

`"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]}