Question

For what values of x, if any, does `f(x) = 1/((x-3)(x-7))` have vertical asymptotes?

Answer 1

`x=3` vertical asymptote
`x=7` vertical asymptote

graph{1/((x-3)(x-7)) [-10, 10, -5, 5]}

Explanation:

Vertical asymptotes could be find in the points where the Exist Conditions of `f(x)` are not satisfied

The Field of Existence of `f(x)` is:

`x in ]-oo,3[uu]3,7[uu]7,+oo[`

because

`f(x)=(N(x))/(D(x))`

FE (Field of Existence):

`D(x)!=0`

`(x-3)(x-7)!=0`

`x_1!=3`
`x_2!=7`

Now these values of `x` are vertical asymptotes if:

`lim_(x rarr x_(1,2)^+-)f(x)=+-oo`

`lim_(x rarr 3^-)f(x)=lim_(x rarr 3^-)1/((2.9-3)(2.9-7))=`
`lim_(x rarr 3^-)1/((0^-)(-4.1))=lim_(x rarr 3^-)1/0^+=+oo`

`lim_(x rarr 3^+)f(x)=lim_(x rarr 3^+)1/((3.1-3)(3.1-7))=`
`lim_(x rarr 3^+)1/((0^+)(-3.9))=lim_(x rarr 3^+) 1/(0^-) =-oo`

`lim_(x rarr 7^-)f(x)=lim_(x rarr 7^-)1/((6.9-3)(6.9-7))=`
`lim_(x rarr 7^-)1/((3.9)(0^-))=lim_(x rarr 7^-)1/(0^-)=-oo`

`lim_(x rarr 7^+)f(x)=lim_(x rarr 7^+)1/((7.1-3)(7.1-7))=`
`lim_(x rarr 7^+)1/((4.1)(0^+))=lim_(x rarr 7^+)1/(0^+)=+oo`