Question

Describe where the function has a vertical asymptote and how you found your answer. Remember that an asymptote is represented by an equation of a line and not just a single value.

Answer 1

x=-4 is a vertical asymptote

Explanation:

A vertical asymptote of the graph of a rational function f(x) is a line x=a, such that one of of these statements is fulfilled

`\displaystyle \lim _(x\to a^(+))f(x)=\pm \infty`

`\displaystyle \lim _(x\to a^(+))f(x)=\pm \infty`

Our function is

`(x^2+7x+10)/(x^2+9x+20)`

To find the candidate values of a, we set the denominator to zero

`x^2+9x+20=0`

Factoring

`(x+4)(x+5)=0`

Which gives us two possible vertical asymptotes: x=-4 or x=-5

We now must confirm if one of the two conditions are true for each value of a

`\displaystyle \lim _(x\to -4^(-))(x^2+7x+10)/(x^2+9x+20)`

The numerator can be factored as

`x^2+7x+10=(x+2)(x+5)`

So our limit is

`\displaystyle \lim _(x\to -4^(-))((x+2)(x+5))/((x+4)(x+5))`

Simplifying:

`=\displaystyle \lim _(x\to -4^(-))((x+2))/((x+4))=+\infty`

We can see x=-4 is a vertical asymptote

Checking with x=-5, and using the simplified limit:

`\displaystyle \lim _(x\to -5^(-))((x+2))/((x+4))=3`

`\displaystyle \lim _(x\to -5^(+))((x+2))/((x+4))=3`

The limit exists and is 3, so x=-5 is NOT a vertical asymptote

The only vertical asymptote of the function is x=4