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

Answer:

x=-4 is a vertical asymptote

Step-by-step 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

$\frac{x^2+7x+10}{x^2+9x+20}$

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

$\displaystyle \lim _{x\to -4^{-}}\frac{x^2+7x+10}{x^2+9x+20}$

The numerator can be factored as

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

So our limit is

$\displaystyle \lim _{x\to -4^{-}}\frac{(x+2)(x+5)}{(x+4)(x+5)}$

Simplifying:

$=\displaystyle \lim _{x\to -4^{-}}\frac{(x+2)}{(x+4)}=+\infty$

We can see x=-4 is a vertical asymptote

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

$\displaystyle \lim _{x\to -5^{-}}\frac{(x+2)}{(x+4)}=3$

$\displaystyle \lim _{x\to -5^{+}}\frac{(x+2)}{(x+4)}=3$

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

The only vertical asymptote of the function is x=4