Answer:
x = -4
Step-by-step explanation:
Vertical asymptotes are found where the denominator has a zero that is not cancelled by a numerator zero. Here, the expression simplifies to ...
[tex]f(x)=\dfrac{x^2+7x+10}{x^2+9x+20}=\dfrac{(x+5)(x+2)}{(x+5)(x+4)}=\dfrac{x+2}{x+4} \qquad x\ne -5[/tex]
The function is undefined at x=-5, but has a vertical asymptote at x=-4.
