Answer:
x=-1 and x=-2
Step-by-step explanation:
Given
[tex]f(x)=\dfrac{1}{(x+1)(x+2)}[/tex]
This function is undefined when the denominator becomes 0. Find values of x for which the denominator is 0:
[tex](x+1)(x+2)=0\\ \\x+1=0\ \text{or}\ x+2=0\\ \\x=-1\ \text{or}\ x=-2[/tex]
Thus, the lines with equations [tex]x=-1[/tex] and [tex]x=-2[/tex] represent vertical asymptotes (see attached graph)
