Question

At which values of x does the graph of the function F(x) have a vertical asymptote? Check all that apply f(x)=1/(x+1)(x+2)

Answer 1

Answer:

x=-1 and x=-2

Step-by-step explanation:

Given

$f(x)=\dfrac{1}{(x+1)(x+2)}$

This function is undefined when the denominator becomes 0. Find values of x for which the denominator is 0:

$(x+1)(x+2)=0\\ \\x+1=0\ \text{or}\ x+2=0\\ \\x=-1\ \text{or}\ x=-2$

Thus, the lines with equations $x=-1$ and $x=-2$ represent vertical asymptotes (see attached graph)