we are given
vertical asymptote at x=-3
so, there will be x+3 in the denominator
we are given zeros at x=-2
so, there will x+2 in the numerator
so, we can set up function as
[tex]f(x)=\frac{a(x+2)}{(x+3)}[/tex]
we can select any one point
and then we can find 'a'
(0,1)
so, we can plug f(x)=1 and x=0
and find 'a'
[tex]1=\frac{a(0+2)}{(0+3)}[/tex]
[tex]a=\frac{3}{2}[/tex]
now, we can plug back
and we get
[tex]f(x)=\frac{\frac{3}{2}(x+2)}{(x+3)}[/tex]..............Answer
