Answer
$f(x)=\frac{x^3}{4(x+2)^2}$ matches with the graph labeled (d).
Work Step by Step
The vertical asymptote is:
$4(x+2)^2=0$
$(x+2)^2=0$
$x+2=0$
$x=-2$
The x-intercept:
$f(x)=\frac{x^3}{4(x+2)^2}=0$.
$x^3=0$
$x=0$
The graph passes through the origin.
The degree of the numerator is greater than the degree of the denominator. The graph has no horizontal asymptotes.
