Calculus: Early Transcendentals, Chapter 4, 4.6, Section 4.6, Problem 14

f(x)=((2x+3)^2(x-2)^5)/(x^3(x-5)^2)
Vertical asymptotes are the undefined points, also known as zeros of denominator.
Let's find the zeros of denominator of the function,
x^3(x-5)^2=0
x^3=0 ,(x-5)^2=0
x=0, x=5
Vertical Asymptotes are x=0 , x=5
For Horizontal Asymptotes
Degree of Numerator of the function=7
Degree of Denominator of the function=5
Degree of Numerator> 1+Degree of Denominator
:. There is no Horizontal Asymptote
Now let's find intercepts
x intercepts can be found when f(x)=0
(2x+3)^2(x-2)^5=0
2x+3=0 , x-2=0
x=-3/2 , x=2
So x intercepts are -1.5 and 2.
Since x is undefined at x=0 , there are no y intercepts.
See the attached graph and link.
From the graph,
Local Minimum
f(-1.5)=0
f(8) ~~ 600
No Local Maximum