sketch the graph and show any asymptotes: f(x)= x^3/2(x-1)^2(x+1)
Plug in x=0, 1, 2, 3 etc and see what values of f(x) come out. Then, you can graph based on that.
ok, do you know if the Horizontal aymptptote is y=1 and the vertical is x=1 and x=-1?
When x =1, y will be infinity. Similarly, when x =-1, y will be infinity. So, yes.
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