Question

f(x)=x(2+x)/2x-x^2
How can I find the hole (if any) in the graph?

Answer 1

Here's my work so far:
x(2+x)/x(2-x)
the x's cancel
2+x/2-x
x = 2 (vertical asymptote)
I think the hole is 0, not sure though.

Answer 2

Because x=0

Answer 3

Well, cannot equal 0.

Answer 4

Yes, the hole would be at x=0

Answer 5

Thank you!