posted by Meron .
What is the zero of x^3+1/x^2-1 (my answer is -1. Is that correct?) please help
If you meant
then if x = -1, your expression becomes 0/0 which is "indeterminate".
notice that (x^3+1)/(x^2-1)
= (x+1)(x^2 + x + 1)/((x+1)(x-1))
= (x^2 + x + 1)/(x-1)
I don't know if you know anything about limits, but
the limit of the above , as x --> - 1
= (1 - 1 + 1)/(-2) = -1/2
There are actually no zeros for this graph,
will show you.
the expression actually reduces to
y = (x^2 + x + 1)/(x-1) , which has a vertical asymptote at x=1
the original function
y = (x^3 + 1)(/(x^2-1) looks exactly the same except it would have a "hole" or missing point at (-1, -1/2)
Thank u so much!