missmathy88 said:
Its attached, thank you
And please correct any mistakes, and I was also stuck on some parts, so please help.
missmathy88,
so you don't miss any information, factor the numerator and the denominator so they have real coefficients:
\(\displaystyle \frac{(x^2 - 1)^2}{x^4 - 1} \ = \\)
\(\displaystyle \frac{(x^2 - 1)(x^2 - 1)}{(x^2 - 1)(x^2 + 1)} \ = \\)
\(\displaystyle \frac{(x - 1)(x + 1)(x - 1)(x + 1)}{(x - 1)(x + 1)(x^2 + 1)}\)
For holes in the graph, one of the ways is where you have the form \(\displaystyle \frac{0}{0}\), which is an indeterminate form.
This happens at \(\displaystyle x = -1 \ and \ at \ x = 1.\)
Except at these two places where there are holes, the graph is continuous (as in the form given by galactus),
so there are no vertical asymptotes as you would find at places in other graphs where there are any
jump discontinuities.
The zeroes are where any x-values in the numerator are equal to \(\displaystyle 0,\) providing those x-values are allowed.
Because the both (all) of the solutions to the numerator set equal to \(\displaystyle 0\) are not part of the domain,
then there are no zeroes.
