The holes in this function are the points where you can cancel factors from the numerator and denominator.
We factor the function.
[tex]y= \dfrac{3x^2-7x+4}{x^2-1} = \dfrac{3x^2-3x-4x+4}{x^2-1}= \dfrac{3x(x-1)-4(x-1)}{x^2-1} \\\\\\ y = \dfrac{(3x-4)(x-1)}{(x+1)(x-1)} = \dfrac{3x-4}{x+1}, \ x \neq -1, \ x \neq 1[/tex]
We canceled (x-1) from the numerator and denominator, so when this factor is equal to zero, there is a hole on the graph, that is, at the point [tex](1, \frac{-1}{2})[/tex].
