Question

How do I find the holes of this function?

Answer 1

The holes in this function are the points where you can cancel factors from the numerator and denominator.

We factor the function.

$y= \dfrac{3x^2-7x+4}{x^2-1} = \dfrac{3x^2-3x-4x+4}{x^2-1}= \dfrac{3x(x-1)-4(x-1)}{x^2-1} \\\\\\ y = \dfrac{(3x-4)(x-1)}{(x+1)(x-1)} = \dfrac{3x-4}{x+1}, \ x \neq -1, \ x \neq 1$

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 $(1, \frac{-1}{2})$.

Answer 2

Factor the numerator and then the denominator as well. If there are any holes in the function, they will be where the numerator and denominator are the same... (x-1)