Question

How do I find the holes of this function?

Answer 1

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)

Answer 2

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

We factor the function.


`y= (3x^2-7x+4)/(x^2-1) = (3x^2-3x-4x+4)/(x^2-1)= (3x(x-1)-4(x-1))/(x^2-1) \\\\\\ y = ((3x-4)(x-1))/((x+1)(x-1)) = (3x-4)/(x+1), \ x \\eq -1, \ x \\eq 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, (-1)/(2))`.