Question

How do you find the removable discontinuity(hole) for the graph of `y=(x^2 - 9x -10)/ (2x^2 - 2)`?

Answer 1

Factoring the numerator and denominator reveals a common factor of `(x+1)` which can be removed so the (modified) fraction is no longer discontinuous at `x=-1`

Explanation:

`y=(x^2-9x-10)/(2x^2-2)`

Factoring:
`color(white)("XXXX")` `y = ((x-10)(x+1))/(2(x^2-1))`

Continue by factoring the difference of squares in the denominator
`color(white)("XXXX")` `y = ((x-10)(x+1))/(2(x+1)(x-1))`

Divide the numerator and denominator by `(x+1)`
`color(white)("XXXX")` `=y = ((x-10)cancel((x+1)))/(2cancel((x+1))(x-1))`

`color(white)("XXXX")` `y = (x-10)/(2(x-1))`

Note that the discontinuity at `x=1` (which would cause an attempt to divide by zero) can not be removed.