What is the discontinuity of (x^2 + 3x - 4) / (x^2 + x - 12) ?

Firstly, factorise both the numerator and denominator to simplify it.

Let y = (x^2 + 3x - 4) / (x^2 + x - 12)

y = (x^2 + 3x - 4) / (x^2 + x - 12)

= (x - 1) (x + 4) / (x - 3) (x + 4)

= (x - 1) / (x - 3)

Then, by long division,

(x - 1) / (x - 3) = 1 + [2 / (x - 3) ]

For vertical asymptote, when y tends to infinity, (x - 3) will tend to 0. Hence, x = 3.

For horizontal asymptote, when x tends to infinity, y = 1.