What are the vertical asymptotes of the function f (x) = the quantity of 4 x plus 8, all over x squared plus 3 x minus 4? x = - 1 and x = - 2 x = - 1 and x = 2 x = - 1 and x = - 4 x = 1 and x = - 4

To find a rational function's vertical asymyptotes, you need to find when the denominator is zero. So when is the denominator x² + 3x - 4 zero? We set that denominator equal to zero.

x² + 3x - 4 = 0

This second degree equation factors. We want a pair of numbers that multiply to 4, and there are only two pairs: 2 and 2, 1 and 4. We want the sum to 3, and one must be negative. - 2 and 2 add to zero, - 4 and 1 add to - 3, - 1 and 4 add to 3. The last one is the pair.

x² + 3x - 4 = 0

(x + 4) (x - 1) = 0

x = - 4 or x = 1

So the function's vertical asymptotes are at x = 1 and x = - 4.