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.

Substitute the coordinate points value in the formula.

From the slope-intercept form y = mx+b where m = slope and b = y-intercept. Therefore slope = m = -1