Question

What are the vertical asymptotes of the function f(x) = 4x+8/x^2+3x-4

Answer 1

Answer:

vertical asymptote at x=-4, 1

Step-by-step explanation:

$f(x) =\frac{4x+8}{x^2+3x-4}$

To find vertical asymptote we set the denominator =0 and solve for x

$x^2+3x-4=0$

Product is -4 and sum is +3

WE find out two factors whose sum is +3 and product is -4

$4 \ times \ (-1)=-4$

$4-1= 3$

$x^2+3x-4=0$

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

Set each factor =0 and solve for x

$x+4=0, x=-4$

$x-1=0, x=1$

vertical asymptote at x=-4, 1

Answer 2

Vertical asymptotes occur  when the denominator = 0.

f(x) = 4x + 8 / (x + 4)(x - 1)

when x = -4 and x = 1 the denominator = 0

so the required asymptotes are x = -4 and x = 1