Question

ASAP!What are the vertical asymptotes of the function f(x) = the quantity of 3x plus 9, all over x squared plus 4x minus 12?

A) x = −6 and x = −2
B) x = −6 and x = 2
C) x = 1 and x = −2
D) x = 1 and x = 2

Answer 1

Answer:

C

Step-by-step explanation:

Answer 2

Answer:

B) x = -6  and  x = 2

Step-by-step explanation:

$\frac{3x+9}{x^{2}+4x-12}$

can be rewritten as

$\frac{3x+9}{(x+6)(x-2)}$

because

-2 · 6 = -12

-2 + 6 = -4

so thats why the denominator is (x+6)(x-2)

A vertical asymptote is where the denominator equals 0, so,

-6 + 6 = 0

2 + -2 = 0

So the Answer is:

B) x = -6  and  x = 2