Question

What are the excluded values for x for x+4/-3x^2+12x+36

Answer 1

we are given

$\frac{(x+4)}{-3x^2+12x+36}$

we know that

values of x excluded when that value of x makes f(x) undefined

and any function is undefined when it's denominator becomes 0

so, we set denominator =0

and then we solve for x

$-3x^2+12x+36=0$

we can factor out

$-3(x^2-4x-12)=0$

$(x^2-4x-12)=0$

$(x-6)(x+2)=0$

we get

$(x-6)=0$

$x=6$

$(x+2)=0$

$x=-2$

so, excluded values are

$x=-2,x=6$..............Answer

Answer 2

Answer:

C) -2, x=6

Step-by-step explanation: