At which value(s) of x does the graph of the function F(x) have a vertical asymptote? Check all that apply. F(x) = x/(x+2)(x-1)

o x=10
o x=-2
o x=0
o x=-1
o x=2
o x=1

x = - 2, x = 1

Step-by-step explanation:

Given

f(x) =

The denominator cannot be zero as this would make f(x) undefined.

Equating the denominator to zero and solving for x gives the values that x cannot be and if the numerator is non zero for these values then they are vertical asymptotes.

(x + 2)(x - 1) = 0

x + 2 = 0 ⇒ x = - 2

x - 1 = 0 ⇒ x = 1

Vertical asymptotes occur at x = - 2 and x = 1

answer: ouch. thats difficult

step-by-step explanation:

2(2x + 3) = -4 + 2(2x + 5)

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 6

4x - 6 + 6 = 4x - 6 + 6

4x = 4x

x = x

unlimited solutions.