What are the vertical asymptotes of the function f(x) = 2x + 8 / x2 + 5x + 6?

I'm assuming the function is f(x) = (2x+8)/(x^2+5x+6). If so, make sure to use parenthesis to indicate that you're dividing all of "2x+8" over all of "x^2+5x+6" as one big fraction. Otherwise, things are ambiguous and it leads to confusion.

Side Note: x^2 means "x squared"

Factor the numerator: 2x+8 = 2(x+4)
Factor the denominator: x^2+5x+6 = (x+2)(x+3)

There are no common factors between the numerator and denominator. So there is nothing to cancel out. 

Recall that you cannot divide by zero. Something like 1/0 is undefined. 
We need to find the x values that cause the denominator to be zero.
Set the denominator equal to zero and solve for x
x^2+5x+6 = 0
(x+2)(x+3) = 0
x+2 = 0 or x+3 = 0
x = -2 or x = -3

The x values x = -2 or x = -3 will lead to the denominator being zero. This means that the vertical asymptotes are x = -2 or x = -3 as shown by the blue dashed vertical lines in the attached image.

put like terms together.orremember that the 45 becomes negative when moved to the other side.orthe minuses cancel out.

(1/x + 1)^2 + 6 (1/x + 1) +5