# Vertical Asymptotes

• Dec 15th 2010, 06:13 PM
Vertical Asymptotes
Sorry, the latex help is down so I don't know how to write this any other way...

G(x) = (x+2)(2x+8)/(x^2 + 4x)

I am asked to find how many vertical and horizontal asymptotes are in this equation. I find the one horizontal asymptote without any issues, however I am having an issue with the vertical asymptote.

I know that square roots and denominators give me vertical asymptotes. In the absence of a square root I just took the denominator and set it to equal zero. This is my logic:

x^2 + 4x = 0
x(x+4) = 0
x = 0 and x+4 = 0
x = 0 and x = -4

So by my thinking there are 2 vertical asymptotes in this equation, namely at -4 and 0. I check this by plugging it in:

G(0) = (0+2)(2(0)+8)/(0^2 + 4(0))
G(0) = (2)(8)/(0 + 0)
G(0) = 16/0
G(0) = undefined!

G(-4) = (-4+2)(2(-4)+8)/((-4)^2 + 4(-4))
G(-4) = (-2)(-8+8)/(16 - 16)
G(-4) = 0/0
G(-4) = undefined! (I think...

So this looks right to me, but my prof posted the answers and did something different. His logic follows:

G(x) = (x+2)(2)(x+4)/x(x+4)
G(x) = 2(x+2)/x where x != -4 //divide both top and bottom by (x+4)

So he says we have a vertical asymptote at x = 0 but no vertical asymptote at x = -4 since the function is undefined at -4 but isn't it also undefined at 0? I'm confused.

Thanks!
• Dec 15th 2010, 06:19 PM
skeeter
there is a removable discontinuity at x = -4 ... also known as a "hole".
• Dec 15th 2010, 06:22 PM