there is a removable discontinuity at x = -4 ... also known as a "hole".
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.
For the vertical asymptote, you need to make sure the function goes towards infinity at that point.
This means to check the limit at that point, and you do this by dividing out the factor like your teacher showed.
Here is an example.
Does the graph y = x have any vertical asymptotes? No right?
What about y = x^2 / x?
x = 0 makes it undefined, but it is the same as y = x at other values (it just has a "hole" at 0 but no "spike"). So you have to make sure you divide out as many factors as possible before concluding a point has a vertical asymptote. Does this make sense?