Calculus one help?
I need help finding the vertical and horizontal asymptotes of this function. I'm mostly stuck with the vertical asymptotes but I would appreciate help with both!!
- Wayne DeguManLv 71 month ago
g(x) = 2x³/(x³ + 8)
Dividing through by x³ we get:
g(x) = 2/(1 + 8/x³)
As x --> ±∞, 8/x³ --> 0
so, g(x) --> 2/(1 + 0)...i.e. 2
Hence, g(x) = 2 is a horizontal asymptote
Vertical asymptotes occur when g(x) does not exist
i.e. when x³ + 8 = 0
so, when x = -2...vertical asymptote
A sketch is below.
- no sea naboLv 61 month ago
Remember that (x³+8)=(x+2))(x²-2x+4) and g(-2)=∄ because (x+2)=0 in the denominator
So your answer is x=-2⁻ and x=-2⁺