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you mean |dw:1378150481156:dw| 5/(x+2) not 5/x+2
Yes, my bad.
anyhoo to find V asymptote you have to find x+2=0 cuzz that would make the denominator so little it will cause the graph to "shoot up" vertically and therefore it is a V asymptote.
Can't it have an H asymptote also?
Now to find the H asymptote you plug in a very large number at x. typically chose 100, 1000, 100000 and you will see the graph get smaller and smaller. In fact it will reach zero and 0 is the H asymptote.
Is there a way i can find the H without plugging anything in?
y=0 is h asymptote x= -2 v asymptote
Why does y=0?
polynomial division is a simple way to find horizontal or oblique asymptotes |dw:1378150869977:dw| ignore the remainder and the quotient is the horizontal asymptote
well 5/(x+2) take that to be 5/a as you plug in a larger and largr number at x you get 5/a and a is larger and larger dividing 5 by a large number gets you 0
@campbell_st that is too complicated lol
lol... easier that limits...
and it works everytime
here is a rule (ax)/(bx) H asymptote will be be a/b EXAMPLE: 2x/(1x) = 2/1 ---- (ax)/(bx^2) Asymptote will be 0 THe above case states if the bottom polynomial is of a higher degree then asymptote will alwasy be 0 1000000x/.4x^2 = 0
so what happens when the degree of the numerator is greater than the degree of the denominator..?
Then there is no Horizontal Asymptote.
ummm interesting ... as the method for finding the oblique is the graph shown is the same as that used for finding the horizontal... division here is an example |dw:1378193922233:dw| the graph is attached ignore the remainder and the quotient is the horizontal or oblique asymptote