Asymptote Questions?

Question

I am not quite getting this...

F(x) = (x^2-3x-10)/(x^2-4)

The question for this function is For which values of x would the graph F have an open circle??

LOL..It's driving me Koo Koo for Cocoa Puffs...HELP!

Answer 1

You are looking for discontinuities. This will occur whenever the denominator is zero. Hence set x^2-4=0. This factors as (x-2)(x+2)=0. So if x=2 or x=-2, you will have a discontinuity.

Answer 2

If you factor the numerator
x^2-3x-10 = 0
x=((3+-sqrt(49)/2
so x = 5 and x= -2
so you can write your expression as
F(x)=(x-5)(x+2)/(x-2)(x+2) which for x not-2 can be written as
F(x) =(x-5)/(x-2) so at x=-2 you have an avoidable discontinuity
defining F(-2)= 7/4
at x=2 F is discontinuos and x=2 is an asymptote as lim F x=<2 is infinity
y=1 is also an asymptote as lim F x=> infinity is 1

Answer 3

You are looking for hole in the graph (a.k.a. "removable discontinuity"). That happens when there is a common factor in the numerator and denominator. You have to factor to determine if and where this occurs.

[(x-5)(x+2)] / [(x-2)(x+2)]

Since x+2 is a common factor, there is a hole in the graph at x=-2. You are now working with a more simplified function:

f(x) = (x-5)/(x-2) with a hole at (-2, ?)

So where is the hole located? Plug in x= -2 into the new simplified function to get the hole's location at (-2, 7/4).