f(x) = (3x^2 - x - 10) / (x^2 - 4)
ok, so the vertical asymptote is x = +/- 2, but when i plug the equation in my graph calculator, i can only see the graph approaches x = -2 but not x = 2. Why is this?
And how do you know whether an equation approaches a diffrerent asymptote, like y= -x +1, ..., besides horizonal asymptote and vertical? i still struggle with these, please help.
2007-03-28 14:28:45 · 4 answers · asked by 7 in Science & Mathematics ➔ Mathematics
You should factor before you do your analysis of asymptotes.
3x^2-x-10=(x-2)(3x+5), so you get f(x)=(3x+5)/(x+2) except for x=2.
You find oblique asymptotes by dividing out: In this case,
(3x+5)/(x+2)=3 - 1/(x+2), so as x-->infty, f(x)-->3. In the case of oblique asymptotes, there would be a linear term plus something going to zero, not just a constant term plus something going to zero.
2007-03-28 14:37:00 · answer #1 · answered by mathematician 7 · 1⤊ 0⤋
Your denominator's zeros could be either VAs or holes (officially known as removable discontinuities). You must factor to know because a factor that's in both the numerator and denominator and cancels out is a hole. If it remains in the denominator, it's a VA.
f(x) = (3x+5)(x-2) / (x-2)(x+2) --> (3x+5) / (x+2)
This means there is a hole in the graph at x = 2 and a VA at x = -2. The actual location of the hole is (2, 11/4) because f(2) = 11/4 (that's the new f(x) after the x-2 factor cancels).
Your second question refers to an oblique or slant asymptote. You will know this exists if the highest degree of the numerator is greater than the highest degree of the denominator. You find the equation of this slant asymptote by performing long or synthetic division and using the quotient without its remainder.
2007-03-28 22:49:21 · answer #2 · answered by Kathleen K 7 · 0⤊ 0⤋
As was stated,
f(x)=(3x+5) / (x+2) after factoring
So we've got the vertical, x = -2
The other asymptotes go like this:
m = degree of denominator
n = degree of numerator
If m > n
then, the asymptote is the horizontal line, y= 0
If m =n
then, the asymptote is the ratio of the highest-power terms:
As in your case:
3x / x = 3; y=3 is the asymptote
If m < n
then, the asymptote is an end-behavior asymptote:
i.e. say it was (3x^2 +5) / (x+2)
and 3x^2 / x = 3x
y = 3x is the asymptote
2007-03-28 21:45:04 · answer #3 · answered by Anonymous · 0⤊ 0⤋
My dear friend, good question, x does not give an asymptote at x=+2 because when you plug-in x=+2 in the numerator, it gives a 0. So, 0/0 is also a 0!!!
Also, if the degree of the numerator is lesser than the degree of the denominator, you get a slant asymptote.
2007-03-28 21:37:08 · answer #4 · answered by Sdrawkcab 2 · 1⤊ 0⤋