f(x)=x-2/ x^2 -3x -10...
I get x=-2, X=5... Is that correct??
Thanks
2006-07-04 11:10:09 · 6 answers · asked by c2pre 2 in Science & Mathematics ➔ Mathematics
Remember that your vertical asymptote(s) is/are the excluded value(s) of x. Your denominator is (x - 5)(x + 2), correct? With that, you know that your denominator canNOT be zero, otherwise the statement is empty set. Your vertical asymptotes are x=-2 and x=5. Yes, you are correct. :]
To show that you are correct, when you plug in -2 or 5 for your value, your denominator will be 0 and that isn't possible. The vertical asymptote cuts off what your answer cannot be on the graph. As I said earlier, the vertival asymptote(s) are the excluded value(s) for x, also known as the domain.
2006-07-04 11:15:01 · answer #1 · answered by tingaling 4 · 0⤊ 0⤋
x^2 - 3x - 10
(x - 5)(x + 2)
You are correct.
2006-07-04 11:41:39 · answer #2 · answered by Sherman81 6 · 0⤊ 0⤋
Here you will find the answer you are looking for...
http://cs.gmu.edu/cne/modules/dau/calculus/limits/limits7_bdy.html
2006-07-04 11:13:59 · answer #3 · answered by Anonymous · 0⤊ 0⤋
only one of the answers is valid
2006-07-04 11:15:57 · answer #4 · answered by alchemthis 2 · 0⤊ 0⤋
For goodness sake, do your own homework.
Moron.
2006-07-04 11:13:22 · answer #5 · answered by Anonymous · 0⤊ 0⤋
Sorry... I'm better with women's figures
2006-07-04 11:17:28 · answer #6 · answered by AL 6 · 0⤊ 0⤋