English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

which graph includes no aysmptotes?

a. f(x) = 1/(x2 + 1)

b. g(x) = 1/(x - 1)

c. h(x) = 1/(x2 - 1)

d. p(x) = 1/[(x2 - 1)(x - 1)]

e. q(x) = x3/[(x2 - 3)(x2 - 5)]

f. r(x) = x3/(x2 + 1)

2007-04-01 08:22:26 · 4 answers · asked by kat 3 in Science & Mathematics Mathematics

4 answers

(f), since the denominator is always positive -- never zero. (a) has no vertical asymptotes, but it approaches a horizontal asymptote of zero. Since the numerator of (f) is of higher degree than the denominator, there is no horizontal asymptote.

(I'm guessing your teacher doesn't buy the concept of a "slant asymptote"?)

2007-04-01 08:31:08 · answer #1 · answered by tedfischer17 3 · 0 1

I presume you are referring to horizontal asymptotes.

f) has no horizontal but it has a diagonal or slant asymptote.

a) to e) have asymptotes y = 0.

Now, about vertical asymptotes,

a) none
b) x =1
c) x^2 = 1 or x = +1 or -1
d) x=1, -1
e) x = sqrt (3) or sqrt(5)

2007-04-01 15:35:38 · answer #2 · answered by Aldo 5 · 0 0

Hi. I'm not great at math. But I notice that some equations have a 'divide by zero' when x=1.

2007-04-01 15:31:14 · answer #3 · answered by Cirric 7 · 0 0

the answer is f

a) has a horizonal asymptote
b) has a vertical asymptote
c) both horizonal and vertical asymptote
d) both horizonal and vetical asymptotes
e) vertical asymtote
f) none

2007-04-01 15:39:54 · answer #4 · answered by      7 · 0 0

fedest.com, questions and answers