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The function f(x) = x / (x2 - 4) has asymptotes at?

2007-07-08 14:19:28 · 5 answers · asked by Sarah K 1 in Science & Mathematics Mathematics

5 answers

at x = +/- 2


f(x) = x / (x^2 - 4)
= x / (x-2)(x+2)
because the denominator is the difference of two squares

Asymptotes occur particularly when the denominator = 0 and therefore x = 2 or -2

Alternatively
x^2 - 4 = 0 using your original denominator
x^2 = 4
x = +/- 2

and y does not = 0 at the asymptote ... oh - I see what they mean. Horizontal asymptote at 0 as x tends towards infinity

2007-07-08 14:21:10 · answer #1 · answered by Orinoco 7 · 0 0

Asymptotes are parts of the graph where the function is undefined because of division by zero, so set the denominator = 0 and solve

x^2 - 4 = 0

x^2 = 4

x = +/- 2

2007-07-08 21:35:31 · answer #2 · answered by bamavonb 1 · 0 0

if I assume that it is x^2 then the answer is +2 and - 2. these two numbers will make 0 in the denominator which would be the asymptote.

If you meant 2x, then the answer is just 2.

2007-07-08 21:24:23 · answer #3 · answered by jkim972 3 · 0 0

1) x = 2 or -2 [makes the denominator zero]
2) y = 0 [f(x) approaches 0 as x approaches + or - infinity]

2007-07-08 21:24:03 · answer #4 · answered by TFV 5 · 0 0

x = 2
x = -2
y = 0
(All that assuming "x2" means x squared)

2007-07-08 21:22:03 · answer #5 · answered by talr 4 · 1 0

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