Sunday
May 19, 2013

# Homework Help: Algebra

Posted by Angie on Wednesday, August 5, 2009 at 6:29pm.

I need the horizontal asymptote for

f(x) = (x)/(x^2 -4)

The y-intercept is 0.
The vertical asymptotes are x = -2, x = 2

So now all i need is horizontal asymptote for

f(x) = (x)/(x^2 -4)

• Algebra - bobpursley, Wednesday, August 5, 2009 at 6:42pm

As x>>inf, the f(x) approaches zero, both directions.

• Algebra - Angie, Wednesday, August 5, 2009 at 6:51pm

What does this mean in words please? Thanks!

• Algebra - Angie, Wednesday, August 5, 2009 at 7:12pm

What is the value of the

horizontal asymptote for

f(x) = (x)/(x^2 -4)

This is my question?

• Algebra - MathMate, Wednesday, August 5, 2009 at 7:26pm

For your benefit, I supply you herewith the link to another thread that may possibly give you some more hint.
http://www.jiskha.com/display.cgi?id=1249414112

• Algebra - Angie, Wednesday, August 5, 2009 at 7:40pm

So does this mean that the horizontal asymptote for

f(x) = (x)/(x^2 -4) Is not 0?

A horizontal asymptote is the y-value approaches as x approaches 亇 亣. Does this include 0? y= 0 ???

• Algebra - MathMate, Wednesday, August 5, 2009 at 8:37pm

I do not know if you have read the link I cited earlier. However, the link includes another link to a sketch of the function which gives you a much better idea of how the function behaves as x approches ±∞.

In my school days, we were supposed to make these sketches free-hand by inspection of the function, no calculators, and no calculations. In fact, one of the objectives of the purpose of the posted question is probably to familiarize the student with how functions behave under different conditions, namely identify asymptotes, x and y-intercepts, concavity/convexity, behaviour of the function for x=±∞, etc.

• Algebra - MathMate, Wednesday, August 5, 2009 at 8:38pm

Forgot to supply the link to the sketch, distracted by babbling, sorry.
http://i263.photobucket.com/albums/ii157/mathmate/Jessica-1.png

• Algebra - Angie, Wednesday, August 5, 2009 at 8:43pm

Thanks! So then the final answer is that

(x)/(x^2-4)

(x)/(x-2) (x+2)
Standard form then this = y = 0?

• Algebra - MathMate, Wednesday, August 5, 2009 at 9:04pm

I am not sure in what context to which you refer as standard form.

The form (x)/((x-2)(x+2)) helps you to identify the vertical asymptotes, as they show where the denominator becomes zero, while the original expression of x/(x²-1) makes the horizontal asymptote evident, assuming you are familiar with the l'Hôpital rule.

The expression (x)/(x-2) (x+2) is algebraically inaccurate because the last factor should be part of the denominator. The expression implies that it is in the numerator. Parentheses are required to clarify and rectify it:
(x)/((x-2)(x+2))

Yes, the horizontal asymptote is at y=0, as indicated by Mr.Bob initially.

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