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August 29, 2014

August 29, 2014

Posted by **Anonymous** on Thursday, October 28, 2010 at 9:37am.

It is supposed to be y=0 but then I even know that b/d is the horizontal asymptote, in this since there is no b, it would be 0/1 which is 0

is the above reasoning correct?

- Math - horizontal asymptote of f(x) (check) -
**jai**, Thursday, October 28, 2010 at 9:39amhorizontal asymptote occurs where the value of x is restricted,,

n the given function f(x)=x/(x-1)^2, what value of x is restricted?

- Math - horizontal asymptote of f(x) (check) -
**jai**, Thursday, October 28, 2010 at 9:41amoops i got it wrong

horizontal asymptote occurs where the value of y is restricted,,

so in the given function f(x)= y = x/(x-1)^2, what value of y is restricted?

- Math - horizontal asymptote of f(x) (check) -
**Reiny**, Thursday, October 28, 2010 at 9:48amyour function expanded is

f(x) = x/(x^2 - 2x + 1)

For a horizontal asymptote we look at what happens to the function as x ---> infinity.

Use an intuitive approach ...

as x becomes very large, say x = 1 million, the denominator becomes large much faster than the numerator.

So you have a division by a hugely large number resulting in a number close to zero

So when x ---> + infinity, f(x) ---> +0 (still above the x-axis)

wen x ---> - infinity , f(x) ----> -0 (slightly below the x-axis)

- Math - horizontal asymptote of f(x) (check) -
**Anonymous**, Thursday, October 28, 2010 at 9:55amis it that the y-value is restricted to being positive. sine the horizontal asymptote has an equation of y=0.

- Math - horizontal asymptote of f(x) (check) -
**jai**, Thursday, October 28, 2010 at 9:59amhorizontal asymptote:

y = 0 , as x approaches plus/minus infinity

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