Math - horizontal asymptote of f(x) (check)
posted by Anonymous .
What is the horizontal asymptote of f(x)=x/(x-1)^2
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?
horizontal 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?
oops 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?
your 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)
is it that the y-value is restricted to being positive. sine the horizontal asymptote has an equation of y=0.
y = 0 , as x approaches plus/minus infinity