SOLUTION: whats the VA and HA of this problem? And how is it graphed? 4x^2/x^2+1

Question 944156: whats the VA and HA of this problem? And how is it graphed?
4x^2/x^2+1
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
to find the vertical asymptote, look for the value of x that makes the denominator equal to 0.
since there is no value of x that will make the denominator equal to 0, you don't have a vertical asymptote.
the trick to finding the horizontal asymptote is to divide both numerator and denominator by the value of x that has the highest exponent.
in your problem, that would be x^2.
your equation becomes:
4x^2 / 4x^2 divided by (x^2 + 1) / x^2 which becomes:
4x^2 / x^2 divided by x^2 / x^2 + 1 / x^2
simplify this expression to get:
4 / 1 divided by 1 + 1 / x^2
as x approaches infinity, 1 / x^2 approaches 0, so the expression becomes:
4 / 1 divided by 1 which is equal to 4.
your horizontal asymptote is equal to 4.
summary:
you do not have a vertical asymptote.
your horizontal asymptote is equal to 4.
here's a graph of your equation.
$$$

here's a link to information about vertical and horizontal asymptotes.
http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut40_ratgraph.htm

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