are you sure it is not
y = x/(4-x^2)
?
if so
as x ---> 0 , y---> x/4 which is zero sort of a HA but not really, just an inflection point
as x---> 2 or -2 y---> oo VA
can someon help me find the horzontial and verticle asymptotes for x square root 4-x^2.
For HA someone told me none but I dont understand how
and is the VA -2 and 2.
Please help and show all work
Thank you
3 answers
No it shows that it is f(x)= x sqauare root 4-x^2
I am so confused
I am so confused
in that case, it is clear that there are no asymptotes of any kind. f(x) is undefined at all for x^2 > 4
the graph is here:
http://www.wolframalpha.com/input/?i=x%E2%88%9A(4-x%5E2)
the graph is here:
http://www.wolframalpha.com/input/?i=x%E2%88%9A(4-x%5E2)