Posted by ladybug on Wednesday, October 31, 2012 at 9:07pm.
For the given equation list the intercepts.
Y= 2X/(X^2+1)
I think I Have this right I am not for sure?
y=2x/(x^2+1)
(0)=(2x/x^2+1)
2x/x^2+1=0
2/x+1=0
2/x+1+0
2/x*x+1*x=0*x
2+x=0
x=2
yintercept
y= 2*0/(0)^2+1
y=o/0^2+1
y=0+1
y=0/0+1
xintercept:(2,0)
yintercept:(0,0)

college algebra  Reiny, Wednesday, October 31, 2012 at 9:28pm
no , from your lines
2x/x^2+1=0
2/x+1=0
at the start you had (x^2 +1) in brackets as I just did
so when you divide top and bottom by x, every term has to be divided by x
and you would get
2/(x + 1/x) = 0
but I would not do that
If a fraction = 0 , then the zero can only come from the numerator, and the denominator could never be zero, or else we would be dividing by zero
so in
2x/(x^2+1) = 0
2x = 0
x = 0
so (0,0) is both an x and a y intercept.

college algebra  ladybug, Wednesday, October 31, 2012 at 10:25pm
ok thanks
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