Posted by Jared on Monday, September 13, 2010 at 7:11pm.
You have vertical asymptotes when the denominator is zero
your values of x=1 and x=4 do not result in a zero denominator.
I have a feeling that your function is
y = 1/(2 - √(x^2-3x))
in that case the denominator is zero when x = 4 or x = -1
so your vertical asymptotes are
x = 4 and x = -1
(your horizontal is y = 0)
2. Again, I think your meant your function to b
y = (x^3 - 2x)/(x^2 + 1)
Please confirm
yes, those are what i meant..sorry about the notation
PLease don't switch names.
for y = (x^3 - 2x)/(x^2 + 1)
there is no vertical or horizontal asymptote.
but when you divide it you get
y = x - 3x/(x^2+1)
so there is a "slanted" asymptote of y = x
The area of a circular trampoline is 112.07 square feet. What is the radius
of the trampoline? Round to the nearest hundredth
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