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January 31, 2015

January 31, 2015

Posted by **Jared** on Monday, September 13, 2010 at 7:11pm.

1. y= 1 / 2-√x^2 - 3x

the asymptotes I got were:

vertical: 1 & 4

horizontal: 0

Is this it or are there more?

2. y= x^3-2x / x^2 + 1

I believe there are no asymptotes? Is this correct?

- Math - asymptotes -
**Reiny**, Monday, September 13, 2010 at 7:54pmYou 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

- Math - asymptotes -
**Amie**, Monday, September 13, 2010 at 8:11pmyes, those are what i meant..sorry about the notation

- Math - asymptotes -
**Reiny**, Monday, September 13, 2010 at 8:22pmPLease 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

- math -
**maria**, Thursday, May 26, 2011 at 8:18pmThe 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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