Identify all asymptotes.

Question

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?

Answer 1

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

Answer 2

yes, those are what i meant..sorry about the notation

Answer 3

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

Answer 4

The area of a circular trampoline is 112.07 square feet. What is the radius
of the trampoline? Round to the nearest hundredth