Identify all asymptotes.
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?
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
To find the asymptotes of a function, we need to consider both vertical asymptotes and horizontal asymptotes.
1. For the function y = 1 / (2 - √(x^2 - 3x)), let's analyze the possible asymptotes:
Vertical asymptotes: These occur when the denominator of the function becomes zero. In this case, 2 - √(x^2 - 3x) = 0. Solving this equation, we have √(x^2 - 3x) = 2. Squaring both sides gives x^2 - 3x = 4. Rearranging the equation, we get x^2 - 3x - 4 = 0. Factoring this quadratic equation gives (x - 4)(x + 1) = 0. So the vertical asymptotes occur at x = 4 and x = -1.
Horizontal asymptotes: To find these, we need to analyze the behavior of the function as x approaches positive or negative infinity. By simplifying the expression 1 / (2 - √(x^2 - 3x)), we see that the term involving x^2 has the greatest power. This term dominates as x becomes very large. Therefore, the horizontal asymptote will be y = 0.
Therefore, for the function y = 1 / (2 - √(x^2 - 3x)), the asymptotes are:
- Vertical asymptotes at x = 4 and x = -1.
- Horizontal asymptote at y = 0.
2. For the function y = (x^3 - 2x) / (x^2 + 1), let's analyze the possible asymptotes:
Vertical asymptotes: These occur when the denominator of the function becomes zero. In this case, x^2 + 1 = 0. However, this equation has no real solutions. Therefore, there are no vertical asymptotes for this function.
Horizontal asymptotes: Similar to the previous example, we need to analyze the behavior of the function as x approaches positive or negative infinity. By examining the term involving x, we see that the power of x is greater in the numerator (x^3) than in the denominator (x^2). Thus, this term dominates as x becomes very large. Therefore, the horizontal asymptote will be y = x^3 / x^2 = x.
Therefore, for the function y = (x^3 - 2x) / (x^2 + 1), there are:
- No vertical asymptotes.
- A horizontal asymptote at y = x.
I hope this helps! Let me know if you have any further questions.