- #1

Taylor_1989

- 402

- 14

## Homework Statement

Hi guys, I would like some pointers on how to do this type of question. My sketch was correct but, I want to just check my method on how I came up with the solution. Also if anyone has any other ways, on tackling these problems, advice would be great.

## Homework Equations

$$x^2=\frac{y}{(y-25)^2}$$

**Mod note**: The equation is actually ##x^2 = \frac y {y^2 - 25}##

$$x=\frac{-b+/- \sqrt{b^2-4ac}}{2a}$$

## The Attempt at a Solution

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1. Horizontal asymptotes @ y=-5 and y=5

2. Function is valid:

y<-5 na

-5<y<0 valid

0<y<5 na

y>5 valid

3. Finding the range of values for x: This is the bit I am iffy on because I just noticed it when I rearranged the equations:

$$x^2y^2-y-25x^2=0$$

$$x\ne 0, y=\frac{1+/- \sqrt{1+4(25x^2)(x^2)}}{2x^2}$$

I then use the fact this will be only valid if discriminant is greater the 0

which gave: $$1+100x^4>0$$

Which is true for all values of x so the range of x is $$(-\infty,\infty)$$

I then found when x=0 y=0 ( this will be for the part of the graph -5<y<0 ( suppose to be a greater/equal) so this part of the graph is continuous

4. I then took the limit of the function and found a asymptote at x=0.

So the graph at y>5 looks like a graph of 1/x in the in x<0 and x>0 and a speed hump with a max at (0,0) in the -5<y<0

Thanks in advance.

p.s I am having trouble uploading my graph sketch from my pad, I think hotspot is bad so will upload graph in an hour.

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