Analyse and sketch the given function

[Taylor_1989]

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

[/B]
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.

 

[Mark44]

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

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

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

The Attempt at a Solution

[/B]
1. Horizontal asymptotes @ y=-5 and y=5

No. The denominator is zero if y = 25. The fraction is defined for both y = 5 and y = -5.
Also, the denominator is positive for any y ≠ 25, and the numerator is positive if y > 0. Since the left side is x² which is greater than or equal to zero for all real x, the right side also has to be greater than or equal to zero.

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.

 

[Taylor_1989]

Sorry it's a y^2, that what I get for doing it on a tablet

 

[Taylor_1989]

Also is using the quadratic wrong? I only ask because when I think about it, surley I could use it to find the range of y for example in other functions?

 

[haruspex]

-5<y<0 valid

It is? What would x be at y=-1?

0<y<5 na

Why? And you left out y=0.

when I rearranged the equations:

Working backwards from that, perhaps you need to correct your original post. Is the equation really:
$x^2=\frac{y}{y^2-25}$?

 

[Taylor_1989]

Yes I am sure. The equation is: $$x^2=\frac{y}{(y^2-25)}$$

for the given equation: is I plug -1 in I get $$\frac{1}{24}$$ Which is valid as I can in this case take a square root of a negative.

Have I missed something here, I am now confused? I assumed that as its and $$x^2$$ that is I get a negative results then $$x^2\neq -1$$

Also can someone tell me why I can edit my original post to correct the equation?

 

[Mark44]

Yes I am sure. The equation is: $$x^2=\frac{y}{(y^2-25)}$$

for the given equation: is I plug -1 in I get $$\frac{1}{24}$$ Which is valid as I can in this case take a square root of a negative.

? 1/24 is positive, so you're not taking the square root of a negative number.

Have I missed something here, I am now confused? I assumed that as its and $$x^2$$ that is I get a negative results then $$x^2\neq -1$$

? This makes no sense.
Since the left side of your equation is $x^2$, which is always nonnegative, the right side has to also be nonnegative. IOW, you must have
$\frac y {(y - 5)(y + 5)} \ge 0$
Divide the number line at y = -5, y = 0, y = 5, and determine which of these intervals gives you a quotient that is greater than or equal to zero. Those intervals will be the domain for your relation.

Also can someone tell me why I can edit my original post to correct the equation?

You can't edit your original post after 8 or 9 minutes, I believe. I have edited your post to put in the correct equation.

 

[haruspex]

Yes I am sure. The equation is: $$x^2=\frac{y}{(y^2-25)}$$

Originally you wrote (y-25)² as the denominator. In post #3 you said it should be y squared, implying (y²-25)². Since that did not fit your later working, I asked if it was really (y²-25). Now that you confirm that, I agree with all your results for the part 2, except that you left out y=0.

this will be only valid if discriminant is greater the 0

You need to be careful with the direction of the logic. You are effectively working backwards here, from the thing to be proved to some tautology. What is important for this step is that it is valid if the discriminant is greater than zero, not that it is only valid in that case.
Again, I do not see consideration of x=0 in part 3 in arriving at the range for x. You only mention it later.

Your verbal description of the graph sounds right.

 

[Mark44]

@Taylor_1989, try to be more careful in your problem description in future posts. A lot of time was wasted trying to make sense of the work you showed, given the equation you showed in post #1.

 

[Taylor_1989]

First I apologise for the mix up. I have been told most my life my working are very scattered , I have trouble putting what I think to paper. I do try are re read my posts but always seem to miss stuff out, I will however make sure my next post will be more clearer. @Mark44 in regards to you post about

y(y−5)(y+5)≥0

I understand what you mean, I just went off in a tangent. I did indeed take the square root and that is how I was working out where the graph lies ect.

Once again sorry for the mix you guys, and thank you for your help. I will make sure I do better next time.