Understanding the Graph of an Equation | Calculus Help

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Hi everyone, I'm learning calculus at school. Recently I was taught this equation like. Y = (2X^2)/(9 - X^2)
So the teacher did all by himself. So I came home and now confused. I know there are 3 graphs(sorry if the word is not right ) so I was doing it again. And I'm stuck at where to get the position of curves.

I got,
X = 0 , X = -3, X= 3
They are 3 graph. So I couldn't figure out how to get more positions and draw it. Unfortunately I can't remember what teacher did 100% .
So someone please demonstrate it for me from the steps I have done.

Thanks
 
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I derived the functions and got their maximum /minimum
 
Here
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Bandarigoda said:
Hi everyone, I'm learning calculus at school. Recently I was taught this equation like. Y = (2X^2)/(9 - X^2)
So the teacher did all by himself. So I came home and now confused. I know there are 3 graphs(sorry if the word is not right ) so I was doing it again. And I'm stuck at where to get the position of curves.

I got,
X = 0 , X = -3, X= 3
They are 3 graph. So I couldn't figure out how to get more positions and draw it. Unfortunately I can't remember what teacher did 100% .
So someone please demonstrate it for me from the steps I have done.

Thanks

Your work for the derivative is correct: dy/dx = 36x/(9 - x2)2

Bandarigoda said:
I derived the functions and got their maximum /minimum
No, you differentiated the function and found the values for which f' = 0 or where the derivative is not defined.

If you set dy/dx = 0, the only solution is x = 0. The tangent line is horizontal when x = 0 (at the point (0, 0)).

dy/dx is undefined where the denominator is zero; namely, when x = 3 or x = -3. The original function is also undefined at the numbers. These turn out to be vertical asymptotes. The graph of the function tends to +∞ or -∞ on either side of these asymptotes.

Since there are two of them, they divide the number line into three intervals: (-∞, -3), (-3, 3), and (3, ∞). These intervals correspond to the three graphs you're talking about.
 
Hi, thank you very much Sir!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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