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Sketch the curve $$y=\frac{2x^3}{x^2-2}.$$

Can someone answer this for me as basic as possible. Year 11 extension if possible. Thanks

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closed as off-topic by Carl Mummert, Grigory M, William, apnorton, daw Jul 15 '14 at 11:13

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Instead of saying "Year 11 extension" (which no one from a different school system will understand), say what math you have done. In particular, have you learned about asymptotes? Have you done any Calculus? – Gerry Myerson Jul 15 '14 at 7:29
How might you be able so simplify the equation? And here's a MathJax tutorial :) – Shaun Jul 15 '14 at 7:30
We've done asymptotes and are yet to do calculus – user112817 Jul 15 '14 at 7:36
Please use WolframAlpha responsibly :) – Shaun Jul 15 '14 at 7:39
  1. Firstly, simplify the equation $$\frac{2x^3}{x^2-2}=2x+\frac{4x}{x^2-2}.$$ Now, we can get the asymptotes to be $y=2x$ since as $x\rightarrow\pm\infty$, the second term on the RHS tends to zero. Also, notice that if $x=\pm\sqrt{2}$, the function would be undefined, so we have 3 asymptotes $$y=2x, x=\pm\sqrt{2}.$$

  2. Now, we need to find turning points which can be done by differentiating the function and solving for zeroes (I'll leave them to you).

  3. Find the $x$ and $y$ intercepts by substituting $x=0$ and $y=0$ into the equation.

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But OP hasn't done Calculus, so Step 2 doesn't fly. – Gerry Myerson Jul 16 '14 at 3:42

HINT: Find the zeroes of the denominator and check what happens to $y$ when $x$ becomes close to these. Also check what happens to $y$ when $x$ becomes really big (positive or negative).

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