Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Carl Mummert, Grigory M, William, anorton, daw
If this question can be reworded to fit the rules in the help center, please edit the question.

2  
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 at 7:29
1  
How might you be able so simplify the equation? And here's a MathJax tutorial :) –  Shaun Jul 15 at 7:30
    
We've done asymptotes and are yet to do calculus –  user112817 Jul 15 at 7:36
    
Please use WolframAlpha responsibly :) –  Shaun Jul 15 at 7:39

2 Answers 2

  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.

share|improve this answer
    
But OP hasn't done Calculus, so Step 2 doesn't fly. –  Gerry Myerson Jul 16 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).

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.