## Symmetry of Equation: x^2 = y^3 + 4

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juliagregs
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Joined: Fri Aug 31, 2012 3:16 am
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### Symmetry of Equation: x^2 = y^3 + 4

I am trying to test an equation for symmetry. The equation is x^2 = y^3 + 4.

I am starting by trying to find the intercepts. I think I found the intercepts for x - (2,0), (-2,0).

I am having trouble finding the intercepts for y.

I substitute 0 for X and get 0 + y^3 + 4. I convert that to y^3 = -4. I am not sure where to go from there. Can I get a cube root of four? Solving for y, would my answer be the cube root of four i? Does that mean there is no y intercept?

I've confused myself... Thanks for any help.

LoveMath
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Joined: Mon Sep 03, 2012 4:35 am
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### Re: Symmetry of Equation: x^2 = y^3 + 4

Julia, actually you got pretty far since you got to y^3 = -4. I think the confusion lies in your thinking that you can't take the cube root of a negative number. If your problem were y^3 = -8 then to find the cube root, think of what number cubed gives you a -8 and that would be -2 since (-2)(-2)(-2) = 4(-2)=-8. So, since this is not a perfect cube, it's easiest to use a calculator to get the answer. Off the top of my head, I could only tell you that the answer would have to be between -1 and -2. The actual answer is -1.5874010519681993.... Rounding to the nearest hundreths, you'd get -1.59. The calculator that I'm using right now won't allow me to put the long answer into it, however, to check if this is correct: take 1.5874 and cube it. This gives 3.999992047624 which is very close to 4 and since you are cubing a negative, the answer also is negative, i.e. -3.999992047624. Got it?