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No. If by "yx-2", you mean "y times x minus two", then this will give you "y = 10/x", which is a curve that approaches infinity as x approaches 0, and which approaches 0 as x approaches infinity (dy/dx = -10/x2).

If on the other hand, you mean "y times x to the power of negative two", then solving for y gives you "y = 8/x2", which will give you a similar looking curve, but which approaches 0 at a faster pace (dy/dx = -16/x3)

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15y ago

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