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Are you sure it is 16-Y^2? not X^2 anyway i graphed X^2 and it looks like a bowl upside down Roots:-4,4 y int:4
If the equation is $\displaystyle y=\sqrt{16-x^2}$ Then recall that the equation of a circle is $\displaystyle x^2 + y^2 = k^2$ If you re-arrange for y, then you get your equation, but only the positive half.
The OP fixed the mistake by posting again. It's actually $\displaystyle \displaystyle \begin{align*} x = \sqrt{16 - y^2} \end{align*}$. This is a semicircle centred at the origin of radius 4 units, taking on only positive values for x.
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