thanks :)

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- Feb 23rd 2013, 04:55 PMcityskylinesketch the graph y=√(16-y^2)
thanks :)

- Feb 23rd 2013, 05:00 PMZashmarRe: sketch the graph y=√(16-y^2)
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 - Feb 23rd 2013, 05:04 PMEducatedRe: sketch the graph y=√(16-y^2)
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. - Feb 23rd 2013, 05:37 PMProve ItRe: sketch the graph y=√(16-y^2)
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.