How do you do this problem?
Which points on the graph of y=4-(x^2) are closest to the points (0,2)?
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To find the point on $\displaystyle y=4-x^2$ closest to $\displaystyle (0,2)$, we first do as VonNemo said and use the distance formula to obtain a function of the distance $\displaystyle s$ from a point on the curve to $\displaystyle (0,2)$ as a function of $\displaystyle x$:
$\displaystyle s=\sqrt{(x-0)^2+((4-x^2)-2)^2}=\sqrt{x^2+(2-x^2)^2}.$
We now find the value of $\displaystyle x$ for which $\displaystyle s$ is a minimum. To do this, we differentiate $\displaystyle s$ and find all critical points. As $\displaystyle s$ is differentiable everywhere and has no boundary points, the only critical points will be those at which $\displaystyle s'=0$. As $\displaystyle s$ approaches infinity in both directions of the $\displaystyle x$-axis, at one of these critical points we must find our minimum.
Hope this helps!