• Nov 8th 2009, 12:42 PM
PrincessIsrael
How do you do this problem?

Which points on the graph of y=4-(x^2) are closest to the points (0,2)?
• Nov 8th 2009, 12:44 PM
VonNemo19
Quote:

Originally Posted by PrincessIsrael
How do you do this problem?

Which points on the graph of y=4-(x^2) are closest to the points (0,2)?

Employ the distance formula.
• Nov 8th 2009, 02:18 PM
Scott H
Welcome to the Math Help Forum! :)

To find the point on $y=4-x^2$ closest to $(0,2)$, we first do as VonNemo said and use the distance formula to obtain a function of the distance $s$ from a point on the curve to $(0,2)$ as a function of $x$:

$s=\sqrt{(x-0)^2+((4-x^2)-2)^2}=\sqrt{x^2+(2-x^2)^2}.$

We now find the value of $x$ for which $s$ is a minimum. To do this, we differentiate $s$ and find all critical points. As $s$ is differentiable everywhere and has no boundary points, the only critical points will be those at which $s'=0$. As $s$ approaches infinity in both directions of the $x$-axis, at one of these critical points we must find our minimum.

Hope this helps!