Find the points on the graph of $\displaystyle y=9-x^2$ that is closest to the point $\displaystyle (0, 3)$.
In general, the distance between two points is
$\displaystyle \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$
Use:
$\displaystyle (x_1,y_1)=(x,9-x^2)$ <-- this is a point on the curve
$\displaystyle (x_2,y_2)=(0,3)$
Then we have the distance between a point on the curve and the point $\displaystyle (0,3)$:
$\displaystyle \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$
$\displaystyle =\sqrt{(x-0)^2+(9-x^2-3)^2}$
$\displaystyle =\sqrt{x^2+(6-x^2)^2}$
$\displaystyle =\sqrt{x^2+36-12x^2+x^4}$
$\displaystyle =\sqrt{x^4-11x^2+36}$
Our goal is to find the minimum value of this, so we find the derivative and set it equal to 0 and solve for x.
Can you take it from here?