5. Find the point (x, y) on the graph of y=sqr(x) nearest the point (4, 0).

Answer 1

distance^2=(x-4)^2 + (y)^2

distnace^2=(x-4)^2 + x^2
d distance/x * 2 * distance=0=2(x-4)+ 2x

or -x=x-4 or x=2
and y= sqrt2. Point is (2, sqrt2)