# Calculus

posted by Mark

Find the points on the curve y=x2+2 closest to the point (0,3). Enter the coordinate with the smallest x-value first and round to the nearest 4 decimal places.

1. Steve

y = x^2+2
y' = 2x
So, at (h,k) y'(h) = 2h
So, you want a line with slope -1/(2h) passing through (0,3)

y-3 = -1/(2h) x
h^2+2 - 3 = -1/2
h^2 = 1/2
h = 1/√2
so, (1/√2,5/2) is closest to (0,3)

1/2 +
Or, consider the distance formula. We have

z^2 = (x-0)^2 + (y-3)^2
= x^2 + (x^2+2-3)^2
= x^2 + x^4-2x^2+1
= x^4 - x^2 + 1

2z z' = 4x^3 - 2x
z' = x(2x^2-1)/z
z'=0 at x=0, x = ±1/√2

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