SOLUTION: What point on the graph of the equation y=x^2 is closest to the point (3,1)?

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Question 382720: What point on the graph of the equation y=x^2 is closest to the point (3,1)?
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
We want to minimize the distance between the points (x, x^2) and (3,1). Using the distance formula, this distance z is equal to
z+=+sqrt+%28%28x-3%29%5E2+%2B+%28x%5E2+-+1%29%5E2%29+=+sqrt%28x%5E4+-+x%5E2+-+6x+%2B+10%29
We wish to minimize the value of z. By taking the derivative of z and setting it to zero, we obtain

This occurs when
2x%5E3+-+x+-+3+=+0 and the denominator is nonzero.
I had to use a calculator to determine that the only real root is x = 1.289623 (meaning that a critical point is located there, it can be checked that it minimizes the distance). Therefore the closest point on the graph y = x^2 to (3,1) is (1.289623, 1.663127)