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Home Q & A Board Science-Related Homework Help Mathematics
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quixie quixie
wrote...
solved
8 years ago
Find the point on the curve y=x^2 closest to the point (-3,0).

a. (0,-3)

b. (0,0)

c. (1,0)

d. (-1,1)
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wrote...
#1
Educator
8 years ago
Distance^2 of any point (x,y) from (-3,0) is
(x+3)^2 + y^2

If the point is on the curve y = x^2 the distance^2 is
(x+3)^2 + x^4

At the minimum d/dx = 0
2(x+3) + 4x^3 = 0
4x^3 + 2x + 6 = 0
2x^3 + x + 3 = 0

By inspection, one root is x = -1 since -2 - 1 + 3 = 0
2x^3 + x + 3 = 0
(x+1)(2x^2 - 2x + 3) = 0

2^2 - 4.2.3 < 0 so the quadratic factor has no real roots

So x = -1 and y = x^2 = 1

(D)
quixie Author
wrote...
#2
8 years ago
Thanks for the explanation!
wrote...
#3
Educator
8 years ago
You're welcome.
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