Question 864033: Find the point on the graph of y= 5-x^2 that is closest to the point (0,1) and has positive coordinates. Hint: The distance from the point (x,y) on the graph to (0,1) is square root of ((x-0)^2 +(y-1)^2)) express y in terms of x
Answer: (square root of (3/2), 3/2)
I really need help. Thanks for your support! :)
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Find the point on the graph of y= 5-x^2 that is closest to the point (0,1) and has positive coordinates. Hint: The distance from the point (x,y) on the graph to (0,1) is square root of ((x-0)^2 +(y-1)^2)) express y in terms of x
Answer: (square root of (3/2), 3/2)
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The point on the graph is (x,y) = (x,5-x^2)
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The distance of that point from (0,1) is::
d = sqrt[(x-0)^2 + (5-x^2-1)^2]
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d = sqrt[x^2 + (4-x^2)^2]
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d = sqrt[x^4-7x^2+16]
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d' = (1/2)(4x^3-14x)/sqrt(x^4-7x^2+16)
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Solve d' = 0
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4x^3-14x = 0
2x(2x^2-7) = 0
x = 0 or x = sqrt(7/2)
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y = 5-x^2
f(sqrt(7/2) = 5-(7/2) = 3/2
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Ans: (sqrt(7/2), 3/2)
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Cheers,
Stan H.
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