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Question 562208: What is the range of y=sqrt(16-x^2)
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! What is the range of y=sqrt(16-x^2)
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Sqrt can never be negative.
Solve 16-x^2 >= 0
(x-4)(x+4) <= 0
Draw a number line and plot x = -4 and x = 4
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Test a value in each of the three intervals:
(x-4)(x+4) < 0
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If x = -10 ; -*- < 0 ; false
If x = 0 ; -*+ < 0 ; true, so solutions in [-4 , 4]
If x = +10 ; +*+ < 0 ; false
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Range: -4 <= x <= 4
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Cheers,
Stan H.
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