Question 557449
Range of function sqrt(x^2-2x+5). Why is the range ≥2?, without graphing
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y = sqrt(x^2-2x+5)
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x^2-2x+5 has its minimum point at x = (-b/(2a)) = 2/(2*1) = 1
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f(1) = 1^2-2*1+5 = 4
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Your problem takes the square root of all
the y value of x^2-2x+5.
The lowest of these value is sqrt(4) = 2
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Cheers,
Stan H.
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