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Question 557449: Range of function sqrt(x^2-2x+5). Why is the range ≥2?, without graphing
Found 2 solutions by stanbon, Theo:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 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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Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! i cheated by graphing it.
now i see what's going on.
what is happening is that the range of the function will be >= 2 because the value of y will never get below 2.
the domain of this function is all values of x such that the sqrt(x^2-2x+5) is positive.
that will occur at all values of x.
when x = -5, the value of y is equal to 6.3245553
when x = 0, the value of y is equal to 2.236068
when x = 5, the value of y is equal to 4.472136
the minimum value of y occurs when x = = 1
y never gets below 2.
y can go as high as you want, depending on the value of x.
for example:
when x = -500, y = 501.00399
when x = 500, y = 499.00401
when x = -5000, y = 5001.0004
when x = 5000, y = 49999.0004
the greater the value of x, the closer y will get to the absolute value of x.
so, if x can go infinite in each direction, this means that y can go infinite in the positive direction, because y will always be positive.
on the reverse side, there is no value of x that will provide a value of y smaller than 2.
that is why the range of the function is >= 2.
we are talking about the range of the function which is the value of y or is the value of f(x) which is equal to y.
the domain is the value of x which can be any real number.
the graph of this equation, by the way, looks like this:
as x gets more negative and as x gets more positive, the value of y will get more positive.
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