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Question 706046: please help me find the x and y intercepts and the range of the following: f(x)=x^2-6x+4
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! find the x and y intercepts and the range of the following:
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f(x)=x^2-6x+4
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y-int:
Let x = 0, then y = 4
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x-int:
Let y = 0,
x = [6 +- sqrt(36-4*1*4)]/2
x = [6 +- sqrt(20)]/2
x = [6 +- 2sqrt(5)]/2
x = [3+-sqrt(5)]
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To find the range, find the vertex:
vertex occurs at x = -b/(2a) = 6/2 = 3
f(3) = 3^2-6*3+4 = 9-18+4 = -5
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So, you have a parabola opening upward from the
point (3,-5)
Range = All Real Numbers >= -5
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Cheers,
Stan H.
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