Question 724015
Finding the maximum y-value on the graph of y=f(x). 
My problem is f(x)= -x^2+8x+7
Vertex occurs at x = -b/(2a) = -8/(2*-1) = 4
Max at f(4) = -16+32+7 = 16+7 = 23
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Using quadratic formula to find any x-intercepts on graph of equation. My problem is y=x^2+6x-1.
Let y = 0
x^2+6x-1 = 0
x = [-6+-sqrt(36-4*-1)]/2
x = [-6+-sqrt(40)]/2
x = [-3+-sqrt(10)]
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and 
To solve by completing the square what value should be added to each side of equation with problem x^2+16x=-4.
x^2+16x+64 = -4+64
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(x+8)^2 = 60
x+8 = +-2sqrt(15)
x = -8+-2sqrt(15)
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Cheers,
Stan H.
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