SOLUTION: find the minimum value of y in the equation y=3x^2-36x+112
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Question 730987
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find the minimum value of y in the equation y=3x^2-36x+112
Answer by
lwsshak3(11628)
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find the minimum value of y in the equation
y=3x^2-36x+112
complete the square:
y=3(x^2-12x+36)-108+112
y=3(x-6)^2+4
This is an equation of a parabola that opens upward.
Vertex:(6,4)
minimum value of y=4