SOLUTION: find the vertex,symmetry, maximum/minumum choose max/min f(x)=3x^2-12x+16

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Question 342647: find the vertex,symmetry, maximum/minumum choose max/min
f(x)=3x^2-12x+16

Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
find the vertex,symmetry, maximum/minumum choose max/min
f(x)=3x^2-12x+16
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Since the coefficient of x^2 is positive, the function has a minimum
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Minimum occurs at x = -b/(2a) = 12/(2*3) = 2
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Vertex at (2,f(2))
f(2) = 3*4-12*2+16 = 4
Vertex: (2,4)
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graph%28400%2C300%2C-10%2C10%2C-10%2C20%2C3x%5E2-12x%2B16%29
Cheers,
Stan

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Given a quadratic function in standard form, namely:



The -coordinate of the vertex is given by:



The -coordinate of the vertex is given by:



The axis of symmetry is given by:



which is to say:



If the lead coefficient is greater than zero, the parabola opens upward and therefore the vertex is a minimum. Otherwise, the parabola opens downward and the vertex is a maximum.

John

My calculator said it, I believe it, that settles it