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) (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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Cheers,
Stan
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Given a quadratic function in standard form, namely:
\ =\ ax^2\ +\ bx +\ c)
The 
-coordinate of the vertex is given by:
The 
-coordinate of the vertex is given by:
 = f\left(\frac{-b}{2a}\right))
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
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