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December 22, 2014

December 22, 2014

Posted by **Allison A** on Monday, November 28, 2011 at 7:11pm.

- Math -
**Steve**, Monday, November 28, 2011 at 7:25pmA parabola has either a minimum or a maximum, but not both.

The axis of symmetry, which contains the vertex, is the line

x = -b/2a = -2/(-4) = 1/2

You should now be able to get the max/min and thus the vertex

- Math -
**MathMate**, Monday, November 28, 2011 at 7:32pmExpress the function in canonical form by completing squares:

if f(x)=a(x-h)²+k,

then

a>0 => the parabola is concave up, hence a minimum exists

a<0 => concave down, hence a maximum.

The location of maximum/minimum is given by the point (h,k).

The line of symmetry is x=h.

To proceed with completing the squares, extract and factor out the coefficient of x²:

f(x)=-2(x²-x) + 2

=-2[(x-1/2)²-(1/2)²]+2

=-2[(x-1/2)²]+5/2

So the curve is concave down, h=1/2, k=5/2 and (h,k)=(1/2,5/2) is a maximum.

The line of symmetry is x=1/2.

- Math -
**Allison A**, Monday, November 28, 2011 at 9:08pmHow do I find the Y coordinate and the x coordinates

is the vertix 1/2, 5/2?

- Math -
**MathMate**, Monday, November 28, 2011 at 10:23pmBy completing squares, we get the two parameters h and k which represent the x- and y-coordinates of the vertex.

In this case, they are (1/2,5/2), as you can see from:

f(x) = -2[(x-1/2)²]+5/2

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