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December 19, 2014

December 19, 2014

Posted by **joe** on Monday, February 4, 2013 at 1:36pm.

f(x)=x^2=2x-9

- algebra -
**Reiny**, Monday, February 4, 2013 at 1:44pmYour parabola opens up because of +x^2, so it has a minimum

to find min, 3 methods:

1. by Calculus,

f ' (x) = 2x + 2

= 0 for a min of f(x)

2x = -2

x=-1

f(-1) = 1 - 2 - 9 = -10

2. complete the square to find the vertex:

f(x) = x^2 + 2x + 1 -1 -9

=(X+1)^2 - 10

the vertex is (-1,-10)

so there is a min of -10 when x = -1

3. by formula

if f(x)= ax^2 + bx + c, the x of the vertex is -b/2a

x of vertex = -2/(2(1)) = -1

f(-1) = -10 , as above

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