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