algebra
posted by joe .
determine the given quadratic function has a minimum valueor maximum vale. Then find the coordinates of the minimum or maximum point.
f(x)=x^2=2x9

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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