Question 549030
The vertex of a parabola is the point where the derivative is equal to 0.
a) f(x) = (x+3)^2 + 3
df/dx = 0 = 2(x+3)(1) -> x = -3
f(-3) = (0)^2 + 3 = 3
So the vertex is (-3,3)
The other vertices are found in a similar fashion
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f(x) = (x+3)^2 - 4
Since this parabola opens upward, it will have a minimum. 
The minimum will be obtained when the 1st term=0: (x+3)^2 = 0 -> x = -3
So the minimum value is -4
f(x) = -x^2 - 18x - 83
Take the derivative and set=0:
df/dx = 0 = -2x - 18 -> x = -9
f(-9) = -(-9)^2 - 18(-9) - 83 = -2
This will be a maximum since the parabola opens downward