Question 962349: Write the vertex of the following equation:
f(x)= (x+3)squared + 2
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! f(x)= (x+3)squared + 2
f(x) = x^2 +6x +9 +2
f(x) = x^2 +6x +11
the axis of symmetry for the parabola is
x = -b / 2a = -6 / 2 = -3
so far we have the following for the vertex (-3, y)
substitute for x in the equation
y = f(x) = x^2 +6x +11
y = f(x) = (-3)^2 + (6*-3) + 11
y = f(x) = 9 - 18 + 11
y = f(x) = 2
the vertex of the parabola is (-3, 2)
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