What is the vertices f(x)=x^2+6x+6 ?

The vertices f(x)=x^2+6x+6 is Minimum (-3,-3)

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vertices f(x)=x^2+6x+6

vertices

f (x) = x^{2} + 6 x + 6

Minimum (- 3, - 3)

Solution steps

y = x^{2} + 6 x + 6

The parabola parameters are:

a = 1, b = 6, c = 6

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{6}{2 \cdot 1}

Simplify

x_{v} = - 3

Plug in

x_{v} = - 3

to find the

y_{v}

value

y_{v} = - 3

Therefore the parabola vertex is

(- 3, - 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 3, - 3)

f (x) = - x^{2} - 2 x + 8

- x^{2} + 6 x - 5

y = x^{2} - 9 x + 20

y = x^{2} - 11 x + 13

y = 3 x^{2} - 12 x + 10

What is the vertices f(x)=x^2+6x+6 ?
The vertices f(x)=x^2+6x+6 is Minimum (-3,-3)