What is the vertices f(x)=x^2+6x+9 ?

The vertices f(x)=x^2+6x+9 is Minimum (-3,0)

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vertices f(x)=x^2+6x+9

vertices

f (x) = x^{2} + 6 x + 9

Minimum (- 3, 0)

Solution steps

y = x^{2} + 6 x + 9

The parabola parameters are:

a = 1, b = 6, c = 9

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} = 0

Therefore the parabola vertex is

(- 3, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 3, 0)

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

f (x) = x^{2} + 5 x + 4

+ y^{2} + 12 x + 4 y - 32 = 0

f (x) = - 2 x^{2} - 16 x - 30

x = - 5 y^{2}

What is the vertices f(x)=x^2+6x+9 ?
The vertices f(x)=x^2+6x+9 is Minimum (-3,0)