What is the vertices f(x)=x^2+6x+3 ?

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

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vertices f(x)=x^2+6x+3

vertices

f (x) = x^{2} + 6 x + 3

Minimum (- 3, - 6)

Solution steps

y = x^{2} + 6 x + 3

The parabola parameters are:

a = 1, b = 6, c = 3

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} = - 6

Therefore the parabola vertex is

(- 3, - 6)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 3, - 6)

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

y = x^{2} - 2 x + 5

y = x^{2} - 2 x + 1

x^{2} - y = 0

y = x^{2} + 8 x + 12

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