What is the vertices f(x)=x^2+12x+16 ?

The vertices f(x)=x^2+12x+16 is Minimum (-6,-20)

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vertices f(x)=x^2+12x+16

vertices

f (x) = x^{2} + 12 x + 16

Minimum (- 6, - 20)

Solution steps

y = x^{2} + 12 x + 16

The parabola parameters are:

a = 1, b = 12, c = 16

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

x_{v} = - \frac{12}{2 \cdot 1}

Simplify

x_{v} = - 6

Plug in

x_{v} = - 6

to find the

y_{v}

value

y_{v} = - 20

Therefore the parabola vertex is

(- 6, - 20)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 6, - 20)

x^{2} + 4 x - 5

- x^{2} + 3 x - 5

y = - x^{2} + 8 x

x^{2} + 3 x - 4

y = (x - 1)^{2}

What is the vertices f(x)=x^2+12x+16 ?
The vertices f(x)=x^2+12x+16 is Minimum (-6,-20)