What is the vertices y=x^2+12x+16 ?

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

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vertices y=x^2+12x+16

vertices

y = 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)

f (x) = - 5 x^{2} + 20 x + 60

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

f (x) = - x^{2} - 14 x - 42

y = 2 x^{2} + 4 x + 6

y = 2 x^{2} + 4 x - 5

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