What is the vertices y=x^2+8x+12 ?

The vertices y=x^2+8x+12 is Minimum (-4,-4)

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vertices y=x^2+8x+12

vertices

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

Minimum (- 4, - 4)

Solution steps

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

The parabola parameters are:

a = 1, b = 8, c = 12

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

x_{v} = - \frac{8}{2 \cdot 1}

Simplify

x_{v} = - 4

Plug in

x_{v} = - 4

to find the

y_{v}

value

y_{v} = - 4

Therefore the parabola vertex is

(- 4, - 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 4, - 4)

f (x) = - x^{2} - 2 x - 1

f (x) = - (x + 3) (x + 10)

y^{2} = - 32 x

y = 2 x^{2} + 16 x + 29

f (x) = - (x^{2} + 3 x - 4)