What is the vertices x^2+8x+19 ?

The vertices x^2+8x+19 is Minimum (-4,3)

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vertices x^2+8x+19

vertices

x^{2} + 8 x + 19

Minimum (- 4, 3)

Solution steps

y = x^{2} + 8 x + 19

The parabola parameters are:

a = 1, b = 8, c = 19

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} = 3

Therefore the parabola vertex is

(- 4, 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 4, 3)

y = x^{2} + 7 x + 12

y = x^{2} - 12 x - 20

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

- x^{2} + 3 x + 1

f (x) = x^{2} + 12 x + 35