What is the vertices y=x^2+8x+18 ?

The vertices y=x^2+8x+18 is Minimum (-4,2)

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vertices y=x^2+8x+18

vertices

y = x^{2} + 8 x + 18

Minimum (- 4, 2)

Solution steps

y = x^{2} + 8 x + 18

The parabola parameters are:

a = 1, b = 8, c = 18

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} = 2

Therefore the parabola vertex is

(- 4, 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 4, 2)

x^{2} - 5 x - 6

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

(x - 1)^{2} = 16 (y - 2)

- 0.5 x^{2} - 3

y^{2} - 8 x + 2 y + 17 = 0