What is the vertices x^2-2x-8 ?

The vertices x^2-2x-8 is Minimum (1,-9)

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vertices x^2-2x-8

vertices

x^{2} - 2 x - 8

Minimum (1, - 9)

Solution steps

y = x^{2} - 2 x - 8

The parabola parameters are:

a = 1, b = - 2, c = - 8

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

x_{v} = - \frac{( - 2 )}{2 \cdot 1}

Simplify

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = - 9

Therefore the parabola vertex is

(1, - 9)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (1, - 9)

x^{2} - 6 x + 10

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

f (x) = x^{2} + 10 x + 9

y = - 3 x^{2} - 12 x - 9

x^{2} = 8 (y + 4)