What is the vertices y=x^2-2x-2 ?

The vertices y=x^2-2x-2 is Minimum (1,-3)

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vertices y=x^2-2x-2

vertices

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

Minimum (1, - 3)

Solution steps

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

The parabola parameters are:

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

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} = - 3

Therefore the parabola vertex is

(1, - 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (1, - 3)

f (x) = 4 x^{2} + 8 x + 3

x = 7 y^{2}

y = - x^{2} + 10 x - 28

+ y^{2} = 28 x

x^{2} - 5 x + 4