What is the vertices x^2-2x-3 ?

The vertices x^2-2x-3 is Minimum (1,-4)

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vertices x^2-2x-3

vertices

x^{2} - 2 x - 3

Minimum (1, - 4)

Solution steps

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

The parabola parameters are:

a = 1, b = - 2, c = - 3

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} = - 4

Therefore the parabola vertex is

(1, - 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (1, - 4)

y = \frac{- 3 x ^{2} - 2 x + 6}{5}

(x + 5)^{2}

y = 4 x^{2} - 4 x^{1}

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

1 - x - x^{2}