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

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

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vertices-x^2-2x+3

vertices

- x^{2} - 2 x + 3

Maximum (- 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 ( - 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

Maximum (- 1, 4)

y^{2} = 6 x - 32

f (x) = - x^{2} + 2 x + 5

2 x^{2} - 4 x

f (x) = 2 (x + 1)^{2} - 3

y = - 2 x^{2} - 8 x + 3