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

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

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vertices-x^2+2x-3

vertices

- x^{2} + 2 x - 3

Maximum (1, - 2)

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

- \frac{2}{2 ( - 1 )} : 1

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = - 2

Therefore the parabola vertex is

(1, - 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (1, - 2)

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

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

f (x) = 8 x^{2} + 4 x - 1

f (x) = - x^{2} - 14 x - 48

x^{2} + 10 x - 2 y + 23 = 0