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)

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

+ x^{2} - 36