What is the vertices x^2+2x ?

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

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vertices x^2+2x

vertices

x^{2} + 2 x

Minimum (- 1, - 1)

Solution steps

y = x^{2} + 2 x

The parabola parameters are:

a = 1, b = 2, c = 0

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} = - 1

Therefore the parabola vertex is

(- 1, - 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 1, - 1)

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

5 x^{2} + 20 x - 17

y = 2 x^{2} + 24 x - 6

- x^{2} + 4

y^{2} - 6 y = 8 x - 1