What is the vertices 2(x+1)^2 ?

The vertices 2(x+1)^2 is Minimum (-1,0)

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vertices 2(x+1)^2

vertices

2 (x + 1)^{2}

Minimum (- 1, 0)

Solution steps

y = 2 (x + 1)^{2}

y = 2 (x + 1)^{2}

The parabola parameters are:

a = 2, m = - 1, n = - 1

x_{v} = \frac{m + n}{2}

x_{v} = \frac{( - 1 ) + ( - 1 )}{2}

Simplify

x_{v} = - 1

Plug in

x_{v} = - 1

to find the

y_{v}

value

y_{v} = 0

Therefore the parabola vertex is

(- 1, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- 1, 0)

t^{2} + 12 t - 18

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

10 y^{2} = 22 x

2 x^{2} + 5 x - 7

x^{2} + 12 y = 0