What is the vertices 2x^2+4x-6 ?

The vertices 2x^2+4x-6 is Minimum (-1,-8)

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vertices 2x^2+4x-6

vertices

2 x^{2} + 4 x - 6

Minimum (- 1, - 8)

Solution steps

y = 2 x^{2} + 4 x - 6

The parabola parameters are:

a = 2, b = 4, c = - 6

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{4}{2 \cdot 2}

Simplify

x_{v} = - 1

Plug in

x_{v} = - 1

to find the

y_{v}

value

y_{v} = - 8

Therefore the parabola vertex is

(- 1, - 8)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- 1, - 8)

y = k (x - 0)^{2} + 2

y = 5 x^{2} - 3 x - 7 + 2 x^{2} - x

4 y^{2} - 48 x - 20 y = 71

y = x^{2} + 2 x - 15

f (x) = - 4 (x + 1)^{2} - 5