What is the vertices f(x)=2x^2+4x-6 ?

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

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vertices f(x)=2x^2+4x-6

vertices

f (x) = 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)

f (x) = 10 x^{2} - 17

3 x^{2} - 5 x + 2

y = - 3 x^{2} + 12 x - 8

x^{2} + 8 x + 18

x = - 2 (y + 3)^{2} + 4

What is the vertices f(x)=2x^2+4x-6 ?
The vertices f(x)=2x^2+4x-6 is Minimum (-1,-8)