What is the vertices y=2x^2+4x+6 ?

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

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vertices y=2x^2+4x+6

vertices

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

Minimum (- 1, 4)

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} = 4

Therefore the parabola vertex is

(- 1, 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- 1, 4)

y = 2 x^{2} + 4 x - 5

- (x + 5)^{2} + 4

y = 2 x^{2} + 4 x - 8

f (x) = - 2 x^{2} + 8 x - 11

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