What is the vertices-2x^2+12x-14 ?

The vertices-2x^2+12x-14 is Maximum (3,4)

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vertices-2x^2+12x-14

vertices

- 2 x^{2} + 12 x - 14

Maximum (3, 4)

Solution steps

y = - 2 x^{2} + 12 x - 14

The parabola parameters are:

a = - 2, b = 12, c = - 14

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

x_{v} = - \frac{12}{2 ( - 2 )}

Simplify

- \frac{12}{2 ( - 2 )} : 3

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = 4

Therefore the parabola vertex is

(3, 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 2

Maximum (3, 4)

f (x) = 2 x^{2} + 3 x - 1

- 2 x^{2} + 12 x

f (x) = 4 x^{2} - 7 x - 2

2 y^{2} + 8 y + x + 1 = 0

y^{2} - 12 y - 4 x + 28 = 0