What is the vertices y=-3x^2+12x-8 ?

The vertices y=-3x^2+12x-8 is Maximum (2,4)

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vertices y=-3x^2+12x-8

vertices

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

Maximum (2, 4)

Solution steps

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

The parabola parameters are:

a = - 3, b = 12, c = - 8

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

x_{v} = - \frac{12}{2 ( - 3 )}

Simplify

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

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = 4

Therefore the parabola vertex is

(2, 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 3

Maximum (2, 4)

x^{2} + 8 x + 18

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

x^{2} + 5 x - 14

f (x) = x^{2} + 8 x + 7

3 x^{2} + 3 x + 2 y = 0

What is the vertices y=-3x^2+12x-8 ?
The vertices y=-3x^2+12x-8 is Maximum (2,4)