What is the vertices y=-3x^2+6x-1 ?

The vertices y=-3x^2+6x-1 is Maximum (1,2)

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vertices y=-3x^2+6x-1

vertices

y = - 3 x^{2} + 6 x - 1

Maximum (1, 2)

Solution steps

y = - 3 x^{2} + 6 x - 1

The parabola parameters are:

a = - 3, b = 6, c = - 1

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

x_{v} = - \frac{6}{2 ( - 3 )}

Simplify

- \frac{6}{2 ( - 3 )} : 1

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = 2

Therefore the parabola vertex is

(1, 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 3

Maximum (1, 2)

y = (x + 5)^{2}

y = x^{2} + 5 x - 14

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

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

+ \frac{x ^{2} + x}{2}