What is the vertices f(x)=-3x^2+6x-4 ?

The vertices f(x)=-3x^2+6x-4 is Maximum (1,-1)

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vertices f(x)=-3x^2+6x-4

vertices

f (x) = - 3 x^{2} + 6 x - 4

Maximum (1, - 1)

Solution steps

y = - 3 x^{2} + 6 x - 4

The parabola parameters are:

a = - 3, b = 6, c = - 4

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} = - 1

Therefore the parabola vertex is

(1, - 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 3

Maximum (1, - 1)

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

f (x) = - 2 x^{2} - 12 x - 20

x^{2} + 6 x - 8 y + 1 = 0

y^{2} - 6 y - 8 x + 17 = 0

y = x^{2} + 3 x - 10

What is the vertices f(x)=-3x^2+6x-4 ?
The vertices f(x)=-3x^2+6x-4 is Maximum (1,-1)