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

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

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vertices f(x)=-3x^2+6x-1

vertices

f (x) = - 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)

\frac{( y + 2 ) ^{2}}{16} - \frac{( x - 3 ) ^{2}}{9} = 1

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

x^{2} - y = 4

16 x^{2} + y^{2} = 16

y = x^{2} - 6 x

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