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

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

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

vertices

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

Maximum (- \frac{1}{3}, \frac{4}{3})

Solution steps

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

The parabola parameters are:

a = - 3, b = - 2, c = 1

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

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

Simplify

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

x_{v} = - \frac{1}{3}

Plug in

x_{v} = - \frac{1}{3}

to find the

y_{v}

value

y_{v} = \frac{4}{3}

Therefore the parabola vertex is

(- \frac{1}{3}, \frac{4}{3})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 3

Maximum (- \frac{1}{3}, \frac{4}{3})

y = 3 (x - 3)^{2} - 1

y = (2 x - 1)^{2}

- 3 x^{2} - 6 x + 2

f (x) = 5 x^{2} - 10 x + 4

(y - 9)^{2} = - 4 (x - 2)

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