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

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

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vertices f(x)=-3x^2+4x-1

vertices

f (x) = - 3 x^{2} + 4 x - 1

Maximum (\frac{2}{3}, \frac{1}{3})

Solution steps

y = - 3 x^{2} + 4 x - 1

The parabola parameters are:

a = - 3, b = 4, c = - 1

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

x_{v} = - \frac{4}{2 ( - 3 )}

Simplify

- \frac{4}{2 ( - 3 )} : \frac{2}{3}

x_{v} = \frac{2}{3}

Plug in

x_{v} = \frac{2}{3}

to find the

y_{v}

value

y_{v} = \frac{1}{3}

Therefore the parabola vertex is

(\frac{2}{3}, \frac{1}{3})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 3

Maximum (\frac{2}{3}, \frac{1}{3})

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

f (x) = - 9.5 x^{2} - 47.5 x + 63

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

f (x) = - \frac{1}{250} x^{2} + \frac{4}{5} x

y = 2 x^{2} - 8 x + 4

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