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

The vertices f(x)=-3x^2+4x+4 is Maximum (2/3 , 16/3)

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

vertices

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

Maximum (\frac{2}{3}, \frac{16}{3})

Solution steps

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

The parabola parameters are:

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

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{16}{3}

Therefore the parabola vertex is

(\frac{2}{3}, \frac{16}{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{16}{3})

f (x) = 2 (x - 5) (x - 1)

- x^{2} + 4 x - 6

y = \frac{6 x}{7} \cdot \frac{4 x}{3}

y^{2} - 6 y - 4 x + 5 = 0

5 x (2 x + 4) - 3 x

What is the vertices f(x)=-3x^2+4x+4 ?
The vertices f(x)=-3x^2+4x+4 is Maximum (2/3 , 16/3)