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

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

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

vertices

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

Maximum (1, 3)

Solution steps

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

The parabola parameters are:

a = - 3, b = 6, c = 0

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} = 3

Therefore the parabola vertex is

(1, 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 3

Maximum (1, 3)

y = 14 (x - 4) (x + 10)

x^{2} + 8 x - 80

f (x) = x^{2} - 6 x - 16

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

(x + 2)^{2} = - 6 (y - 1)

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