What is the vertices f(x)=12x-2x^2 ?

The vertices f(x)=12x-2x^2 is Maximum (3,18)

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vertices f(x)=12x-2x^2

vertices

f (x) = 12 x - 2 x^{2}

Maximum (3, 18)

Solution steps

y = 12 x - 2 x^{2}

Rewrite

y = 12 x - 2 x^{2}

in the form

y = a x^{2} + b x + c

y = - 2 x^{2} + 12 x

The parabola parameters are:

a = - 2, b = 12, c = 0

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

x_{v} = - \frac{12}{2 ( - 2 )}

Simplify

- \frac{12}{2 ( - 2 )} : 3

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = 18

Therefore the parabola vertex is

(3, 18)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 2

Maximum (3, 18)

+ 1 + x^{2}

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

x^{2} - 5 x - 36

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

100 (1 - \frac{x}{40})^{2}

What is the vertices f(x)=12x-2x^2 ?
The vertices f(x)=12x-2x^2 is Maximum (3,18)