What is the vertices y=3x-x^2 ?

The vertices y=3x-x^2 is Maximum (3/2 , 9/4)

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vertices y=3x-x^2

vertices

y = 3 x - x^{2}

Maximum (\frac{3}{2}, \frac{9}{4})

Solution steps

y = 3 x - x^{2}

Rewrite

y = 3 x - x^{2}

in the form

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

y = - x^{2} + 3 x

The parabola parameters are:

a = - 1, b = 3, c = 0

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

x_{v} = - \frac{3}{2 ( - 1 )}

Simplify

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

Plug in

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

to find the

y_{v}

value

y_{v} = \frac{9}{4}

Therefore the parabola vertex is

(\frac{3}{2}, \frac{9}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (\frac{3}{2}, \frac{9}{4})

y = 3 x^{2} + 12 x - 3

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

f (x) = 3 x^{2} - 5

f (x) = 2 x^{2} - 12 x + 20

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