What is the vertices-2x^2+25x ?

The vertices-2x^2+25x is Maximum (25/4 , 625/8)

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vertices-2x^2+25x

vertices

- 2 x^{2} + 25 x

Maximum (\frac{25}{4}, \frac{625}{8})

Solution steps

y = - 2 x^{2} + 25 x

The parabola parameters are:

a = - 2, b = 25, c = 0

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

x_{v} = - \frac{25}{2 ( - 2 )}

Simplify

x_{v} = \frac{25}{4}

Plug in

x_{v} = \frac{25}{4}

to find the

y_{v}

value

y_{v} = \frac{625}{8}

Therefore the parabola vertex is

(\frac{25}{4}, \frac{625}{8})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 2

Maximum (\frac{25}{4}, \frac{625}{8})

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

y = x^{2} - 2 x - 1

3 x^{2} - 2 x + 1

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

x (2 x + 3)