What is the vertices y=-x^2+10x ?

The vertices y=-x^2+10x is Maximum (5,25)

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vertices y=-x^2+10x

vertices

y = - x^{2} + 10 x

Maximum (5, 25)

Solution steps

y = - x^{2} + 10 x

The parabola parameters are:

a = - 1, b = 10, c = 0

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

x_{v} = - \frac{10}{2 ( - 1 )}

Simplify

- \frac{10}{2 ( - 1 )} : 5

x_{v} = 5

Plug in

x_{v} = 5

to find the

y_{v}

value

y_{v} = 25

Therefore the parabola vertex is

(5, 25)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (5, 25)

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

+ 1 + x^{2}

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

x^{2} - 5 x - 36

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