What is the vertices y=-x^2+8x ?

The vertices y=-x^2+8x is Maximum (4,16)

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vertices y=-x^2+8x

vertices

y = - x^{2} + 8 x

Maximum (4, 16)

Solution steps

y = - x^{2} + 8 x

The parabola parameters are:

a = - 1, b = 8, c = 0

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

x_{v} = - \frac{8}{2 ( - 1 )}

Simplify

- \frac{8}{2 ( - 1 )} : 4

x_{v} = 4

Plug in

x_{v} = 4

to find the

y_{v}

value

y_{v} = 16

Therefore the parabola vertex is

(4, 16)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (4, 16)

x^{2} + 3 x - 4

y = (x - 1)^{2}

(x - 3)^{2} = 8 (y + 5)

f (x) = - 3 x^{2} + 6 x - 4

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