What is the vertices y=-x^2+8x-15 ?

The vertices y=-x^2+8x-15 is Maximum (4,1)

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vertices y=-x^2+8x-15

vertices

y = - x^{2} + 8 x - 15

Maximum (4, 1)

Solution steps

y = - x^{2} + 8 x - 15

The parabola parameters are:

a = - 1, b = 8, c = - 15

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} = 1

Therefore the parabola vertex is

(4, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (4, 1)

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

f (x) = - 2 x^{2} + 8 x - 9

f (x) = - 5 x^{2} + 10 x - 6

y = x^{2} - 4 x - 1

(x - 4)^{2} + 2