What is the vertices y=-x^2+8x-7 ?

The vertices y=-x^2+8x-7 is Maximum (4,9)

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vertices y=-x^2+8x-7

vertices

y = - x^{2} + 8 x - 7

Maximum (4, 9)

Solution steps

y = - x^{2} + 8 x - 7

The parabola parameters are:

a = - 1, b = 8, c = - 7

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} = 9

Therefore the parabola vertex is

(4, 9)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (4, 9)

- 3 x^{2} + 6 x + 5

f (x) = - x^{2} + 12 x - 35

f (x) = 5 x^{2} - 2 x - 1

- 4 x^{2} + 6 x - 1

f (x) = x^{2} - 8 x + 18