What is the vertices f(x)=-x^2+6x-8 ?

The vertices f(x)=-x^2+6x-8 is Maximum (3,1)

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vertices f(x)=-x^2+6x-8

vertices

f (x) = - x^{2} + 6 x - 8

Maximum (3, 1)

Solution steps

y = - x^{2} + 6 x - 8

The parabola parameters are:

a = - 1, b = 6, c = - 8

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

x_{v} = - \frac{6}{2 ( - 1 )}

Simplify

- \frac{6}{2 ( - 1 )} : 3

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = 1

Therefore the parabola vertex is

(3, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (3, 1)

- x^{2} + 5 x - 7

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

y = - \frac{1}{16} (x + 2)^{2} - 6

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

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

What is the vertices f(x)=-x^2+6x-8 ?
The vertices f(x)=-x^2+6x-8 is Maximum (3,1)