What is the vertices f(x)=-x^2+10x-16 ?

The vertices f(x)=-x^2+10x-16 is Maximum (5,9)

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vertices f(x)=-x^2+10x-16

vertices

f (x) = - x^{2} + 10 x - 16

Maximum (5, 9)

Solution steps

y = - x^{2} + 10 x - 16

The parabola parameters are:

a = - 1, b = 10, c = - 16

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} = 9

Therefore the parabola vertex is

(5, 9)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (5, 9)

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

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

y = 3 x^{2} - 3 x - 6

f (x) = x^{2} - 3 x + 10

- 2 x^{2} + 12 x - 20

What is the vertices f(x)=-x^2+10x-16 ?
The vertices f(x)=-x^2+10x-16 is Maximum (5,9)