What is the vertices f(x)=-x^2+8x-15 ?

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

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vertices f(x)=-x^2+8x-15

vertices

f (x) = - 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)

y = x^{2} + 6

y = x^{2} + 2

x^{2} + 2 x - 8

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

- 2 x^{2} + 25 x

What is the vertices f(x)=-x^2+8x-15 ?
The vertices f(x)=-x^2+8x-15 is Maximum (4,1)