What is the vertices f(x)=-x^2+2x-5 ?

The vertices f(x)=-x^2+2x-5 is Maximum (1,-4)

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vertices f(x)=-x^2+2x-5

vertices

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

Maximum (1, - 4)

Solution steps

y = - x^{2} + 2 x - 5

The parabola parameters are:

a = - 1, b = 2, c = - 5

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

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

Simplify

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

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = - 4

Therefore the parabola vertex is

(1, - 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (1, - 4)

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

f (x) = 2 x^{2} + 4 x - 6

f (x) = 10 x^{2} - 17

3 x^{2} - 5 x + 2

y = - 3 x^{2} + 12 x - 8

What is the vertices f(x)=-x^2+2x-5 ?
The vertices f(x)=-x^2+2x-5 is Maximum (1,-4)