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

The vertices f(x)=-x^2+2x+5 is Maximum (1,6)

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

vertices

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

Maximum (1, 6)

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} = 6

Therefore the parabola vertex is

(1, 6)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (1, 6)

2 x^{2} - 4 x

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

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

x^{2} - 2 x + 1

y^{2} = 5 x + 3

What is the vertices f(x)=-x^2+2x+5 ?
The vertices f(x)=-x^2+2x+5 is Maximum (1,6)