What is the vertices-x^2+2x+4 ?

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

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vertices-x^2+2x+4

vertices

- x^{2} + 2 x + 4

Maximum (1, 5)

Solution steps

y = - x^{2} + 2 x + 4

The parabola parameters are:

a = - 1, b = 2, c = 4

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} = 5

Therefore the parabola vertex is

(1, 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (1, 5)

4 x^{2} - 8 x - 2

2 x (x - 9) - (x - 9)

y = x^{2} + 2 x + 4

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

y = - 2 x^{2} - 4 x - 7