What is the vertices-x^2+2x+3 ?

The vertices-x^2+2x+3 is Maximum (1,4)

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vertices-x^2+2x+3

vertices

- x^{2} + 2 x + 3

Maximum (1, 4)

Solution steps

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

The parabola parameters are:

a = - 1, b = 2, c = 3

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)

2 x^{2} + 4 x - 1

(x + 2)^{2} = - 12 (y - 5)

2 x^{2} - 8 x + 6

f (x) = x^{2} + 14 x - 72

f (x) = x^{2} + 8 x