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

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

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vertices x^2+2x+3

vertices

x^{2} + 2 x + 3

Minimum (- 1, 2)

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 \cdot 1}

Simplify

x_{v} = - 1

Plug in

x_{v} = - 1

to find the

y_{v}

value

y_{v} = 2

Therefore the parabola vertex is

(- 1, 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 1, 2)

y = - x^{2}

x^{2} - 4 x + 4

t^{2}

f (x) = - x^{2} - 6 x + 3

x^{2} - 2 x - 3