What is the vertices f(x)=x^2+2x+3 ?

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

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vertices f(x)=x^2+2x+3

vertices

f (x) = 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)

f (x) = x^{2}

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

f (x) = - 3 x^{2} + 5 x

(x - 3)^{2} = 8 (y - 5)

(x - 2) = 3 (y - 5)^{2}

What is the vertices f(x)=x^2+2x+3 ?
The vertices f(x)=x^2+2x+3 is Minimum (-1,2)