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

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

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vertices (x+2)^2+3

vertices

(x + 2)^{2} + 3

Minimum (- 2, 3)

Solution steps

y = (x + 2)^{2} + 3

Rewrite

y = (x + 2)^{2} + 3

in the form

y = a x^{2} + b x + c

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

The parabola parameters are:

a = 1, b = 4, c = 7

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{4}{2 \cdot 1}

Simplify

x_{v} = - 2

Plug in

x_{v} = - 2

to find the

y_{v}

value

y_{v} = 3

Therefore the parabola vertex is

(- 2, 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 2, 3)

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

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

- x^{2} - x + 2

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

y = + 16 - (20 x + 15)^{2}