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

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

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vertices y=(x+2)^2-3

vertices

y = (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 + 1

The parabola parameters are:

a = 1, b = 4, c = 1

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) = 2 x^{2} - 8 x - 5

f (x) = 2 x^{2} - 8 x + 6

y = + 0.5 x^{2} - 6 x + 21

f (x) = x^{2} + 2 x - 4

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