What is the vertices (x+2)^2-9 ?

The vertices (x+2)^2-9 is Minimum (-2,-9)

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vertices (x+2)^2-9

vertices

(x + 2)^{2} - 9

Minimum (- 2, - 9)

Solution steps

y = (x + 2)^{2} - 9

Rewrite

y = (x + 2)^{2} - 9

in the form

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

y = x^{2} + 4 x - 5

The parabola parameters are:

a = 1, b = 4, c = - 5

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} = - 9

Therefore the parabola vertex is

(- 2, - 9)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 2, - 9)

y = x^{2} - 8 x + 7

f (x) = - x^{2} + 7

y = - 4 x^{2} + 8 x + 2

y^{2} + 12 x - 2 y - 11 = 0

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