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

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

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

vertices

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

Minimum (- 3, - 2)

Solution steps

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

Rewrite

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

in the form

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

y = x^{2} + 6 x + 7

The parabola parameters are:

a = 1, b = 6, c = 7

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

x_{v} = - \frac{6}{2 \cdot 1}

Simplify

x_{v} = - 3

Plug in

x_{v} = - 3

to find the

y_{v}

value

y_{v} = - 2

Therefore the parabola vertex is

(- 3, - 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 3, - 2)

f (x) = x^{2} + 4 x + 6

f (x) = 2 x^{2} - 6 x + 4

x^{2} - 9 x + 20

x (x - 4)

f (x) = x^{2} - 8 x + 9

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