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

The vertices f(x)=(x+3)^2 is Minimum (-3,0)

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

vertices

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

Minimum (- 3, 0)

Solution steps

y = (x + 3)^{2}

y = (x + 3)^{2}

The parabola parameters are:

a = 1, m = - 3, n = - 3

x_{v} = \frac{m + n}{2}

x_{v} = \frac{( - 3 ) + ( - 3 )}{2}

Simplify

x_{v} = - 3

Plug in

x_{v} = - 3

to find the

y_{v}

value

y_{v} = 0

Therefore the parabola vertex is

(- 3, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 3, 0)

f (x) = - 9.5 x^{2} - 47.5 x + 63

f (x) = 3 (x - 1) (x + 2)

f (x) = - \frac{1}{250} x^{2} + \frac{4}{5} x

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

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