What is the vertices f(x)=(x+4)^2 ?

The vertices f(x)=(x+4)^2 is Minimum (-4,0)

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vertices f(x)=(x+4)^2

vertices

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

Minimum (- 4, 0)

Solution steps

y = (x + 4)^{2}

y = (x + 4)^{2}

The parabola parameters are:

a = 1, m = - 4, n = - 4

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

x_{v} = \frac{( - 4 ) + ( - 4 )}{2}

Simplify

x_{v} = - 4

Plug in

x_{v} = - 4

to find the

y_{v}

value

y_{v} = 0

Therefore the parabola vertex is

(- 4, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 4, 0)

f (x) = x^{2} + 3 x + \frac{1}{4}

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

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

f (x) = x^{2} + 6 x + 3

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