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

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

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

vertices

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

Minimum (- 2, 0)

Solution steps

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

The parabola parameters are:

a = 1, b = 4, c = 4

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} = 0

Therefore the parabola vertex is

(- 2, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 2, 0)

x = 4 y^{2}

9 x^{2} + 4 y^{2} = 36

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

3.5 (x + 2.2)^{2} + 3.5 (y - 11.1)^{2} - 21 = 0

y^{2} - 4 y + 4 x = - 16

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