What is the vertices y=x^2+4x+4 ?

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

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vertices y=x^2+4x+4

vertices

y = 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 - 7 - 7

\frac{ln ( x ) + 1}{ln ( x ) - 1}

f (x) = x - 7 - 7

(6, 8), (5, 1)

\frac{x - 2}{x - 1}