What is the vertices 4x^2+8x+4 ?

The vertices 4x^2+8x+4 is Minimum (-1,0)

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vertices 4x^2+8x+4

vertices

4 x^{2} + 8 x + 4

Minimum (- 1, 0)

Solution steps

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

The parabola parameters are:

a = 4, b = 8, c = 4

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

x_{v} = - \frac{8}{2 \cdot 4}

Simplify

x_{v} = - 1

Plug in

x_{v} = - 1

to find the

y_{v}

value

y_{v} = 0

Therefore the parabola vertex is

(- 1, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 4

Minimum (- 1, 0)

y = (x + 2)^{2} + 1

3 x^{2} + 6 x + 2

2 x^{2} - 3 x

f (x) = 4 (x - 2)^{2} + 11

f (x) = (2 x + 2) (x - 5)