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

The vertices 4x^2-8x-2 is Minimum (1,-6)

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vertices 4x^2-8x-2

vertices

4 x^{2} - 8 x - 2

Minimum (1, - 6)

Solution steps

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

The parabola parameters are:

a = 4, b = - 8, c = - 2

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} = - 6

Therefore the parabola vertex is

(1, - 6)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 4

Minimum (1, - 6)

2 x (x - 9) - (x - 9)

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

y = x^{2} + 2 x + 5

y = - 2 x^{2} - 4 x - 7

f (x) = x^{2} - 12 x