What is the vertices y=4x+4x-2+x^2-3 ?

The vertices y=4x+4x-2+x^2-3 is Minimum (-4,-21)

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vertices y=4x+4x-2+x^2-3

vertices

y = 4 x + 4 x - 2 + x^{2} - 3

Minimum (- 4, - 21)

Solution steps

y = 4 x + 4 x - 2 + x^{2} - 3

Rewrite

y = 4 x + 4 x - 2 + x^{2} - 3

in the form

y = a x^{2} + b x + c

y = x^{2} + 8 x - 5

The parabola parameters are:

a = 1, b = 8, c = - 5

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

x_{v} = - \frac{8}{2 \cdot 1}

Simplify

x_{v} = - 4

Plug in

x_{v} = - 4

to find the

y_{v}

value

y_{v} = - 21

Therefore the parabola vertex is

(- 4, - 21)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 4, - 21)

y = 3 - x^{2}

y = 25 x^{2} - 36

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

y^{2} = 4 x

y = x^{2} - 2 x - 3

What is the vertices y=4x+4x-2+x^2-3 ?
The vertices y=4x+4x-2+x^2-3 is Minimum (-4,-21)