What is the vertices+x^2+4x-1 ?

The vertices+x^2+4x-1 is Minimum (-2,-5)

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vertices+x^2+4x-1

vertices

+ x^{2} + 4 x - 1

Minimum (- 2, - 5)

Solution steps

y = x^{2} + 4 x - 1

The parabola parameters are:

a = 1, b = 4, c = - 1

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} = - 5

Therefore the parabola vertex is

(- 2, - 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 2, - 5)

y^{2} - 12 y - 4 x = - 28

24000 x^{2} + 900 x - 150

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

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

f (x) = x^{2} - 12 x + 27