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

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

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vertices y=x^2+4x

vertices

y = x^{2} + 4 x

Minimum (- 2, - 4)

Solution steps

y = x^{2} + 4 x

The parabola parameters are:

a = 1, b = 4, c = 0

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} = - 4

Therefore the parabola vertex is

(- 2, - 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 2, - 4)

f (x) = (x - 4)^{2} - 3

f (x) = 2 x^{2} - 3 x - 5

y = - 2 x^{2} - 16 x - 37

x = 3 y^{2}

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