What is the vertices f(x)=2x^2+4x ?

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

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vertices f(x)=2x^2+4x

vertices

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

Minimum (- 1, - 2)

Solution steps

y = 2 x^{2} + 4 x

The parabola parameters are:

a = 2, b = 4, c = 0

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

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

Simplify

x_{v} = - 1

Plug in

x_{v} = - 1

to find the

y_{v}

value

y_{v} = - 2

Therefore the parabola vertex is

(- 1, - 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- 1, - 2)

y = 3 x^{2} - 6 x + 4

y = 3 x^{2} + 2 x - 1

x^{2} - x

y = - x^{2} + 6 x - 1

- (x - 30)^{2} + 40

What is the vertices f(x)=2x^2+4x ?
The vertices f(x)=2x^2+4x is Minimum (-1,-2)