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)

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

f (x) = 2 x^{2} - 12 x + 20

f (x) = 3 x^{2} + 2 x + 1

f (x) = 9 x^{2} - x + 1

f (x) = - 2 x^{2} - 4 x + 6