What is the vertices y=4x^2-4x^1 ?

The vertices y=4x^2-4x^1 is Minimum (1/2 ,-1)

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vertices y=4x^2-4x^1

vertices

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

Minimum (\frac{1}{2}, - 1)

Solution steps

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

Rewrite

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

in the form

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

y = 4 x^{2} - 4 x

The parabola parameters are:

a = 4, b = - 4, c = 0

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

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

Simplify

- \frac{- 4}{2 \cdot 4} : \frac{1}{2}

x_{v} = \frac{1}{2}

Plug in

x_{v} = \frac{1}{2}

to find the

y_{v}

value

y_{v} = - 1

Therefore the parabola vertex is

(\frac{1}{2}, - 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 4

Minimum (\frac{1}{2}, - 1)

y = - x^{2} + 2 x - 10

1 - x - x^{2}

f (x) = x^{2} - 5 x + 4

x^{2} + 4

y = x^{2} - 4 x - 5

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