What is the vertices x^2-4x+12 ?

The vertices x^2-4x+12 is Minimum (2,8)

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vertices x^2-4x+12

vertices

x^{2} - 4 x + 12

Minimum (2, 8)

Solution steps

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

The parabola parameters are:

a = 1, b = - 4, c = 12

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} = 8

Therefore the parabola vertex is

(2, 8)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (2, 8)

y = x^{2} - x

f (x) = 6 x^{2} - 1

f (x) = - 2 x^{2} + 16 x - 31

f (x) = - x^{2} - 6 x - 10

y = 2 x (x - 4) + 7