What is the vertices x^2-12x+44 ?

The vertices x^2-12x+44 is Minimum (6,8)

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vertices x^2-12x+44

vertices

x^{2} - 12 x + 44

Minimum (6, 8)

Solution steps

y = x^{2} - 12 x + 44

The parabola parameters are:

a = 1, b = - 12, c = 44

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

x_{v} = - \frac{( - 12 )}{2 \cdot 1}

Simplify

x_{v} = 6

Plug in

x_{v} = 6

to find the

y_{v}

value

y_{v} = 8

Therefore the parabola vertex is

(6, 8)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (6, 8)

2 y^{2} - 8 x - 24 y + 56 = 0

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

f (x) = x^{2} + 3 x + \frac{1}{4}

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

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