What is the vertices y=x^2-12x+2 ?

The vertices y=x^2-12x+2 is Minimum (6,-34)

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vertices y=x^2-12x+2

vertices

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

Minimum (6, - 34)

Solution steps

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

The parabola parameters are:

a = 1, b = - 12, c = 2

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} = - 34

Therefore the parabola vertex is

(6, - 34)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (6, - 34)

f (x) = - x^{2} + 7 x + 18

y = (2 x - 1) (5 x - 6)

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

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

x^{2} - 6 x