What is the vertices y=x^2-12x-20 ?

The vertices y=x^2-12x-20 is Minimum (6,-56)

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vertices y=x^2-12x-20

vertices

y = x^{2} - 12 x - 20

Minimum (6, - 56)

Solution steps

y = x^{2} - 12 x - 20

The parabola parameters are:

a = 1, b = - 12, c = - 20

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} = - 56

Therefore the parabola vertex is

(6, - 56)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (6, - 56)

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

- x^{2} + 3 x + 1

f (x) = x^{2} + 12 x + 35

y^{2} + 14 y + 4 x + 45 = 0

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

What is the vertices y=x^2-12x-20 ?
The vertices y=x^2-12x-20 is Minimum (6,-56)