What is the vertices x^2-2x-80 ?

The vertices x^2-2x-80 is Minimum (1,-81)

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vertices x^2-2x-80

vertices

x^{2} - 2 x - 80

Minimum (1, - 81)

Solution steps

y = x^{2} - 2 x - 80

The parabola parameters are:

a = 1, b = - 2, c = - 80

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

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

Simplify

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = - 81

Therefore the parabola vertex is

(1, - 81)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (1, - 81)

y^{2} + 2 y - 4 x + 9 = 0

f (x) = 2 x^{2} + 5 x - 12

y = 3 x^{2} - 6 x + 3

(y + 6)^{2} = 12 (x + 1)

y = 3 x^{2} + 2 x + 6