What is the vertices y=+x^2-2x-6 ?

The vertices y=+x^2-2x-6 is Minimum (1,-7)

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vertices y=+x^2-2x-6

vertices

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

Minimum (1, - 7)

Solution steps

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

The parabola parameters are:

a = 1, b = - 2, c = - 6

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} = - 7

Therefore the parabola vertex is

(1, - 7)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (1, - 7)

f (x) = x^{2} - 8 x

x^{2} + 16 y = 0

+ 3 x^{2} + x + 5

y = 3 x^{2} + x - 1

28 x^{2} + 47 x + 15