What is the vertices x^2-4x-6 ?

The vertices x^2-4x-6 is Minimum (2,-10)

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vertices x^2-4x-6

vertices

x^{2} - 4 x - 6

Minimum (2, - 10)

Solution steps

y = x^{2} - 4 x - 6

The parabola parameters are:

a = 1, b = - 4, c = - 6

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

x_{v} = - \frac{( - 4 )}{2 \cdot 1}

Simplify

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = - 10

Therefore the parabola vertex is

(2, - 10)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (2, - 10)

y = 84 x - x^{2}

\frac{x ^{2} x ^{1}}{x}

y = x^{2} - 2 x

f (x) = 3 x^{2} + 6 x + 4

f (x) = 5 x^{2}