What is the vertices f(x)=x^2-6x-4 ?

The vertices f(x)=x^2-6x-4 is Minimum (3,-13)

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Find the Vertex f(x)=x^2-6x-4

vertices

f (x) = x^{2} - 6 x - 4

Minimum (3, - 13)

Solution steps

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

The parabola parameters are:

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

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

x_{v} = - \frac{( - 6 )}{2 \cdot 1}

Simplify

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = - 13

Therefore the parabola vertex is

(3, - 13)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (3, - 13)

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

h (x) = x^{2} - 14 x + 6

8 x^{2} - 48 x + 65

x^{2} + y^{2} - 8 x - 6 y + 21 = 0

f (x) = x^{2} - 14 x + 6

What is the vertices f(x)=x^2-6x-4 ?
The vertices f(x)=x^2-6x-4 is Minimum (3,-13)