What is the vertices f(x)=x^2-6x+8 ?

The vertices f(x)=x^2-6x+8 is Minimum (3,-1)

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vertices f(x)=x^2-6x+8

vertices

f (x) = x^{2} - 6 x + 8

Minimum (3, - 1)

Solution steps

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

The parabola parameters are:

a = 1, b = - 6, c = 8

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} = - 1

Therefore the parabola vertex is

(3, - 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (3, - 1)

x^{2} - y^{2} = 10 (x - y) + 1

\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1

F = \frac{m v ^{2}}{r}

(x - 0)^{2} + (y - 20)^{2} = 400

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

What is the vertices f(x)=x^2-6x+8 ?
The vertices f(x)=x^2-6x+8 is Minimum (3,-1)