What is the vertices x^2-6x+8 ?

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

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vertices x^2-6x+8

vertices

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)

y = - 4 x^{2} - 24 x - 40

f (x) = x^{2} + 6 x + 9

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

f (x) = x^{2} + 5 x + 4

+ y^{2} + 12 x + 4 y - 32 = 0