What is the vertices x^2-4x+8 ?

The vertices x^2-4x+8 is Minimum (2,4)

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vertices x^2-4x+8

vertices

x^{2} - 4 x + 8

Minimum (2, 4)

Solution steps

y = x^{2} - 4 x + 8

The parabola parameters are:

a = 1, b = - 4, c = 8

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} = 4

Therefore the parabola vertex is

(2, 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (2, 4)

y = 3 x^{2} + 30 x + 65

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

y = (x - 8)^{2}

y = x^{2} - 8 x + 12

f (x) = (x + 2) (x + 3)