What is the vertices x^2-5x+6 ?

The vertices x^2-5x+6 is Minimum (5/2 ,-1/4)

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vertices x^2-5x+6

vertices

x^{2} - 5 x + 6

Minimum (\frac{5}{2}, - \frac{1}{4})

Solution steps

y = x^{2} - 5 x + 6

The parabola parameters are:

a = 1, b = - 5, c = 6

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

x_{v} = - \frac{( - 5 )}{2 \cdot 1}

Simplify

x_{v} = \frac{5}{2}

Plug in

x_{v} = \frac{5}{2}

to find the

y_{v}

value

y_{v} = - \frac{1}{4}

Therefore the parabola vertex is

(\frac{5}{2}, - \frac{1}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (\frac{5}{2}, - \frac{1}{4})

2 x^{2} + 3

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

- 3 x^{2} + 5 x

y = 4 x + 4 x - 2 + x^{2} - 3

y = 3 - x^{2}