What is the vertices f(x)=x^2-5x+6 ?

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

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vertices f(x)=x^2-5x+6

vertices

f (x) = 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})

- x^{2} + 4 x - 3

y = 2 x^{2} - 1 - (5 x - 6)

y = x^{2} - x - 20

\frac{- 3 x ^{2} - 2 x + 6}{5}

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

What is the vertices f(x)=x^2-5x+6 ?
The vertices f(x)=x^2-5x+6 is Minimum (5/2 ,-1/4)