What is the vertices x^2-5x+4 ?

The vertices x^2-5x+4 is Minimum (5/2 ,-9/4)

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vertices x^2-5x+4

vertices

x^{2} - 5 x + 4

Minimum (\frac{5}{2}, - \frac{9}{4})

Solution steps

y = x^{2} - 5 x + 4

The parabola parameters are:

a = 1, b = - 5, c = 4

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{9}{4}

Therefore the parabola vertex is

(\frac{5}{2}, - \frac{9}{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{9}{4})

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

y = 7 x^{2} - 10 x + 5

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

(x + 2)^{2} - 9

y = x^{2} - 8 x + 7