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

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

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vertices y=x^2-4x+5

vertices

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

Minimum (2, 1)

Solution steps

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

The parabola parameters are:

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

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} = 1

Therefore the parabola vertex is

(2, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (2, 1)

f (x) = 3 x^{2}

y = 4 x^{2} - 8 x + 7

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

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

y = 2 x^{2}