What is the vertices y=(x-2)^2+1 ?

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

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vertices y=(x-2)^2+1

vertices

y = (x - 2)^{2} + 1

Minimum (2, 1)

Solution steps

y = (x - 2)^{2} + 1

Rewrite

y = (x - 2)^{2} + 1

in the form

y = a x^{2} + b x + c

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)

x = - 3 y^{2} + 5

f (x) = x^{2} + 10 x + 6

- 3 x^{2} - 12 x - 9

y = x^{2} - 2 x 5

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