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

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

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

vertices

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

Minimum (1, 2)

Solution steps

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

Rewrite

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

in the form

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

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

The parabola parameters are:

a = 1, b = - 2, c = 3

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

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

Simplify

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = 2

Therefore the parabola vertex is

(1, 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (1, 2)

f (x) = x^{2} + 12 x + 16

x^{2} + 4 x - 5

- x^{2} + 3 x - 5

y = - x^{2} + 8 x

x^{2} + 3 x - 4

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