What is the vertices f(x)=(x-3)^2+2 ?

The vertices f(x)=(x-3)^2+2 is Minimum (3,2)

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vertices f(x)=(x-3)^2+2

vertices

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

Minimum (3, 2)

Solution steps

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

Rewrite

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

in the form

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

y = x^{2} - 6 x + 11

The parabola parameters are:

a = 1, b = - 6, c = 11

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

x_{v} = - \frac{( - 6 )}{2 \cdot 1}

Simplify

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = 2

Therefore the parabola vertex is

(3, 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (3, 2)

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

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

(x + 2)^{2} + 3

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

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

What is the vertices f(x)=(x-3)^2+2 ?
The vertices f(x)=(x-3)^2+2 is Minimum (3,2)