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

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

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

vertices

y = (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)

2 x^{2} + 5 x - 3

x^{2} - 2 x - 24

f (x) = 2 x^{2} + 18

(r - 14)^{2} - 56.25

y = 7 - 6 x^{2}