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

The vertices y=+(x-3)^2 is Minimum (3,0)

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vertices y=+(x-3)^2

vertices

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

Minimum (3, 0)

Solution steps

y = (x - 3)^{2}

y = (x - 3)^{2}

The parabola parameters are:

a = 1, m = 3, n = 3

x_{v} = \frac{m + n}{2}

x_{v} = \frac{3 + 3}{2}

Simplify

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = 0

Therefore the parabola vertex is

(3, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (3, 0)

3 (x - 6)^{2} + 1

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

f (x) = x^{2} + 4 x + 6

f (x) = 2 x^{2} - 6 x + 4

x^{2} - 9 x + 20