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

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

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

vertices

f (x) = (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)

x^{2} + 9 y^{2} + 6 x - 90 y + 225 = 0

x^{2} = - 4 y

x^{2} + y^{2} + 10 x + 12 y + 25 = 0

x^{2} = 16 y

y^{2} = - 10 x