What is the vertices f(x)=(x-5)^2 ?

The vertices f(x)=(x-5)^2 is Minimum (5,0)

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vertices f(x)=(x-5)^2

vertices

f (x) = (x - 5)^{2}

Minimum (5, 0)

Solution steps

y = (x - 5)^{2}

y = (x - 5)^{2}

The parabola parameters are:

a = 1, m = 5, n = 5

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

x_{v} = \frac{5 + 5}{2}

Simplify

x_{v} = 5

Plug in

x_{v} = 5

to find the

y_{v}

value

y_{v} = 0

Therefore the parabola vertex is

(5, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (5, 0)

f (x) = x^{2} - 10 x + 9

y = - x^{2} - 8

x^{2} - 7 x + 12

y = - x^{2} + 1

f (x) = - 2 x^{2} + 20 x + 48