What is the vertices y=(x-1)^2 ?

The vertices y=(x-1)^2 is Minimum (1,0)

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vertices y=(x-1)^2

vertices

y = (x - 1)^{2}

Minimum (1, 0)

Solution steps

y = (x - 1)^{2}

y = (x - 1)^{2}

The parabola parameters are:

a = 1, m = 1, n = 1

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

x_{v} = \frac{1 + 1}{2}

Simplify

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = 0

Therefore the parabola vertex is

(1, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (1, 0)

(x - 3)^{2} = 8 (y + 5)

f (x) = - 3 x^{2} + 6 x - 4

f (x) = x^{2} - 3 x + 1

f (x) = - 2 x^{2} - 12 x - 20

x^{2} + 6 x - 8 y + 1 = 0