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

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

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vertices y=(x-1)^2-2

vertices

y = (x - 1)^{2} - 2

Minimum (1, - 2)

Solution steps

y = (x - 1)^{2} - 2

Rewrite

y = (x - 1)^{2} - 2

in the form

y = a x^{2} + b x + c

y = x^{2} - 2 x - 1

The parabola parameters are:

a = 1, b = - 2, c = - 1

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{( - 2 )}{2 \cdot 1}

Simplify

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = - 2

Therefore the parabola vertex is

(1, - 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (1, - 2)

y = - 4 x^{2} - 12 x - 8

y = + x^{2} + 8 x + 9

25 x^{2} - 36

y = 4 x^{2} - 40 x + 98

x = 2 y^{2}