What is the vertices y=3(x-1)^2-2 ?

The vertices y=3(x-1)^2-2 is Minimum (1,-2)

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vertices y=3(x-1)^2-2

vertices

y = 3 (x - 1)^{2} - 2

Minimum (1, - 2)

Solution steps

y = 3 (x - 1)^{2} - 2

Rewrite

y = 3 (x - 1)^{2} - 2

in the form

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

y = 3 x^{2} - 6 x + 1

The parabola parameters are:

a = 3, b = - 6, c = 1

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

x_{v} = - \frac{( - 6 )}{2 \cdot 3}

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 = 3

Minimum (1, - 2)

y = - x^{2} + 2 x - 5

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

y = x^{2} - 5 x - 14

x^{2} + 14 x - 8 y + 65 = 0

x^{2} - 4 x - 4