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

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

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

vertices

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

Minimum (3, - 1)

Solution steps

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

Rewrite

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

in the form

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

y = 3 x^{2} - 18 x + 26

The parabola parameters are:

a = 3, b = - 18, c = 26

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

x_{v} = - \frac{( - 18 )}{2 \cdot 3}

Simplify

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = - 1

Therefore the parabola vertex is

(3, - 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (3, - 1)

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

- 3 x^{2} - 6 x + 2

f (x) = 5 x^{2} - 10 x + 4

(y - 9)^{2} = - 4 (x - 2)

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