What is the vertices 3(x-6)^2+1 ?

The vertices 3(x-6)^2+1 is Minimum (6,1)

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vertices 3(x-6)^2+1

vertices

Minimum (6, 1)

Solution steps

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

Rewrite

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

in the form

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

y = 3 x^{2} - 36 x + 109

The parabola parameters are:

a = 3, b = - 36, c = 109

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

x_{v} = - \frac{( - 36 )}{2 \cdot 3}

Simplify

x_{v} = 6

Plug in

x_{v} = 6

to find the

y_{v}

value

y_{v} = 1

Therefore the parabola vertex is

(6, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (6, 1)