What is the vertices y=3x^2-6x+5 ?

The vertices y=3x^2-6x+5 is Minimum (1,2)

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vertices y=3x^2-6x+5

vertices

y = 3 x^{2} - 6 x + 5

Minimum (1, 2)

Solution steps

y = 3 x^{2} - 6 x + 5

The parabola parameters are:

a = 3, b = - 6, c = 5

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)

x^{2} - 2

f (x) = - \frac{1}{4} (x + 2)^{2} - 2

y = (x + 4)^{2}

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

3 x^{2} - 10