What is the vertices f(x)=3x^2-12x+13 ?

The vertices f(x)=3x^2-12x+13 is Minimum (2,1)

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vertices f(x)=3x^2-12x+13

vertices

f (x) = 3 x^{2} - 12 x + 13

Minimum (2, 1)

Solution steps

y = 3 x^{2} - 12 x + 13

The parabola parameters are:

a = 3, b = - 12, c = 13

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

x_{v} = - \frac{( - 12 )}{2 \cdot 3}

Simplify

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = 1

Therefore the parabola vertex is

(2, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (2, 1)

y = x^{2} - x - 6

f (x) = - x^{2} + 6 x - 8

- x^{2} + 5 x - 7

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

y = - \frac{1}{16} (x + 2)^{2} - 6

What is the vertices f(x)=3x^2-12x+13 ?
The vertices f(x)=3x^2-12x+13 is Minimum (2,1)