What is the vertices f(x)=3x^2-6x+5 ?

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

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vertices f(x)=3x^2-6x+5

vertices

f (x) = 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)

2 x^{2} + 4 x - 6

y = k (x - 0)^{2} + 2

y = 5 x^{2} - 3 x - 7 + 2 x^{2} - x

4 y^{2} - 48 x - 20 y = 71

y = x^{2} + 2 x - 15

What is the vertices f(x)=3x^2-6x+5 ?
The vertices f(x)=3x^2-6x+5 is Minimum (1,2)