What is the vertices f(x)=3x^2-6x+4 ?

The vertices f(x)=3x^2-6x+4 is Minimum (1,1)

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vertices f(x)=3x^2-6x+4

vertices

f (x) = 3 x^{2} - 6 x + 4

Minimum (1, 1)

Solution steps

y = 3 x^{2} - 6 x + 4

The parabola parameters are:

a = 3, b = - 6, c = 4

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} = 1

Therefore the parabola vertex is

(1, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (1, 1)

16 x^{2} - 9 y^{2} = 144

y^{2} = 8 x

\frac{( x - 2 ) ^{2}}{36} + \frac{( y + 1 ) ^{2}}{25} = 1

\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1

\frac{x ^{2}}{9} + \frac{y ^{2}}{25} = 1

What is the vertices f(x)=3x^2-6x+4 ?
The vertices f(x)=3x^2-6x+4 is Minimum (1,1)