What is the vertices f(x)=3x^2-12x+4 ?

The vertices f(x)=3x^2-12x+4 is Minimum (2,-8)

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vertices f(x)=3x^2-12x+4

vertices

f (x) = 3 x^{2} - 12 x + 4

Minimum (2, - 8)

Solution steps

y = 3 x^{2} - 12 x + 4

The parabola parameters are:

a = 3, b = - 12, c = 4

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} = - 8

Therefore the parabola vertex is

(2, - 8)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (2, - 8)

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

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

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

x^{2} + 4 y^{2} = 16

y^{2} = - 28 x

What is the vertices f(x)=3x^2-12x+4 ?
The vertices f(x)=3x^2-12x+4 is Minimum (2,-8)