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

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

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vertices f(x)=3x^2-12x+8

vertices

f (x) = 3 x^{2} - 12 x + 8

Minimum (2, - 4)

Solution steps

y = 3 x^{2} - 12 x + 8

The parabola parameters are:

a = 3, b = - 12, c = 8

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} = - 4

Therefore the parabola vertex is

(2, - 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (2, - 4)

x^{2} + y^{2} + 4 x - 2 y = - 1

\frac{x ^{2}}{49} + \frac{y ^{2}}{16} = 1

9 x^{2} + 16 y^{2} - 144 = 0

x^{2} + y^{2} = 49

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

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