Question

Please help me ASAP 50 points

Answer 1

Answer:

y>= -8

Step-by-step explanation:

Answer 2

Answer:

x=2

Vertex (2,-8)

Intercepts: 2 ±2/3 sqrt(6)

Because the parabola opens upwards, it has a minimum at the vertex

Domain:  all real numbers

Range: y>= -8

Step-by-step explanation:

y= 3x^2 -12x+4

To find the axis of symmetry, we use the formula

h= -b/2a  where ax^2 +bx+c

h = -(-12)/2*3 = 12/6=2

The axis of symmetry is x=2

The x coordinate of the vertex is at the axis of symmetry. To find the y coordinate, substitute x into the equation

y= 3(2)^2 -12(2) +4

y = 3*4-24+4

  = 12-24+4

  = -8

The vertex is at (2,-8)

We can use the quadratic formula to find the x intercepts

x = -b ±sqrt(b^2-4ac)

    -----------------------

           2a

  =  12 ± sqrt(12^2 - 4*3*4)

  ---------------------------------

    2*3

=   12± sqrt(144-48)

 ----------------------

  6

= 12±sqrt(96)

---------------

6

=2 ±1/6 * 4sqrt(6)

= 2 ±2/3 sqrt(6)

Because the parabola opens upwards, it has a minimum at the vertex.

The domain is the values that x can take.

X can be all real numbers

The range is the values that y can take.  Since there is a minimum, y must be greater than that minimum

Range: y>= -8