Anyone help me solve for y.
2 answers:
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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
