Math
posted by Becky .
Give vertex, axis, xintercepts and yintercepts to parabola
y= 3x^26x+4

Y = 3X^2  6X + 4.
Vertex Form: y = a(x  h)^2 + k.
h = Xv = b/2a = 6/6 = 1.
Substitute 1 for x in the given Eq:
k = 3(1)^2  6(1) + 4=3 + 6 + 4 = 7.
V(h , k) = V(1 , 7).
Vertex Form: y = 3(x + 1)^2 + 7.
Axis = h =Xv = 1.
The axis is a vertical line and equals
1 for all values of y.
Let x = 0 and solve for y:
y = 3*0^2  6*0 + 4 = 4 = YInt.
The solutions,xIntercepts, and roots
are all the same.
x = (6 + sqrt(6^2 + 48)) / 6,
x = (6 + sqrt(84)) / 6,
x = (6 + 9.17) / 6,
x = (6 + 9.17) /6 =  2.5275.
x = (6  9.17) / 6 = 0.5275.
Solution set: x = 2.5275, x = 0.5275.