f(x)= 3x^2 + 4x + 2
= 3(x^2 + 4x/3) + 2
= 3(x^2 + 4x/3 + 4/9) + 2 - 3(4/9)
= 3(x+2/3)^2 + 2/3
f(x) = a(x-h)^2 + k where a=3, h=-2/3 and k = 2/3
vertex = (h,k) = (-2/3, 2/3) = minimum point on the parabola
y intercept =2 or (0,2)
x intercepts when 3x^2 +4x + 2 = 0
x = -4/6 + or - (1/6)sqr(16-4(12)(2))
there are no real x intercepts, just imaginary as the discriminate <0
axis of symmetry is x = -2/3
It's an upward opening parabola
