It's not multiple of 2 because 73 does not have 2 instead of 3:-)
F(x) = 3x^2+6x-2
= 3(x^2+2x-2/3)
= 3(x^2 +2x + 1 - 1 - 2/3)
= 3((x+1)^2 - 1 - 2/3)
= 3((x+1)^2 -5/3))
= 3(x+1)^2 - 5
So our function is in the form:
f(x) = 3(x+1)^2 - 5
where a = 3, h = -1, k = -5
Substitute this into the parabolic equation,
We're told the line 
intersects 
twice, which means the quadratic above has two distinct real solutions. Its discriminant must then be positive, so we know
We can tell from the quadratic equation that has its vertex at the point (3, 6). Also, note that
and
so the furthest to the right that extends is the point (5, 2). The line passes through this point for 
. For any value of 
, the line passes through either only once, or not at all.
So 
; in set notation,
