The parabolas defined by the equations $y=x^2+4x+6$ and \(y=\frac{1}{2}x^2+x+6\) intersect at points $(a,b)$ and $(c,d)$, where \(c\ge a\). What is c-a?
Set the functions equal
(1/2)x^2 + x + 6 = x^2 + 4x + 6 rearrange as
(1/2)x^2 + 3x = 0
Factor
x [ (1/2)x + 3 ] = 0
Setting both factors equal to 0 and solving for x, we have that
x = 0 and x = -6
c = 0 and a = - 6
So c - a = 0 - -6 = 6
