2015 AMC 12A Problem 12

The $y$-intercepts of the two parabolas are $-2$ and $4.$ To intersect the $x$-axis, the first parabola opens upward and the second opens downward, so their $x$-intercepts are $\pm t$ for some $t \gt 0.$ The kite has one diagonal of length $4 - (-2) = 6$ along the $y$-axis and the other of length $2t.$ Its area is $\dfrac{1}{2}\cdot 6\cdot 2t = 6t = 12,$ so $t = 2.$ Thus the $x$-intercepts are $\pm 2.$ For the first parabola, $0 = a(2)^2 - 2$ gives $a = \dfrac{1}{2};$ for the second, $0 = 4 - b(2)^2$ gives $b = 1.$ Therefore $a + b = 1.5.$ Thus, the correct answer is B.