The area of the triangle formed by points of intersection of parabola y=a(x-3)(x+2) with the coordinate axes is 12. Find a if it is known that parabola opens upward.
+308 If the parabola opens upwards, then a must be positive.
The roots will always be 3 and -2. This gives a base of 5.
5h/2 = 12
h = 24/5
However, since the parabola opens upwards, the intersection with the y-axis will be negative.
The graph of the parabola must have a y intersect at -24/5. Plugging this point in:
-24/5=a(-3)(2)
-24/5=-6a
4/5 = a
