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.

Guest Oct 23, 2022

#1**+1 **

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

WhyamIdoingthis Oct 23, 2022