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.
+36916 a ( x^2 -x-6) no matter what value 'a' is, the zeroes are still going to be 3 and -2
so we know one leg of the triangle is +3
triangle area = 1/2 b * h
1/2 3 * h = 12
shows the other leg (the y-axis intercept) needs ot be | 8 |
|a (-6) | = 8
a = - 1 1/3 or + 1 1/3
so
a = + 1 1/3 ( since the parabola opens upward)
