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.
When x = 0 you will find the y - axis intercept
a (-3)(2) = -6 a = y
when y = 0 you will find the x-axis intercept
0 = a (x-3)(x+2) x = 3 and -2 ( or a = 0 , but we will ignore that...we know this is a parabola) )
Try: y = -6a x = 3
1/2 ( -6a)(3) = 12
-6a = 8
a = - 6/8 but this would result in downward opening parabola ( there would be a negative x^2 coefficient)
Try: y = - 6a x = - 2
1/2 (-6a)(-2) = 12
a = 2
Here is a graph......note we are looking at the LEFT side of the parabola.....