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.
+128631 The roots are -2 and 3
So the base of the triangle = 3 - -2 = 5
So
We need to find a y intercept such that (1/2)(5) (h) = 12 ....... so
(5/2) h = 12
h = 12 * 2/5 = 24 / 5
Expand the polynomial
a ( x + 2) (x -3) =
a [x^2 -x - 6 ]
ax^2 - ax - 6a
Now....when x = 0 the y intercept = -6a
And we need to have this = 24/5
So
-6a = 24/5
a = -24/ 30 = - 4/5
But.....we need abs (a) = 4/5 since the parabola turns upward
So
a = 4/5
