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 Jul 23, 2021

#1**+1 **

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)

ElectricPavlov Jul 23, 2021