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 Feb 22, 2022

#1**+1 **

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.....

https://www.desmos.com/calculator/kot9e2fc8v

ElectricPavlov Feb 22, 2022