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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.

 Feb 22, 2022
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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                 

 Feb 22, 2022

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