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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.

 Jul 23, 2021
 #1
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+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)

 Jul 23, 2021
edited by ElectricPavlov  Jul 23, 2021

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