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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.

 Oct 23, 2022
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If the parabola opens upwards, then a must be positive.

 

The roots will always be 3 and -2. This gives a base of 5.

 

5h/2 = 12

h = 24/5

 

However, since the parabola opens upwards, the intersection with the y-axis will be negative.

The graph of the parabola must have a y intersect at -24/5. Plugging this point in:

-24/5=a(-3)(2)

-24/5=-6a

4/5 = a

 Oct 23, 2022

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