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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.

 May 6, 2022
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The roots are     -2  and 3 

 

So  the base of the triangle =  3  - -2 =   5

 

So

 

We need to  find  a    y intercept  such  that   (1/2)(5) (h) =  12 ....... so

 

(5/2) h = 12

 

h = 12 * 2/5  =    24 / 5

 

Expand the polynomial

 

a ( x + 2) (x -3)  =

 

a [x^2 -x - 6 ]

 

ax^2  - ax   -  6a

 

Now....when x  = 0   the y intercept =  -6a

 

And we need to have this  =   24/5

 

So

 

-6a =  24/5

a = -24/ 30 =  - 4/5

 

But.....we need   abs (a) =   4/5   since the parabola turns upward

 

So

 

a = 4/5

 

cool cool cool

 May 6, 2022

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