Consider the two parabolas :y1=2x^2-3x-1 and y2=x^2+7x+20. a-Find the points of intersection of the parabolas and decide which one is greater than the other between the intersection points(b)...

 Consider the two parabolas :y1=2x^2-3x-1 and y2=x^2+7x+20. 

a-Find the points of intersection of the parabolas and decide which one is greater than the other between the intersection points(b) Compute the area enclosed by the two parabolas. 

 

 

Asked on by soud

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thilina-g | College Teacher | (Level 1) Educator

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To find the intersection point you can equate the two expressions.

`y(1) = y(2)`

`2x^2-3x-1 = x^2+7x+20`

`x^2-10x-21 = 0`

Now if we solve for x in this equation, we would be able to find the intersection points.

`(x-11.7823)(x+1.7823) = 0`

therefore, x = 11.7823 or x = -1.7823.

so the two parabolas intersect each other at -1.7823 and 11.7823.

 

To find which one is greater in this (-1.7823,11.7823) we can simply substitute a value in between the two intersection poin and compare the value of the two parabolas.

easily, we can select x = 0, since it is in (-1.7823,11.7823).

so at x=0,

`y(1) = 2x^2-3x-1 = -1`

`y(2) = x^2+7x+20 = 20`

So between the two intersection points `y(2) = x^2+7x+20` is greater.

 

To find the area, you can integrate the difference function from -1.7823 to 11.7823

 

`A = int_-1.7823^11.7823(y(2) - y(1))dx`

`A = int_-1.7823^11.7823(x^2+7x+20 - (2x^2-3x-1))dx`

`A = int_-1.7823^11.7823(-x^2+10x+21)dx`

`int(-x^2+10x+21)dx = (-x^3)/3+(10x^2)/2+21x`

`int(-x^2+10x+21)dx = (-x^3)/3+5x^2+21x`

Therefore,

`A = int_-1.7823^11.7823(-x^2+10x+21)dx = ((-11.7823^3)/3+5*11.7823^2+21*11.7823)-((-(-1.7823)^3)/3+5(-1.7823)^2+10(-1.7823))`

`A = int_-1.7823^11.7823(-x^2+10x+21)dx =396.3248 -(-19.6581) = 415.9829`

 

The area enclosed by two parabolas is 415.9829

 

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