# 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.

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### 1 Answer

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