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`
`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