# Determine the points that are on the curves y=x^2+x+1 and y=-x^2-2x+6.

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To determine the points that lie on both the curves y=x^2+x+1 and y=-x^2-2x+6, we have to equate the two. Doing this gives

x^2 + x + 1 = -x^2 - 2x + 6

=> x^2 + x^2 + x + 2x + 1 - 6 = 0

=> 2x^2 + 3x - 5 = 0

=> 2x^2 + 5x - 2x - 5 =0

=> x ( 2x + 5) - 1( 2x + 5) = 0

=> ( x -1)( 2x + 5) = 0

This gives x = 1 and x = -5/2

At x = 1, y = x^2 + x + 1 = 1 + 1 + 1 = 3

At x = -5/2 , y = (-5/2)^2 - 5/2 + 1 = 19/4

**Therefore the points that lie on both the curves are ( 1, 3) and ( -5/2 , 19/4).**

Supposing that you want to determine the intercepting point of the given curves, we'll have to solve the system formed from the equations of the curves.

We'll put

x^2 + x + 1 = -x^2 - 2x + 6

We'll move all terms to the left side:

2x^2 + 3x - 5 = 0

We'll apply the quadratic formula:

x1 = [-3+sqrt(9 + 40)]/4

x1 = (-3 + 7)/4

x1 = 1

x2 = (-3 - 7)/4

x2 = -5/2

We'll find the corresponding y coordinates:

y1 = x1^2 + x1 + 1

y1 = 1 + 1 + 1

y1 = 3

y2 = x2^2 + x2 + 1

y2 = 25/4 - 5/2 + 1

y2 = (25 - 10 + 4)/4

y2 = 19/4

**The intercepting points are: (1 ; 3) and (-5/2 ; 19/4). **