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

justaguide | College Teacher | (Level 2) Distinguished Educator

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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).

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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

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). ﻿﻿﻿