Find algebraically the points of intersection of the parabolas with equations y= 2x^2 + 3x - 1 and y= 5 - x^2
Eliminate the y's:
2x2 + 3x -1 = 5 - x2
Add x2 - 5 to both sides
3x2 + 3x - 6 = 0
Factorize
3(x+2)(x-1) = 0
so x = -2 and x = 1
when x = -2 y = 5 - (-2)2 = 1
when x = 1 y = 5 - 12 = 4
So the intersections occur at (-2, 1) and (1, 4)
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