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If: y = 2 -2x -x^2 and y = 3x^2 +10x +11

Then: 3x^2 +10x +11 = 2 -2x -x^2

So it follows that: 4x^2 +12x +9 = 0

Using the quadratic equation formula: x = -3/2 and x also = -3/2

Therefore by substitution the point of contact is at: (-3/2, 11/4)

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Q: What is the point of contact that the parabolas of y equals 2 -2x -x squared and y equals 3x squared plus 10x plus 11 make with each other showing work?
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