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If: y = x^2 -4x +8 and y = 8x -x^2 -14

Then: x^2 -4x+8 = 8x -x^2 -14

Transposing terms: 2x^2 -12x+22 = 0

Divide all terms by 2: x^2 -6x +11 = 0

Using the discriminant b^2 -4(ac): 36 -4(1*11) = -8

Therefore it follows that there are no points of intersection because the discriminant is less than zero.

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Q: What if any are the points of intersection of the parabolas of y equals x2 -4x plus 8 and y equals 8x -x2 -14?
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