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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.

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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They intersect at the point of: (-3/2, 11/4)

We believe that those equations have no real solutions, and that their graphs therefore have no points of intersection.

x2-x3+2x = 0 x(-x2+x+2) = 0 x(-x+2)(x+1) = 0 Points of intersection are: (0, 2), (2,10) and (-1, 1)

If: y = 4x2-2x-1 and y = -2x2+3x+5 Then: 4x2-2x-1 = -2x2+3x+5 So: 6x2-5x-6 = 0 Solving the quadratic equation: x = -2/3 or x = 3/2 Points of intersection by substitution: (-2/3, 19/9) and (3/2, 5)

If: y = 4x^2 -2x -1 and y = -2x^2 +3x +5 Then: 4x^2-2x-1 = -2x^2+3x+5 =>6x^2-5x-6 = 0 Solving the above quadratic equation: x = -2/3 or x = 3/2 Therefore by substitution the points of intersection are: (-2/3, 19/9) and (3/2, 5)

Related questions

They intersect at the point of: (-3/2, 11/4)

The points are (-1/3, 5/3) and (8, 3).Another Answer:-The x coordinates work out as -1/3 and 8Substituting the x values into the equations the points are at (-1/3, 13/9) and (8, 157)

We believe that those equations have no real solutions, and that their graphs therefore have no points of intersection.

If: y = x2+20x+100 and x2-20x+100 Then: x2+20x+100 = x2-20x+100 So: 40x = 0 => x = 0 When x = 0 then y = 100 Therefore point of intersection: (0, 100)

x2-x3+2x = 0 x(-x2+x+2) = 0 x(-x+2)(x+1) = 0 Points of intersection are: (0, 2), (2,10) and (-1, 1)

If: y = x2-4x+8 and y = 8x-x2-14 Then: x2-4x+8 = 8x-x2-14 So: 2x2-12x+22 = 0 Discriminant: 122-(4*2*22) = -32 Because the discriminant is less than 0 there is no actual contact between the given parabolas

If: y = 4x2-2x-1 and y = -2x2+3x+5 Then: 4x2-2x-1 = -2x2+3x+5 So: 6x2-5x-6 = 0 Solving the quadratic equation: x = -2/3 or x = 3/2 Points of intersection by substitution: (-2/3, 19/9) and (3/2, 5)

If: y = -2x^2 +3x +5 and y = 4x^2 -2x -1 Then: 4x^2 -2x -1 = -2x^2 +3x +5 So it follows: 6x^2 -5x -6 = 0 Using the quadratic equation formula: x = -2/3 or x = 3/2 Therefore points of intersection by substitution are at: (-2/3, 19/9) and (3/2, 5)

If: y = 4x^2 -2x -1 and y = -2x^2 +3x +5 Then: 4x^2-2x-1 = -2x^2+3x+5 =>6x^2-5x-6 = 0 Solving the above quadratic equation: x = -2/3 or x = 3/2 Therefore by substitution the points of intersection are: (-2/3, 19/9) and (3/2, 5)

If: y = 4x^2 -2x -1 and y = -2x^2+3x+5 Then: 4x^2 -2x -1 = -2x^2+3x+5 => 6x^2-5x-6 = 0 Solving the above quadratic equation: x = -2/3 or x = 3/2 Therefore the points of intersection by substitution are: (-2/3, 19/9) and (3/2, 5)

If: y = x^2 -2x +4 and y = 2x^2 -4x +4 Then: 2x^2 -4x +4 = x^2 -2x +4 Transposing terms: x^2 -2x = 0 Factorizing: (x-2)(x+0) => x = 2 or x = 0 Therefore by substitution points of intersect are at: (2, 4) and (0, 4)

If: y = 4x2-2x-1 and y = -2x2+3x+5 Then: 4x2-2x-1 = -2x2+3x+5 And so: 6x2-5x-6 = 0 Using the the quadratic equation formula: x = -2/3 and x = 3/2 Substitution: when x = -2/3 then y = 19/9 and when x = 3/2 then y = 5 Points of intersection: (-2/3, 19/9) and (3/2, 5)