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Q: What are the points of intersection of the parabolas y equals x squared -2x plus 4 and y equals 2x squared -4x plus 4?

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The points of intersection are: (7/3, 1/3) and (3, 1)

Points of intersection work out as: (3, 4) and (-1, -2)

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)

The points of intersection of the equations 4y^2 -3x^2 = 1 and x -2 = 1 are at (0, -1/2) and (-1, -1)

Straight line: 3x-y = 5 Curved parabola: 2x^2 +y^2 = 129 Points of intersection works out as: (52/11, 101/11) and (-2, -11)

They work out as: (-3, 1) and (2, -14)

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)

You need two, or more, curves for points of intersection.

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 = 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 = -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 2x^2 = 3x -y +5 Then: 4x^2 -2x -1 = -2x^2 +3x +5 Transposing terms: 6x^2 -5x -6 = 0 Factorizing: (3x+2)(2x-3) = 0 => x = -2/3 or x = 3/2 By substitution points of intersection are at: (-2/3, 19/9) and (3/2, 5)

If: y = 5x^2 -2x +1 and y = 6 -3x -x^2Then: 5x^2 -2x +1 = 6 -3x -x^2Transposing terms: 6x^2 +x -5 = 0Factorizing: (6x -5)(x +1) = 0 => x = -1 or x = 5/6Through substitution points of intersect are at: (-1, 8) and (5/6, 101/36)

If: y = x^2 +3x -10 and y = -x^2 -8x -15 Then: x^2 +3x -10 = -x^2 -8x -15 Transposing terms: 2x^2 +11x +5 = 0 Factorizing the above: (2x +1)(x +5) = 0 meaning x = -1/2 or -5 Therefore by substitution points of intersection are at: (-1/2, -45/4) and (-5, 0)

If: y = 5x^2 -2x +1 and y = 6 -3x -x^2 Then: 5x^2 -2x +1 = 6 -3x -x^2 Transposing terms: 6x^2 +x -5 = 0 Factorizing the above: (6x -5)(x +1) = 0 meaning x = 5/6 or x = -1 By substitution points of intersection are at: (5/6, 101/36) and (-1, 8)

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 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)

(3/4, 0) and (5/2, 0) Solved with the help of the quadratic equation formula.

If: x+y = 7 and x2+y2 = 25 Then: x = 7-y and so (7-y)2+y2 = 25 => 2y2-14y+24 = 0 Solving the quadratic equation: y = 4 and y = 3 By substitution points of intersection: (3, 4) and (4, 3)

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)

Improved Answer:-If: 2x+y = 5 and x^2 -y^2 = 3Then by rearranging: y = 5 -2x and -3x^2 -28+20x = 0Solving the above quadratic equation: x = 2 and x = 14/3By substitution points of intersection are: (2, 1) and (14/3, -13/3)

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: 3x-y = 5 and 2x2+y2 = 129 Then: 3x-y = 5 => y = 3x-5 And so: 2x2+(3x-5)2 = 129 => 11x2-30x-104 = 0 Using the quadratic equation formula: x = 52/11 and x = -2 By substitution points of intersection are: (52/11, 101/11) and (-2, -11)

If: x-2y = 1 and 3xy-y2 = 8 Then: x =1+2y and so 3(1+2y)y-y2 = 8 => 3y+5y2-8 = 0 Solving the quadratic equation: y = 1 or y = -8/5 Points of intersection by substitution: (3, 1) and (-11/5, -8/5)

x2+y2+4x+6y-40 = 0 and x = 10+y Substitute the second equation into the first equation: (10+y)2+y2+4(10+y)+6y-40 = 0 2y2+30y+100 = 0 Divide all terms by 2: y2+15y+50 = 0 (y+10)(y+5) = 0 => y = -10 or y = -5 Substitute the above values into the second equation to find the points of intersection: Points of intersection are: (0, -10) and (5, -5)