If: y = 4x^2 -2x -1 and y = -2x^2 +3x+5
Then: 4x^2 -2x-1 = -2x^2 +3x+5 or 6x^2 -5x -6 = 0
Factorizing the above: (3x+2)(2x-3) = 0 meaning x = -2/3 or x = 3/2
By substitution points of contact are at: (-2/3, 19/9) and (3/2, 5)
They intersect at the point of: (-3/2, 11/4)
They touch each other at (0, 100) on the x and y axis.
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)
They intersect at points (-2/3, 19/9) and (3/2, 5) Solved by combining the two equations together to equal nought and then using the quadratic equation formula to find the values of x and substituting these values into the equations to find the values of y.
If: y = 4x^2 -12x -3 and y = x^2 +11x +5 Then: 4x^2 -12x -3 = x^2 +11x +5 Transposing terms: 3x^2 -23x -8 = 0 Factorizing: (3x+1)(x-8) = 0 => x = -1/3 or x = 8 Therefore the x coordinates are: -1/3 and 8
They intersect at the point of: (-3/2, 11/4)
Because when collated together the discriminant of b2-4ac = -32 i.e. 144-(4*2*22) = -32 In order for the parabolas to make contact with each other the discriminant must equal zero or be above zero.
They touch each other at (0, 100) on the x and y axis.
If: y = 4x2-2x-1 and y = -2x2+3x+5 Then the length of the line works out as: 65/18 or 3.6111 ... recurring decimal 1
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 -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)
There is no connection between the given curves because when they are combined into a single quadratic equation the discriminant of the equation is less than zero which means they share no valid roots.
The graphs of the two equations will intersect when x² + 20x + 100 = y = x² - 20x + 100 Subtracting x² +100 from both sides you get 20x = -20x that will only be true when x = 0. At x = 0, y = 100 for both equations - so the point of contact would be (0,100)
They intersect at points (-2/3, 19/9) and (3/2, 5) Solved by combining the two equations together to equal nought and then using the quadratic equation formula to find the values of x and substituting these values into the equations to find the values of y.
If: y = 4x^2 -12x -3 and y = x^2 +11x +5 Then: 4x^2 -12x -3 = x^2 +11x +5 Transposing terms: 3x^2 -23x -8 = 0 Factorizing: (3x+1)(x-8) = 0 => x = -1/3 or x = 8 Therefore the x coordinates are: -1/3 and 8
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)
If: y = 3x^2 +10x +11 and y = 2 -2x -x^2 Then: 3x^2 +10x +11 = 2 -2x -x^2 Transposing terms: 4x^2 +12x +9 = 0 Factorizing: (2x +3)(2x +3) = 0 => x = -3/2 and also x = -3/2 Therefore by substituting the point of contact is at: (-3/2, 11/4)