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)
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 intersect at the point of: (-3/2, 11/4)
They touch each other at (0, 100) on the x and y axis.
If: y = 4x^2 -2x -1 and y = -2x^2 +3x +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 the points of contact are at: (-2/3, 19/9) and (3/2, 5)
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
(b + 2c)(b - c)
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.
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)
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)
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)
It is (-0.3, 0.1)
If: y = 2 -2x -x^2 and y = 3x^2 +10x +11 Then: 3x^2 +10X +11 = 2 -2x -x^2 Or: 4x^2 +12x +9 = 0 Factorizing the above: (2x+3)(2x+3) = 0 meaning x = -3/2 When x = -3/2 then by substitution y = 11/4 Therefore point of contact is at: (-3/2, 11/4)