Equation: y = 8x^2 -26x+15
Equation when factorized: y = (4x-3)(2x-5)
When x = 0 then y = (0, 15) which is the point of intersection on the y axis
When y = 0 then x = (3/4, 0) and (5/2, 0) which are the points of intersection on the x axis
The points of intersection are: (7/3, 1/3) and (3, 1)
Points of intersection work out as: (3, 4) and (-1, -2)
Any metric or non-metric units can be represented by points on the plotted line.
If 3x -5y = 16 and xy = 7 then by combining both equations into a single quadratic equation and solving it then the points of intersection are at (-5/3, -21/5) and (7, 1)
They work out as: (-3, 1) and (2, -14)
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 of intersection of the equations 4y^2 -3x^2 = 1 and x -2 = 1 are at (0, -1/2) and (-1, -1)
Any metric or non-metric units can be represented by points on the plotted line.
If 3x -5y = 16 and xy = 7 then by combining both equations into a single quadratic equation and solving it then the points of intersection are at (-5/3, -21/5) and (7, 1)
They work out as: (-3, 1) and (2, -14)
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)
We believe that those equations have no real solutions, and that their graphs therefore have no points of intersection.
If it is plotted correctly, there is no reason for it to be anything but accurate.
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)
Scatter Graph
It works out that the points of intersection between the equations of 2x+5 = 5 and x^2 -y^2 = 3 are at: (14/3, -13/3) and (2, 1)