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.
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 it is plotted correctly, there is no reason for it to be anything but accurate.
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)