Best Answer

So by definition both equations equal one another:

5x2-2x+1 = 6-3x-x2

5x2+x2-2x+3x+1-6 = 0

6x2+x-5 = 0

Factor the above equation with the help of the quadratic equation formula:

(6x-5)(x+1) = 0

When x = -1 y = 8

When x = 5/6 y = 101/36

Therefore the curves intersect at points: (-1, 8) and (5/6, 101/36)

Q: Where are the points of intersection on the curves y equals 5x squared minus 2x plus 1 and y equals 6 minus 3x minus x squared?

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

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. The coordinates of such points will be the solutions to the simultaneous equations representing the curves.

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

Related questions

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

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)

The points of intersection. The coordinates of such points will be the solutions to the simultaneous equations representing the curves.

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)

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)

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: 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)

There are none. Those two curves do not intersect.

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