How do you find the point of intersection(s) for x = 2y2 + 3y + 1 and 2x + 3y2 = 0
A) You cannot find points of intersections for non-functions.
B) Plug in 0 for x into both equations and solve for y. Then plug that answer back into the other equation to find the corresponding x-coordinate.
C) Solve both equations for x and set them equal to each other. This will give you the y-coordinates of the points of intersection. Then plug back into one of the equations to find the corresponding x-coordinates.
D) Solve both equations for y and set them equal to each other. This will give you the x-coordinates of the points of intersection. Then plug back into one of the equations to find the corresponding y-coordinates.
OptionC) Solve both equations for x and set them equal to each other. This
will give you the y-coordinates of the points of intersection. Then plug
back into one of the equations to find the corresponding x-coordinates.
x = 2y^2 + 3y + 1 x = - 3y^2 / 2
2y^2 + 3y + 1 = - 3y^2 / 2
4y^2 + 6y +2 = -3y^2
4y^2 + 3y^2 + 6y +2 = 0
7y^2 +6y + 2 = 0
Now you apply the quadratic formula and find y-values. Then use those values to find the x -values.
