LucyNeedsMathHelp
contestada

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.

Respuesta :

Edufirst
x = 2y2 + 3y + 1 and 2x + 3y2 = 0


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.



Otras preguntas