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.
I think its C
You must have meant:
x = 2y^2 + 3y + 1 and 2x + 3y^2 = 0
so you are intersecting two parabolas both with a horizontal axis.
I ran it through Wolfram, and it shows that they do not intersect.
http://www.wolframalpha.com/input/?i=plot+x+%3D+2y%5E2+%2B+3y+%2B+1+and+2x+%2B+3y%5E2+%3D+0+from+-3+to+3
let's attempt to solve them anyway:
plug x from the first into the second
2(2y^2 + 3y + 1) + 3y^2 = 0
7y^2 + 6y + 2 = 0
y = (-6 ±√-20)/14 <----- not real
so there is no intersection
check your typing of the equations
Yes, you are correct. The correct option to find the point of intersection(s) for the given equations is option C.
To find the points of intersection, you need to 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, you can plug these y-coordinates back into one of the original equations to find the corresponding x-coordinates.
Let's go through the steps to find the points of intersection:
1. Start with the given equations:
Equation 1: x = 2y^2 + 3y + 1
Equation 2: 2x + 3y^2 = 0
2. Solve Equation 1 for x:
x = 2y^2 + 3y + 1
3. Solve Equation 2 for x:
2x = -3y^2
x = -3y^2 / 2
4. Set the expressions for x from both equations equal to each other:
2y^2 + 3y + 1 = -3y^2 / 2
5. Simplify the equation:
4y^2 + 6y + 2 = -3y^2
6. Rearrange the equation and combine like terms:
7y^2 + 6y + 2 = 0
7. Solve this quadratic equation using factoring, completing the square, or using the quadratic formula. In this case, the equation factors as:
(7y + 2)(y + 1) = 0
Set each factor equal to zero and solve for y:
7y + 2 = 0 --> y = -2/7
y + 1 = 0 --> y = -1
8. Plug these y-coordinates back into one of the original equations to find the corresponding x-coordinates. Let's use Equation 1:
For y = -2/7:
x = 2(-2/7)^2 + 3(-2/7) + 1
= 4/7 - 6/7 + 1
= 5/7
For y = -1:
x = 2(-1)^2 + 3(-1) + 1
= 2 - 3 + 1
= 0
9. So, the points of intersection are (5/7, -2/7) and (0, -1).
Therefore, option C is the correct method to find the point of intersection(s) for the given equations.