Finding an equation of a parabola given three points.

In summary: The three points are not on the same parabola -- at least they're not consistent with being on the same parabola if it's of the form: y = 2 x2 + b x + c . If you're given a parabola and want to find the points where it intersects another parabola, you can use the intersecting lines theorem. In summary, the author attempted to solve an equation for a parabola given three points, but was not successful. They did mention that if the equation is in standard form, y=a(x-h)2+k, then completing the square will give the equation.
  • #1
starchild75
100
1

Homework Statement



Find an equation of a parabola given three points without a vertex point.

Homework Equations



y=a(x-h)+k

The Attempt at a Solution



The parabola is upside down to I know a is negative.
 
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  • #2
Any thoughts about the problem? What does the general form of the equation of a parabola look like?
 
  • #3
y=ax^2+bx+c
 
  • #4
I don't know how to find the variables given the three points.
 
  • #5
starchild75 said:
I don't know how to find the variables given the three points.
The general form will give you your method. If you have three points, then you can form three equations in general form. The unknowns will be the coefficients. You can simply solve the system of linear equations. This would be introductory level algebra.
 
  • #6
if you have 3 points and you know the general equation of a parabola what could you make?
 
  • #7
the 3 points are: (-2,2), (0,1), (1,-2.5)

I substituted the numbers, giving three equations.

2=4a-2b+c

1=0x^2+0b+C

-5/2=a+b+c

so c=1

-2b=-4a+1

b=(4a+1)/2


I can't get the numbers to work.


This is actually in my calculus book. chapter 1
 
  • #8
I got the numbers to work. The equation in general form is 2x^2-11/2x+1. How would I convert that to standard form. What is the trick to solving these? I feel like I got lucky to get the answer.
 
  • #9
To solve these you will just need to know algebra to solve a system of equations.

if instead of (0,1) you were given some other point you would go ahead and combine 2 equations into 1 hopefully eliminating an x^2 or x or c term, and then working with that one and the remaining equation try to solve for a,b,or c, and using that to solve for the other co-efficients.

to get that equation to the standard form you need to complete the square, but you first need to remove the 2 in front of x^2 first.
 
  • #10
starchild75 said:
the 3 points are: (-2,2), (0,1), (1,-2.5)

I substituted the numbers, giving three equations.

2=4a-2b+c

1=0x^2+0b+C

-5/2=a+b+c

so c=1

-2b=-4a+1

b=(4a+1)/2


I can't get the numbers to work.


This is actually in my calculus book. chapter 1
You said you had 3 equations- you've only written 2 there!
The three equations you get are:
(-2,2) 4a- 2b+ c= 2
(0, 1) c= 1
(1, -2.5) a+ b+ c= -2.5

Yes, from the second equation, you get c= 1. Putting that into the other two equations, 4a- 2b= 1 and a+ b= -3.5. Can you solve those two equations for a and b?

starchild75 said:
I got the numbers to work. The equation in general form is 2x^2-11/2x+1. How would I convert that to standard form. What is the trick to solving these? I feel like I got lucky to get the answer.
Sorry, but it's too early to feel lucky! That's not at all right. Putting a= 2, b= 11/2, c= 1 into the first equation, you get 4(2)- 2(11/2)+ 1= 2- 11+ 1= -8, not 2.

Once you do have the correct equation, if by "standard form" you mean y= a(x- h)2+ k, you get that form by completing the square.
 
  • #11
I had 3 equations.

-2=4a+2b+c
-5/2=a+b+c
1=0+0+c
 
  • #12
the solutions I got were 2, -11/2, and 1. They worked in all 3 equations.
 
  • #13
Every time I do this problem, I get different answers. I have always had trouble with these. I can handle trig easily compared to these problems. I am so frustrated.
 
  • #14
I think I switched the signs around. Now I get
2=4a-2b+c
1=0+0+c
-5/2=a+b+c

and got

a=-1
b=-5/2
c=1

Does that sound better?
 
  • #15
starchild75 said:
I think I switched the signs around. Now I get
2=4a-2b+c
1=0+0+c
-5/2=a+b+c

and got

a=-1
b=-5/2
c=1

Does that sound better?

looks good to me.
 
  • #16
If I am given the points (-6; -2), (4; -2) and (1; -8) with the equation y = 2 x^2 + bx + c, CAN I use any of the two points two simultaneously solve for b and c?

By the way I tried and get different values with different points. The only reasonable solution is if I solve the problem using the fact that x = -1 is an axis of symmetry and then use the general equation y = 2(x + 1)^2 + q. Then use the point (1; -8) to find q. Rewriting the equation gives b = 4 and c = -14.

Why don't I get the same when I solve simultaneously?
 
Last edited:
  • #17
The three points are not on the same parabola -- at least they're not consistent with being on the same parabola if it's of the form: y = 2 x2 + b x + c .
 

What is the Process for Finding the Equation of a Parabola with Three Given Points?

Finding the equation of a parabola given three points involves following these steps:

  1. Identify the three points: Determine the coordinates of the three points (x₁, y₁), (x₂, y₂), and (x₃, y₃) through which the parabola passes.
  2. Write the general equation: Use the general equation for a parabola, either in vertex form or standard form, depending on the information you have.
  3. Set up a system of equations: Plug the coordinates of the three points into the general equation to create a system of three equations with three variables (a, b, and c).
  4. Solve the system: Solve the system of equations to find the values of a, b, and c.
  5. Write the specific equation: Substitute the determined values of a, b, and c back into the general equation to obtain the specific equation of the parabola.

What Are the General Equations for a Parabola?

The general equations for a parabola are as follows:

  • Vertex Form: \(y = a(x - h)^2 + k\), where (h, k) is the vertex of the parabola.
  • Standard Form: \(y = ax^2 + bx + c\), where a, b, and c are constants that determine the shape, orientation, and position of the parabola.

How Do I Determine the Type of Parabola (Upward/Downward) from Three Points?

The type of parabola (whether it opens upward or downward) can be determined from the sign of the coefficient "a" in the general equation. If "a" is positive, the parabola opens upward; if "a" is negative, it opens downward.

Can I Find the Equation of a Parabola Given Three Points if They Are Not in a Straight Line?

Yes, you can find the equation of a parabola given three non-collinear (not in a straight line) points. Three non-collinear points uniquely determine a parabola. However, if the three points are collinear (lie on a straight line), they do not uniquely define a parabola, as an infinite number of parabolas can pass through them.

What Do I Do if the Three Points Are in a Straight Line?

If the three given points are collinear and lie on a straight line, you cannot uniquely determine a parabola that passes through them. In such cases, you may not be able to find a unique equation for the parabola.

Are There Any Special Cases to Consider When Finding the Equation of a Parabola with Three Points?

One special case to consider is when the three points are all located on the axis of symmetry (vertical line) of the parabola. In this case, the vertex form of the equation can be simplified, making it easier to find the equation. Additionally, if one of the given points is the vertex of the parabola, it simplifies the equation and makes solving for the coefficients a, b, and c more straightforward.

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