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Finding an equation of a parabola given three points.

by starchild75
Tags: equation, parabola, points
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starchild75
#1
Sep15-07, 12:49 PM
P: 100
1. The problem statement, all variables and given/known data

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

2. Relevant equations

y=a(x-h)+k

3. The attempt at a solution

The parabola is upside down to I know a is negative.
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radou
#2
Sep15-07, 12:52 PM
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Any thoughts about the problem? What does the general form of the equation of a parabola look like?
starchild75
#3
Sep15-07, 01:04 PM
P: 100
y=ax^2+bx+c

starchild75
#4
Sep15-07, 01:14 PM
P: 100
Finding an equation of a parabola given three points.

I don't know how to find the variables given the three points.
symbolipoint
#5
Sep15-07, 01:25 PM
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Quote Quote by starchild75 View Post
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.
bob1182006
#6
Sep15-07, 01:32 PM
P: 492
if you have 3 points and you know the general equation of a parabola what could you make?
starchild75
#7
Sep15-07, 02:08 PM
P: 100
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
starchild75
#8
Sep15-07, 02:22 PM
P: 100
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.
bob1182006
#9
Sep15-07, 02:35 PM
P: 492
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.
HallsofIvy
#10
Sep15-07, 04:06 PM
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Quote Quote by starchild75 View Post
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?

Quote Quote by starchild75 View Post
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.
starchild75
#11
Sep15-07, 05:28 PM
P: 100
I had 3 equations.

-2=4a+2b+c
-5/2=a+b+c
1=0+0+c
starchild75
#12
Sep15-07, 05:34 PM
P: 100
the solutions I got were 2, -11/2, and 1. They worked in all 3 equations.
starchild75
#13
Sep15-07, 05:43 PM
P: 100
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.
starchild75
#14
Sep15-07, 05:51 PM
P: 100
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?
learningphysics
#15
Sep15-07, 06:50 PM
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Quote Quote by starchild75 View Post
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.
blackguyza
#16
Aug12-11, 02:18 PM
P: 1
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?
SammyS
#17
Aug12-11, 04:05 PM
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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 .


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