
#1
Sep1507, 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(xh)+k 3. The attempt at a solution The parabola is upside down to I know a is negative. 



#2
Sep1507, 12:52 PM

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P: 3,225

Any thoughts about the problem? What does the general form of the equation of a parabola look like?




#3
Sep1507, 01:04 PM

P: 100

y=ax^2+bx+c




#4
Sep1507, 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.




#5
Sep1507, 01:25 PM

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P: 2,693





#6
Sep1507, 01:32 PM

P: 492

if you have 3 points and you know the general equation of a parabola what could you make?




#7
Sep1507, 02:08 PM

P: 100

the 3 points are: (2,2), (0,1), (1,2.5)
I substituted the numbers, giving three equations. 2=4a2b+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
Sep1507, 02:22 PM

P: 100

I got the numbers to work. The equation in general form is 2x^211/2x+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
Sep1507, 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 coefficients. 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
Sep1507, 04:06 PM

Math
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P: 38,898

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? 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
Sep1507, 05:28 PM

P: 100

I had 3 equations.
2=4a+2b+c 5/2=a+b+c 1=0+0+c 



#12
Sep1507, 05:34 PM

P: 100

the solutions I got were 2, 11/2, and 1. They worked in all 3 equations.




#13
Sep1507, 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.




#14
Sep1507, 05:51 PM

P: 100

I think I switched the signs around. Now I get
2=4a2b+c 1=0+0+c 5/2=a+b+c and got a=1 b=5/2 c=1 Does that sound better? 



#15
Sep1507, 06:50 PM

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#16
Aug1211, 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? 



#17
Aug1211, 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 x^{2} + b x + c .



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