(Solved) How do I find the equation/function of a parabola by looking at the graph?

jules78245

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11 years ago

How do I find the equation/function of a parabola by looking at the graph?

The ordered pairs are (1,12), (2,22), (3,30), (4,36),
(5,40), (6,42), (7,42), (8,40), (9,36), (10,30), (11,22), (12,12). How do I find the equation/function of that graph?

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nursestudent8

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11 years ago

Since you know that it is a parabola, a parabola has two forms:
y = ax^2 + bx + c (standard form)
y = a(x - h)^2 + k (vertex form, where (h,k) is the vertex).

We could use either form, but I choose to use the standard form. We pick three of those ordered pairs, and plug in x and y for the equation y = ax^2 + bx + c. We would then solve for a, b, and c as a system of equations.

For (1, 12), using y = ax^2 + bx + c, x = 12 and y = 1, so
12 = a(1)^2 + b(1) + c
12 = a + b + c

For (2, 22), we do the same thing:
22 = a(2)^2 + b(2) + c
22 = 4a + 2b + c

For (3,30), we get:
30 = a(3)^2 + b(3) + c
30 = 9a + 3b + c

So our three equations and three unknowns are:
12 = a + b + c
22 = 4a + 2b + c
30 = 9a + 3b + c

To solve this, I'm going to use substitution. From the first equation, if 12 = a + b + c, then a = 12 - b - c. I'm going to substitute this into the second and third equations.

Plugging a = 12 - b - c into the second equation:

22 = 4(12 - b - c) + 2b + c
22 = 48 - 4b - 4c + 2b + c
22 = 48 - 2b - 3c
-26 = -2b - 3c
26 = 2b +3c

Plugging a = 12 - b - c into the third equation:

30 = 9(12 - b - c) + 3b + c
30 = 108 - 9b - 9c + 3b + c
30 = 108 - 6b - 8c
-78 = -6b - 8c
39 = 3b + 4c

So now we have two equations, two unknowns;

26 = 2b +3c
39 = 3b + 4c

We'll solve this using elimination. Multiplying the top equation by
-3 and the bottom equation by 2, we get

-78 = -6b - 9c
78 = 6b + 8c

And adding them together, yields 0 = -c, or c = 0.
Since c = 0, we plug this into any one of the equations, and
78 = 6b + 0
78 = 6b
b = 13

So far, we have b = 13, c = 0. now, all we have to do is get a by plugging in BOTH values into one of the three original equations.

12 = a + b + c
12 = a + 13 + 0
12 = a + 13
a = -1

a = -1, b = 13, c = 0. Therefore, the function is

y = ax^2 + bx + c
y = -x^2 + 13x

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Lena

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11 years ago

I can do it with calculus, but I'm guessing based on the question you haven't gotten that far, and plus I don't feel like typing it all out... have fun.

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jujubee

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11 years ago

The general equation for a parabola is y = ax^2 + bx + c. To solve, all you need to do is pick any three points and plug them in. Now you have three linear equations in three unknowns. This you can solve. Picking the three points will give you answers for your x and y variables, leaving you to solve for the constants a, b, and c. Picking any two equations will allow you to treat a,b, and c as variables for this part only. Once solved, they retain the value you obtain for them and go back to being constants. I'll choose the first three points:
12 = a * 1^2 + b * 1 + c; 22 = a * 2^2 + b * 2 + c; and 30 = a * 3^2 + b * 3 + c. Now regroup: 12 = a + b + c; 22 = 4a + 2b + c; and 30 = 9a + 3b + c. If you pick the first two equations, and subtract (2) from (1) you cancel out the c. Similarly with equations (3) and (1). Now we have two equations in two unknowns. Simple Algebra I. You can take it from here.

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Answer accepted by topic starter

tony9506

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How do I find the equation/function of a parabola by looking at the graph?

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