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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?

2006-12-30 11:08:13 · 6 answers · asked by watergirl286 1 in Science & Mathematics Mathematics

6 answers

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

2006-12-30 11:17:25 · answer #1 · answered by Puggy 7 · 2 0

ok, when you consider that there is not any scale figuring out on numbers and plugging in won't help, you're precise! Now what you will desire to understand is that x^2 is your foundation for assessment, so understand what that feels like, particularly evidently like a grin. Now, the quantity in front of the x determines how skinny or huge the smile is. in case you have a quantity below a million, such as you had a million/2, then it makes the smile wider. And a quantity extra beneficial than a million makes it thinner. So i think you had a million/2x^2, x^2, 2x^2 and 3x^2. So the widest one is the a million/2x^2. And the subsequent widest is the x^2, then 2x^2, then 3x^2. So, a million is c, 2 is b, 3 is a, and seven is d. Now, for the negatives, the same rule applies to cause them to huge or skinny. yet they are in basic terms flipped over the horizontal (x) axis. So, the -a million/3x^2 is the widest one that's the opposite direction up, or a frown, no longer a grin anymore! So 4 is h, 5 is g, 6 is f and eight is e.

2016-12-15 11:53:05 · answer #2 · answered by ? 4 · 0 0

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.

2006-12-30 11:32:35 · answer #3 · answered by flyfisher_20750 3 · 0 0

If you are just interested in the equation, a really quick way to get it is to put three of the points into your calculator and have it do a quadratic regression. Although this is generally used to find the "best fit" quadratic for a sample of points, it can find the only parabola for three points.

On the TI-83, put the 3 x-values in L1, the 3 corresponding y-values in L2. Then hit STAT, choose CALC, the QuadReg.

2006-12-30 11:55:58 · answer #4 · answered by grand_nanny 5 · 0 0

You can use the symmetry of any parabola.

General form,
f(x) = a(x-h)^2 + k

By symmetry,
h = (6+7)/2 = 13/2

f(6) - f(5) = 2 = a(1/2)^2 - a(3/2)^2 = -2a

a = -1

f(x) = -(x-13/2)^2 + k
f(1) = 12 = -121/4 + k
k = 169/4

f(x) = -(x-13/2)^2 + (13/2)^2 = -x^2 + 13x

2006-12-30 11:38:03 · answer #5 · answered by sahsjing 7 · 0 0

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.

2006-12-30 11:23:24 · answer #6 · answered by northwestsk8 1 · 0 2

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