# Quadratics, i was good until here.

1. Sep 16, 2008

### offtheleft

im going to try to make this as short and easy to follow as possible haha

and im just warning, this is probably so simple that i cant even grasp it because im reading too deeply into it.

okay, im working with these three points:
$$A (0, 1) C (2, 6) D (3, 10)$$

i worked them into three equations, the first one is killing me though!

$$a(0)^2 + b (0) + c = 1$$ this is the one im having trouble with, i cant set it up properly because all i have is $$c = 1$$

i have for the other two:

$$C. . ., 4a + 2b +c = 6$$

$$D. . ., 9a + 3b + c = 10$$

now, in order to get my model for $$y = ax^2 + bx + c$$

how would i set this up, other than that im doing alright..

2. Sep 16, 2008

### danago

Well from the first equation you have found that c=1, so just sub that into the following two equations and then you have a system of two equations in two variables (a and b). Why should the first one cause any trouble?

3. Sep 16, 2008

### symbolipoint

$$a(0)^2 + b (0) + c = 1[tex] this is the one im having trouble with, i cant set it up properly because all i have is [tex] c=1$$
You SHOULD get a + 0 + c = 1

Continue from there and you should have no trouble.

4. Sep 16, 2008

### offtheleft

like i said, so simple that i cant eve grasp it.

thanks, ill see how it works out! :)

-------

i got:

$$y = \frac{1}{2}x^2 + 3x + 1$$

Last edited: Sep 16, 2008
5. Sep 17, 2008

### HallsofIvy

Staff Emeritus
No, you shouldn't. 02= 0, not 1. c= 1 is correct.

6. Sep 17, 2008

### offtheleft

i used $$c = 1$$. im quite confident that the answer i got$$( y = \frac{1}{2}x^2 + 3x + 1)$$ is incorrect. when i plug in all the numbers, get the changes from the original y value and square it, i get all sorts of weird answers for this one. usually, its all 0's except for one value but thats not happening.

7. Sep 17, 2008

### danago

Perhaps you solved the system of equations incorrectly. Have a look at your working, and if you still cant figure it out, post your working here

8. Sep 17, 2008

### offtheleft

okay, heres what we did in class:

four points;
$$A (0, 1)$$
$$B (1, 2)$$
$$C (2, 6)$$
$$D (3, 10$$

we went over (what im sure everyone here, including my self, already know. its good to think about and keep fresh in my mind) how two points define a line and for a function such as a half of a parabola, (at least) three points are required.

for our first model, we chose points $$ABC$$.

i have three equations:

$$c =1$$
$$1a + 1b + c = 2$$
$$4a + 2b + c = 6$$

with that info, i dont know how i got from there to here(our first model):

$$\hat{y}_{1} = \frac{3}{2}x^2 - \frac{1}{2}x + 1$$

plug in all the numbers to get the residual (the problem i had last time, but now i can do it in my sleep).

heres what im about to do, hopefully i can elaborate enough.

im going to plug in the numbers from all four points, not just the three that i am using for this model.
point $$A$$

$$\hat{y}_{1} = \frac{3}{2}(0)^2 - \frac{1}{2}(0) + 1 = 1$$

$$(y - \hat{y}_{1})^2 = 0$$ (the residual)

$$(0 - 0)^2 = 0$$

point $$B$$

$$\hat{y}_{1} = \frac{3}{2}(1)^2 - \frac{1}{2}(1) + 1 = 2$$

$$(y - \hat{y}_{1})^2 = 0$$ (the residual)

$$(2 - 2)^2 = 0$$

im posting this now and will continue immediately. this is just to be sure nothing happens and i have to redo the whole thing.

9. Sep 17, 2008

### offtheleft

point $$C$$

$$\hat{y}_{1} = \frac{3}{2}(2)^2 - \frac{1}{2}(2) + 1 = 6$$

$$(y - \hat{y}_{1})^2 = 0$$

$$(0 - 0)^2 = 0$$

heres where it finally changes!

$$\hat{y}_{1} = \frac{3}{2}(3)^2 - \frac{1}{2}(3) + 1 = 13$$

$$(y - \hat{y}_{1})^2 = 9$$

$$(10 - 13)^2 = 0$$

so for model $$ABC$$

$$\hat{y}_{1} = \frac{3}{2}x^2 - \frac{1}{2}x + 1$$

$$\sum(y - \hat{y}_{1})^2 = 9$$

the sum of our residual is nine

10. Sep 17, 2008

### offtheleft

the second model is with points $$ABD$$

heres how it looks: $$\hat{y}_{2} = x^2 + 1$$

(this was done is class as well)

plug in the numbers

everything is 0 except for point $$C$$ which will look like this:

heres how it looks: $$\hat{y}_{2} = 2^2 + 1 = 5$$

$$(6 - 5)^2 = 1$$

so,

$$\sum(y_{1} - \hat{y}_{2})^2 = 1$$

11. Sep 17, 2008

### offtheleft

our third model, well, fourth, the third one is the one i am trying to figure out:

ill keep it short, i demonstrated how everything was done in the first long post. :)

this is done with points $$BCD$$

the model looks like this: $$\hat{y}_{4} = 4x^2 - 2$$

all but the first point were 0's. ill plug in the number

$$\hat{y}_{4} = 4(o)^2 - 2 = -2$$

$$(1 - 2)^2 = 1$$

so for: $$\hat{y}_{4} = 4x^2 - 2; \sum(y - \hat{y}_{4})^2 = 1$$

12. Sep 17, 2008

### HallsofIvy

Staff Emeritus
In your original post you said the points were A(0,1), B(2, 6), C(3, 10)
Writing y= ax2+ bx+ c, those give a02[/sub]+ b0+ c= 1 or c= 1, 4a+ 2b+ c= 6, and 9a+ 3b+ c= 10. Since c= 1, the last two are 4a+ 2b+ 1= c or 4a+ 2b= 5 and 9a+ 3b+ 1= 10 or 9a+ 3b= 9. If you multiply the first of those equations by 3 you get 12a+ 6b= 15 and if you multiply the second by 2 you get 18a+ 6b= 18. Subtracting the first from the second, 6a= 3 so a= 1/2. Then 4a+ 2b= 5 becomes 2+ 2b= 5 so 2b= 3 and b= 3/2, not b= 3.

The equation if y= (1/2)x2+ (3/2)x+ 1.

But later you give A(0, 1), B(1, 2), C(2, 6), D(3, 10). The equation y= (1/2)x2+ (3/2)x+ 1 gives y(0)= (1/2)(0)+ (3/2)(0)+ 1= 1, y(1)= (1/2)(1)+ (3/2)(1)+ 1= 3, y(2)= (1/2)4+ (3/2)2+ 1= 6, and y(3)= (1/2)9+ (3/2)3+ 1= 10. That is correct for all except B(1,2) which was NOT in the original problem. You can, in general, fit a parabola through 3 given points but not 4. There is no quadratic function whose graph will pass through all 4 points.

If you have 4 points, you should try a cubic formula: y= ax3+ bx2+ cx+ d so that the 4 points will give 4 equations for the 4 coefficients.