# Algebra 2 - residual^2 or something ?

1. Sep 11, 2008

### offtheleft

i have a function $$y=f(x)$$ with three points;
$$A (1, 40)$$
$$B (3, 80)$$
$$C (5, 90)$$

the $$x$$ model represents days. and the $$y$$ model represents failed parts on an assembly line.

the question is: how many failed parts will have been identified $$1$$ week from now?

I have no problem with doing the work, getting the slopes, figuring out all the algebraic gymnastics and getting the equations.

heres what i have:
$$A\rightarrow B, y=20x-20$$

$$A\rightarrow C, y=\frac{25}{2}x+\frac{55}{2}$$

$$B\rightarrow C, y=\frac{77}{5}x+26$$

I actually think my equation from $$B\rightarrow C$$ could be incorrect.

now, the part that is getting me is im trying to get the residual? i need to get some medication because my ADHD is seriously out of control. its getting worse as i get older.

i need to construct some table... this is what i have for the first one:

real obverved measures residual
$$x,... y,... y_{1}=20x-20,... (y-y_{1})^2$$
$$1,... 40, ...40, ..................0$$
$$3,... 80, ...80, ..................0$$
$$5,... 90, ...120, ................900$$

real obverved measures residual
$$x,... y,... y_{2}=\frac{25}{2}x+\frac{55}{2},... (y-y_{1})^2$$
$$1,$$
$$2,$$
$$3,$$

not quite sure how to do all of this. :(

2. Sep 11, 2008

### Diffy

You might want to check your equation for A to C as well as from B to C...

I am assuming of course you want to write an equation that goes through both those points in y = mx +b form.

Other than that I can't help you because I don't really understand the question. I know you want to predict what y is when x is 7, but there are many methods of doing this and I am not sure which one you are supposed to be doing.

3. Sep 12, 2008

### offtheleft

Here are my Equations:

$$A\rightarrow B, y=20x-20$$

$$A\rightarrow C, y=\frac{25}{2}x+\frac{55}{2}$$

$$B\rightarrow C, y=5x+65$$

the first two are correct because we did them in class. the third i just corrected, there was a minor mistake in my arithmetic that knocked everything off the rocker.

now, i have to do something with the $$y$$ points $$A(1,40) B(3,80) C(5,90)$$

what i have to do for this table

I have each point

Last edited: Sep 12, 2008
4. Sep 12, 2008

### Diffy

I think A -> B is incorrect!

y = 20x - 20

If this is supposed to go through the point A(1,40) then when I put 1 in for x I should get y = 40.

y = 20(1) - 20 = 20 - 20 = 0

So y = 20x - 20 does not go through the point (1,40) it goes through (1, 0)

It's probably just a typo from you notes from your class, but the correct equation should be y = 20x + 20

5. Sep 12, 2008

### offtheleft

YOURE RIGHT!! i actually got the answer early in the class which was $$y=20x+20$$ i dont know why its copied incorrectly thereafter

"$$y = 20(1) - 20 = 20 - 20 = 0$$"

how would i construct a table just as that for the other two equations... but, there was three parts to each. one for the Y value for each point.

6. Sep 12, 2008

### Diffy

I'm sorry can you elaborate on the table you are supposed to create, I don't really understand that part.

7. Sep 12, 2008

### offtheleft

real obverved measures residual
$$x,... y,... y_{1}=20x+20,... (y-y_{1})^2$$
$$1,... 40, ...40, ..................0$$
$$3,... 80, ...80, ..................0$$
$$5,... 90, ...120, ................900$$

i think i finally understand it. because i had the equation copied wrong it all came out wrong, the numbers didnt match up and it threw me off.

$$y_{1}=20(1)+20=40$$

and the residual thing. i take the real $$y$$ which is $$40$$ and subtract it from the new $$y$$ which is $$40$$: $$y_{1}=20(1)+20=40$$

$$(40-40)^2=0$$

let me try and set this up

my line: $$y=20(1)+20=40, (40-40)^2=0$$

$$y=20(3)+20=80, (80-80)^2=0$$

$$y=20(5)+20=120, (120-90)^2=900$$

8. Sep 12, 2008

### offtheleft

$$B\rightarrow C, y=5x+65$$

$$y=5(1)+65=70, (70-40)^2=900$$

$$y=5(3)+65=80, (80-80)^2=0$$

$$y=5(5)+65=90, (90-90)^2=0$$

now, i have to figure out where i went wrong with the second line.. from A\rightarrow C.

i dont have my notebook with me at the moment though, ugh.

9. Sep 13, 2008

### offtheleft

just worked it out and the second equation is correct

$$A\rightarrow C, y=\frac{25}{2}x+\frac{55}{2}$$

$$y=\frac{25}{2}(1)+\frac{55}{2}=40, (40-40)^2=0$$

$$y=\frac{25}{2}(3)+\frac{55}{2}=65, (80-65)^2=225$$

$$y=\frac{25}{2}(5)+\frac{55}{2}=90, (90-90)^2=0$$

thanks for all the help! i think i may be able to hold it down from here, until class on tuesday :p