# Thread: Fitting data to a model

1. ## Fitting data to a model

given the following set of data where x represents the number of items sold and y(x) represents the profit when x items are sold
X # of items Y(profit)
0 -1350
1 -800
2 -350
3 0
4 250
5 400
6 450
7 400
for letter A i already got the answer the quadratic formula which is y=-50x+600x+350
B. create a specific algebraic model for profit as a function of the number of items sold.show your work and clearly define variables
C. the break-even point is when profit is zero. what is/are the break even points?explain your answer

2. Hyvää on! Mistä häiriöstä?

$y = -50x^{2} + 600x - 1350$

Olkaa varovaisia!

B: Isn't that what you did in A? Don't forget $x \; \epsilon \; [0,7]$

C: Solve! $0 = -50x^{2} + 600x - 1350$

3. Originally Posted by kenkarylle5
given the following set of data where x represents the number of items sold and y(x) represents the profit when x items are sold
X # of items Y(profit)
0 -1350
1 -800
2 -350
3 0
4 250
5 400
6 450
7 400
for letter A i already got the answer the quadratic formula which is y=-50x+600x+350
B. create a specific algebraic model for profit as a function of the number of items sold.show your work and clearly define variables
C. the break-even point is when profit is zero. what is/are the break even points?explain your answer
Aren't A and B one and the same?

Did you mean $-50x^2+600x-1350 = 0$? Otherwise f(0) isn't correct nor is it a quadratic as you've written it.

Cancel out a factor of 50 to make this simpler to solve and move everything to the other side (you don't have to do either of these but I prefer having positive a)

$50x^2-600x+1350 = 0 \rightarrow x^2 - 12x + 27 = 0$

Solve using your favourite method (I get $x=3$ and $x=9$ as solutions)

4. i got B, but im not sure with C, am i going to base my answer on the chart for letter C...

5. If you have a perfect model, which you do, the chart or the equation will do just fine. Since this is a quadratic model, there will be another zero for somewhere x > 7. You will have to decide if this is a valid result. Is the model defined for x > 7? Perhaps only 7 can be produced?

6. so my break-even point when profit is zero is 7 then huh.?
mann im sorry, im really stuck, but thanks for your helps guys appreciate it..

7. Originally Posted by kenkarylle5
so my break-even point when profit is zero is 7 then huh.?
mann im sorry, im really stuck, but thanks for your helps guys appreciate it..
No, 7 may be your upper limit on the domain rather than being a 0 point. It means that you can disregard any solutions above 7.

This won't apply if your domain (x value) can go higher than 7

8. ah okay... tnx, now i will work on 2 more word problems.tks tsk tsk