Please I really need answers to these questions.

Question

1. Lines can be used to approximate a wide variety of functions; often a function can be described using many lines.

If a stock price goes from $10 to $12 from January 1st to January 31, from $12 to $9 from February 1st to February 28th, and from $9 to $15 from March 1st to March 31th is the price change from $10 to $15 a straight line? 

It is clear that in each of the three time intervals mentioned there was a complex daily variation of prices as in an electrocardiogram.

But what would be a simplified solution for a first naive view of the situation?

Would a simple function hold up?

What is the simplest function to represent this situation?

Does your naïve initial and simplified model allow you to predict the behavior of the stock in the next month?

How can I use three “pieces” of lines to describe the price movements from the beginning of January to the end of March? Show the graph for the price movement.

Answer 1

All of these questions can be answered using the formula of a linear equation:

y = mx + b, where m = (y₂ - y₁) / (x₂ - x₁)

So let's begin.

First, we make a table of the dates versus stock prices, and then identify our 3 pieces of line:

___x___ ___y___

Jan 1 $10

} Line A = y_A

Jan 31 $12

______________________________

Feb 1 $12

} Line B = y_B

Feb 28 $ 9

________________________________

Mar 1 $ 9

} Line C = y_C

Mar 31 $15

________________________________

At this point, it's important to rename the x-values from the date to the number of days from Jan 1 in order to make our calculations. Let's set Jan 1 to day 0:

___x___ ___y___

0 10

} Line A = y_A

30 12

______________________________

31 12

} Line B = y_B

58 9

______________________________

59 9

} Line C = y_C

90 15

______________________________

Now let's calculate our 3 line pieces:

Line A : y = mx + b using the points (0,10) and (30,12)

m_A = (12 - 10) / (30 - 0) = 1/15

=> y_A = (1/15) x + b

We find b by substituting one of the 2 points into the equation y_A. Let's use (0,10):

10 = (1/15)(0) + b

10 = b

So, y_A = (1/15) x + 10

Line B : y = mx + b using the points (31,12) and (58,9)

m_B = (9 - 12) / (58 - 31) = - 3/27 = - 1/9

=> y_B = (-1/9) x + b

We find b by substituting one of the 2 points into the equation y_B. Let's use (31,12):

12 = (-1/9)(31) + b

12 = (-31/9) + b

12 + (31/9) = (-31/9) + (31/9) + b

(139/9) = b

So, y_B = (-1/9) x + (139/9)

Line C : y = mx + b using the points (59,9) and (90,15)

m_C = (15 - 9) / (90 - 59) = 6/31

=> y_C = (6/31) x + b

We find b by substituting one of the 2 points into the equation y_B. Let's use (90,15):

15 = (6/31)(90) + b

15 - (540/31) = (540/31) - (540/31) + b

- 75/31 = b

So, y_C = (6/31) x - (75/31)

Now you graph the 3 line pieces, which I don't know how to do here. Stock prices go on the y-axis and the number of days from Jan 1 go on the x-axis. Just plot the 6 points and connect them, and you'll find 3 disconnected lines in the shape of a zig zag (or EKG).

Each disconnected line represents the price change in each month between Jan 1 and Mar 31.

Line A := the price change in stock during Jan

Line B := the price change in stock during Feb

Line C := the price change in stock during Mar

The overall line, we'll call it Y_D, begins at the first point (0,10) and ends at the last point (90, 15). This line represents the change in stock price over all 3 months as one continuous line.

To find Line D, we do the same work as we did above for the 3 line pieces:

m_D = (15 - 10) / (90 - 0) = 5/90 = 1/18

=> y_D = (1/18) x + b

Find b by substituting the point (0,10) into the equation right above:

10 = (1/18)(0) + b

10 = b

So, y_D = (1/18) x + 10

Now let's answer your questions:

(1) ... is the price change from $10 to $15 a straight line?

No, not when we go from month to month. Then it's a zigzag line.

(2) But what would be a simplified solution for a first naive view of the situation?

A linear solution? I'm not sure what your teacher means.

(3) Would a simple function hold up?

Yes, a linear function? Again, not sure what the teacher means.

(4) What is the simplest function to represent this situation?

That would be Line D.

(5) Does your naïve initial and simplified model allow you to predict the behavior of the stock in the next month?

The simplified model, Line D, can help you to predict future changes in stock price, but remember, it would an estimate, not fact.

Answer 2

This question is worded a little confusingly, so I'll try to answer what I think they're asking.

Looking simply from 10 to 15 would be a straight line, if you zoomed out to only have a point every 3 months. That line could be represented by a simple function, a line under y = mx + b. (m is slope, b is intercept). The origin could be January 1st, if there was one graph per year, so you can use that assumption to find b. Just make sure to write where you're putting the origin, and your teacher should accept it. m is found using rise over run. Define your units for both axes! Remember to use the zoomed out scale for this line.

For the prediction ability of your line, it's oversimplified to neglect a number of data points you were given. That pattern (which i guess-timated on a piece of paper by roughly putting numbers where they would be on a graph, to get a quick image) has rises and dips that aren't shown by your model line. If your prediction was supposed to predict those months, it will not allow you to accurately predict the stock market. If it was used to predict the stock market in 3 months however, it might be more accurate because that is the scale it was modeled on.

The three lines you could use to represent the full data you were given is obvious once you put dots on the graph, as there are 4 points which can be connected by 3 lines.

Hope this helped soon enough!