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MAT1201 Discussion Assignment Unit 2
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Discussion assignment meant for peer review and discussion on graphs, linear and quadratic functions of College Algebra course at the University of the People.
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MATH 1201 Unit 1, 4, 7 Notes
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MATH 1201 College Algebra.Discussion Forum Unit 2.
This stock situation for the three months ca be analyzed by representing the three months in
function and graphical form. 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 the price change from $10 to $15 is not an entirely straight line. These will be three
different lines which can be joined to form a graph of unique pattern like that of an
electrocardiograph.
For a first naïve view of the situation, a simplified solution would be to get an average for each
of the three-time intervals and plot only three representative points to easily make sense of an
electrocardiograph scenario.
The general linear function or equation of a straight line is the form: y = 𝑓(𝑥) = 𝑚𝑥 + c, where m
is the slope or gradient of the straight line which can be calculated from any two points using the
formula: 𝑚=Δ𝑦/Δ𝑥. On the other hand, c is the y-intercept of this function.
Let x represent the number of months passed in this stock exchange case, the two points of the
first linear function are: (0, 10) and (1, 12).
Using these points, we can calculate the slope as follows:
𝑥1=0, 𝑥2=1 𝑦1=10, 𝑦2=12.
𝑚=𝑦2 – y1/ 𝑥2−𝑥1
12−10/1−0
m=2
Using one of the points, for instance (1,12) to determine the value of the y-intercept, c:
y = mx +c
12=2∗1 +c
c=10.
The linear function will therefore be: 𝑓(𝑥)=2𝑥+10
The graph from will be as shown below:
1|Brian Wanyonyi-University of the People.
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