Algebra Assignment: Analyzing Lines and Quadratic Functions Solutions

This assignment solution analyzes lines and quadratic functions, addressing concepts such as gradients, y-intercepts, and parallel and perpendicular lines. The solution begins by examining stock price movements and determining whether they can be represented by a straight line. It then explores how to use three "pieces" of lines to describe the price movements and includes the equations for each segment, with a suggestion to use Desmos for graphing. The assignment proceeds to solve problems involving parallel, perpendicular, and neither lines, providing rigorous algebraic solutions. It also addresses a quadratic equation and its solution, applying it to a real-world scenario involving the height of a building. Finally, the solution reflects on the concepts needed to understand lines and quadratic functions, the simplest forms of these functions, and their applications in daily life. It also mentions using Excel as a tool for plotting graphs and calculating gradients.

Assignment
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?
The change is not a straight line since the number of days and distribution of prices in the days
is not equal. For instance, the gradient (the distribution of changes in price against price) for
January is
Gradient= 12−10
31
= 0.065
The gradient for feb :-
Gradient = 9−12
28
= 0.11
Gradient for march:-
Gradient = 15−9
31
= 0.19

The gradients are not equal and therefore there is no way the line could straight. The line is not
straight.
The simplified answer is to get the difference between the first value and the final value without
considering the changes in between the months. A simple function cannot hold up in this
situation, there are so many variables is play here especially now that none of the months have
equal days.
The simplest function to represent this would be
Y=mx +c
Where c is the intercept on the y axis.
The initial naïve prediction does not hold up.
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.
The gradient for the first part
Gradient= 12−10
31
= 0.065
The equation now is

0.065= y−10
x−1
Y=0.065x +(10-0.065)
The gradient for feb :-
Gradient = 9−12
28
= 0.11
0.11= y−1 2
x−1
Y=0.11x +(12 – 0.11)
Gradient for march:-
Gradient = 15−9
31
= 0.19
0.19= y−1 5
x−1

Y=0.19x +(15-0.19)
Your Discussion should be a minimum of 250 words in length and not more than 750 words.
For this written assignment, answer the following questions showing all of your work.
1. Determine whether the lines given by the equations below are parallel, perpendicular, or
neither. Also, find a rigorous algebraic solution for each problem.

a. b. c.
[Suggestion: go to www.desmos.com/calculator, write the two equations and try to conclude the
answer.]
for lines to be parallel, the gradients must be equal. In order to get the gradient, the equations
will be rewritten in form of y=mx+c
where m is the gradient of the equation and c is the y intercept ( but does not matter in this case
since we will be using gradients only)
a. 3y=12-4x
-6y=8x + 1
Rewriting both equations
Y= 3 - 4
3 y
Y=-4/3y -1/6
The gradients are equal and therefore the lines are parallel.
b. 3y=12-x
-y=8x-1
Rewriting the above
Y=- 1/3 x + 12

Y=-8x -1
The gradients ,-1/3 and -8, are not equal and therefore the lines should not be parallel
c.
7y=4x -10
4y=1-7x
The gradients are 4/7 and -7/4. They are not equal and therefore the lines are not parallel
2.
and observe the answers.]
this is a quadratic equation whose solution is given by
-b ± √b2−4 ac
2 a
B= 24
a = -4.9
c= 8
t = -24 ± √ 242− ( 4 x 8 x−4.9 )
2 x−4.9
= 5.211252949500183
Or -0.3132937658267152

Since it is height, the correct answer is 5.211252949500183
The height of the building now is
H(t) = -4.9(5.21)^2 + 24(5.21) + 8
= 0.0334 m
3.
The equations are
75 = 20x + c
76= 17x + c
Using elimination method
-1=-3x
X=1/3
75 = 20/3 + c
C= 68
Reflect on the concept of lines and quadratic functions. What concepts (only the names) did you
need to accommodate the concept of lines and quadratic functions in your mind? What are the

simplest line and quadratic function you can imagine? In your day to day, is there any occurring
fact that can be interpreted as lines and quadratic functions? What strategy are you using to get
the graph of lines and quadratic functions?
The concepts needed when learning about lines is the gradient and y intercept. The gradient is
the ratio of change in y to change in x. for the quadratic equation the coefficients are important
when finding the solution. The simplest line I can imagine is
y= a
where a is a constant
and the simplest quadratic equation imaginable is
y=a x2 +bx+ c
where a, b and c are all constants.
In day to day, the equations can be used to determine the distance covered after breaking while
testing vehicles of the range of a drone.
When plotting the graphs, the excel is one tool that very useful. Using excel, calculating the
gradients and plotting the values is simple and very fast.