Rayen Antillanca.
Final Project
Part
1
Consider rational equations of the form . Explore the mathematical ideas for various
values of the parameter a, b, c, and d.
Investigation
1
Consider the equation where a=c=d=1
and b>0 varies 



Consider where a=1, c=d=1 and b>0
varies 



The hyperbolas cut the yaxis in the
value of b and the vertical asymptote is given by d, because the denominator
is x+b. In this figure, the equation of the vertical asymptote is x=1. 
Consider the equation where a=c=d=1
and b <0 varies 



As b is a negative number, the hyperbolas
are in second and fourth quadrant. The value of d gives the vertival
asymptote. In these equations the vertical asymptote is given by the equation
x= 1. 
Consider the equation where a=1,
c=d=1 and b <0 varies 



Consider a=2, b=4, c=1 and d varies 



As d varies, the value of the vertical asymptote changes 
Summary 
,
then if I take the numerator as a independent equation I get that ,
so ,
in this point is where the hyperbola cuts the xaxe. Now,
if I take the denominator as another independent equation, ,
then, will be the vertical asymptote of this
rational fraction. 
Investigation
2
Now
I am going to analyze what happen when where is a quadratic equation and is a linear equation.
Graph of 

Let and As a rational equation as I can rewrite, after doing the division of
polynomials, as The graph of this rational equation
is the brown curve, which is a hyperbola. The blue line is the equation of which correspond to one asymptote of the
hyperbola. And the other asymptote is which is given by the denominator of the
rational equation, namely .


Another graph for 

Let and .
The rational equation, ,
after doing the division of polynomials I get: The green curve corresponds to the
rational equation which is a hyperbola. The red line is one asymptote of the
hyperbola which is given by the equation ,
and the other asymptote is ,
which is given by the denominator of the rational equation, namely 



Part
2
Square
Inscribed in a Semicircle: Find a ratio
Given a semicircle
with an inscribed square of side . Let be the length on the diameter on each side
of the square. Find 

Consider the triangle AND right in D because it is incribed
in a semicircle. The triangle MAD right in A since AD is high. We can easily see
that triangle MAD is similar to triangle AND, so that we may write the following proportion 
(1) Let , and replace
in (1) . Multiplying by x rewriting as a quadratic equation Solving this quadratic equation The positive value
of x solves this equation, because x is distance, which always must be
positive, then So I find the answer, which is , the Golden Ratio 
Construct
the square inscribed in a semicircle
Construction 
First,
let ABCD any square 

Draw
a segment from E, the midpoint of AB, to C 

Draw
a semicircle with radius EC 

Then,
the square is inside the semicircle. 
Part
3
Construct
a triangle and its medians.
Let ABC be any triangle, where AD, BE and CF are medians 

.
To construct a new triangle with the
medians of triangle ABC 

Construct a parallel line to a median
CF that passes through point E. 
Construct another parallel line to
the median AD that passes through point B. Let G be the intersect point between
these parallel lines. 


The triangle BEG is the triangle
constructed by the three medians of the triangle ABC 

Summary 
The
triangles ABC and BEG are not congruent, their sides are not equal according
to the criteria SSS. What happens with their areas? 
Notice this time we have two pairs of
parallel lines, GF is parallel to EB and GC is parallel to AD. This created
two parallelograms, FBEG and AGCD 

We know that the median AD of
triangle ABC bisects the area of ABC in half. This means that Area triangle ADB is (area triangle ABC). Side AB is the diagonal
of parallelogram AGBD, this means that triangle ADB is half the area of the
parallelogram. Thus Area triangle ADB = Area parallelogram AGBD 
We know triangle AIE is congruent to
triangle FIG by ADA since EI = IG, by the midpoint theorem, angle AIE = angle
FIG by opposite angles and angle AEI = angle FGI by alternate interior angles
(AC is parallel to GF). 


This gives parallelogram FGEC to have
the same area as triangle ACF. Triangle ACF is half the area of triangle ABC
so the smaller parallelogram is half the area of the larger parallelogram and
half the area of triangle ABC. 
Area of triangle AIE = Area triangle
GIF = Area triangle AIG = Area triangle GFH = Area triangle BHF. Triangle GIB
is half the area of the triangle made up of the medians and it has the same
area as triangle GIF + triangle GFH + triangle FHC. 


Summary 
Triangle AGB has the same area as
triangle AGI + triangle GIB. The area of triangle AGB = area of triangle ADB
because they area each half of the area of Parallelogram AGBD. Since triangle
ADB is half the area of the original triangle and triangle GIB is half the
area of the triangle of the medians, we can use this to show that the
triangle made up of the medians has an area that of the original triangle. Triangle GIB
is of triangle AGB since it is made up of 3 of
the 4 triangles with the same area. Since triangle AGB = triangle ADB then
triangle GIB is the area of triangle ADB. Since triangle GIB
is half the area of triangle GEB and triangle ADB is half the area of
triangle ABC, then triangle GEB is the area of triangle ABC. 
Note: All graphs of this webpage were made with
Graphing Calculator 4.0 and the geometric images were made with The
Geometer’s Sketchpad 