The document provides step-by-step solutions to 4 math problems involving rational functions, polynomials, and perimeter/area calculations. The first problem asks whether a farmer has enough fence to enclose a field given the perimeter needed and available fence. The next problems involve simplifying a rational expression, graphing a rational function, and finding the domain and range of a polynomial function. The reflection at the end discusses choosing these problems because they cover a variety of math concepts and allowed the author to feel confident explaining the material to others.

1. DEV PROJECT
ALEX WHITMER

2. Farmer Ted Problem
 A farmer had 1000 feet of fence available to
enclose a filed with, the area he needs to
enclose in 100,000 square feet. The land
borders a river so he only has the make 3
sides. Does this farmer have enough fence to
do so?

3.  First we have to use the equation for
perimeter, P=2x+2y.
 Since there are only 3 sides, we take away
one of the y’s to make the equation P=2x+y.
 We know that the perimeter will be 1000 feet
once it is built so we enter that in for P. Our
equation now looks like this, 1000=2x+y.
 We have to get Y by itself and in order to do
that we have to subtract the 2x from the right
and move it over to the left side. Our
equation is now 1000-2x=y.
 We are done solving this equation but we will
have to use it later.

4.  We now move on to the area equation. A=XxY
 We know what the Y is from the previous slide
so we plug that in and our equation becomes
A=X(1000-2x). We are going to put this aside
for a minute.
 Now we have to figure out how long each
side of the fence will be in order to get the
max amount of land enclosed. We take
1000/4 and that gives us 250, the length of
each side. This number is also our X for
finding area.

5.  Our area from right now looks like this,
A=250(1000-2(250))
 When you plug this into your calculator you
get 125,000. This number is the number of
feet squared you could enclose with 1000 feet
of fence.
 So to answer the question, yes, you would be
able to enclose 100,000 square feet with 1000
feet of fence.

6. Problem #2
Rational over Rational
(x+7) + (x+3)
(x-4) (x+7)
(x) + (3x)
(x+7) (x)

7.  In this rational over a rational the goal is to try
and simplify it as best as possible. We want to
keep this in factored form as well.
 The first thing we do is break it up into 2
problems, a problem on top and a problem on
bottom.
 Once we mentally do this we need to set both
sides equal to each other. We do this by
multiplying the opposite side by each
denominator. So on the “top” problem we would
multiply the left side by (x+7) since the
denominator on the right is (x+7) and we would
multiply the right side by (x-4) since the
denominator on the left is (x-4)

8.  After you do that to both sides you distribute.
So on the top left you would multiply
(x+7)and (x+7) giving you x^2+7x+7x+49.
 Then you would combine like terms, so in this
case combine the 7x’s to make the equation
x^2+14x+49 but then we switch it to factored
form to make it (x+7) and (x+7).
 After we do this process to each side in each
problem our equation looks like this,
 (x+7)(x+7)+(x+3)(x-4)
 (x+7)(x-4)
 3x(x+7)
 x(x+7)

9.  To continue on further with this question we
have to flip the bottom problem and multiply
it by the top one making our equation this,
 (x+7)(x+7)+(x+3)(x-4) x x(x+7)
 (x+7)(x-4) 3x(x+7)
 From here you can cross out like terms.
 On the left side you would cross out 1 (x+7)
from the top and one from the bottom and
the (x-4) and on the right you would cross out
the x and the (x+7).

10.  The simplified equation would then be,
 (x+7)+(x+3)
 3

11. Problem #3
Graphing Rational Functions
 2x(x+5)
 (x-7)(x+5)

12.  First like we did before, we simplify the
equation. We can cross out the (x+5) since
there is one in both the numerator and
denominator.
 When we cross it out though a hole will show
up on the graph and we have to remember to
show that.
 So since we crossed that out our equation is
now, 2x
 x-7

13.  To graph a rational function we need to figure
out a couple things. Vertical asymptote,
Horizontal asymptote, any x-intercepts and
any y-intercepts.
 To find the Vertical asymptote you see where
the denominator equals 0. So in this equation
the V.A. is 7.
 To find the Horizontal asymptote you look
and see how many X’s are in the numerator
and denominator and divide them. So in our
case the number of X’s is the same on the top
and bottom so it would just be 1.

14.  To find the x-intercepts you set the
numerator equal to zero and find x. In our
case there would be no x-intercept.
 To find the y-intercept you set all the X’s
equal to zero and solve. So for our problem
there would be a y-intercept at -7
 So to recap what we have done,
 Hole: -5
 V.A.: 7 H.A.: 2
 Y-int: -7 X-int: none

15.  Now we have to determine what direction
our graph will be going. So you create a graph
that looks like the one below.
 In the green area you put your final equation
and in the orange spot you would put any V.A.
that you have.

16.  You may have noticed the positive and
negative signs next to the 7’s. in order to
figure out what way the graphs go you must
pick a number just to the left of 7 and on just
to the right and enter them in for X.
 Then you just solve the simple problem and if
it ends up being positive then you put and
addition sign and if it is negative then you put
a subtraction sign.

17.  Finally it is time to graph. You now know that
the first part of the graph will be going
towards negative infinity as it approaches Y
and the second part will be going towards
positive infinity as it approaches y so graph it.
 This is what your graph should look like in the
end.

18. Problem #4
Finding Domain and Range of
a Polynomial
 F(x)=x^2+10x+25(x+14)(x-5)

19.  First thing you want to do in this situation is
get everything into factored form. So you
would change the x^2+10x+25 to (x+5) (x+5).
 This then makes the entire problem
(x+5)(x+5)(x+14)(x-5).
 When you graph a polynomial you look and
see if any of the variables are the same. If
there is only one variable then the graph will
go right through that point.
 If the amount of that variable is a factor of 2
then the graph will bounce at that point.
 And if the amount of that variable is a factor
of 3 then it will curve through that point.

20.  In our problem we have (x+5)^2 (x+14) and
(x-5) so our graph will go through the points 5
and -14 but bounce at-5.
 We also have to find out the degree of the
equation by looking at the number of X’s if it
is even then the graph will start and end
going the same direction, if the amount is
odd then they will start and end going in
different directions.
 Our graph has a positive degree of 4 so it will
open up.

21.  Once we graph it, it should look like this.

22. Reflection
The concepts that I chose were finding domain and range of a
polynomial, Graphing rational functions, rational over
rational and a farmer ted problem. I chose these problems
because I felt comfortable with them and believed that I
would be able to explain them in a confident manor. These
problems provide an overview of my best mathematical
capabilities because they cover a wide variety of what we
have learned in class and show that I am confident enough
to teach them. These problems cover polynomials,
rationals, and quadratics. I believe that this was a beneficial
project for me because it secured my knowledge base of
the material in order for me to be able to become the
teacher. I believe that if you learn something, then test on
it, then teach it; it is much more likely to stay with you
longer.