Week 8: Summary of Chapter 3

This week, we had one more lesson to look at till the end of chapter 3, Polynomial and Rational Functions, however we decided to not look at it until a later time due to it being connected to Calculus BC. Since we had a three-day weekend just recently, we only had three days of mathland. We reviewed this week mainly as we had a test on Friday on chapter 3.

To summarize Chapter 3, We began with Polynomial Functions. 

This lesson was simply review of Algebra 2 of End Behavior and Multiplicity. We also looked over even, odd and neither fuctions. 

The most important was multiplicity as it shows how many zeros are there of that one number. Usually this occurs if there is a quadratic function which creates the zero. 

We then looked at Division of Polynomial Functions. 

In this lesson, it is quite obvious we looked at how to divide polynomial functions. To divide you would have the the f(x) and divide it by d(x). To divide it is exactly just as you would divide regular numbers in long division but using x’s and exponents. Below is an example:

Just as shown above the equation’s last exponent on the right side ix x^3. To solve an equation, you would need to put 0’s to fill in the left over pieces. What is needed is shown in red above. You would divide with the number of exponents and number needed to divide. You times the number and put it under the exponent number and then make the numbers negative and you continue till you cannot divide anymore. Another type of division is Synthetic, which is shown below.

You can only divide if the polynomial x has one exponent. ex: x+4

You then take the numbers and plug in any 0’s needed and you divide down and then times the number at the bottom and put that number in the next column and repeat the same process to the end. If the last number is 0 then it is a factor and it has no remainder.

Sometimes you are looking for a remainder and you put the number from the divisor and plug it in for zero. That is the remainder thereom to find the remainder. 

We then looked at Zeros and Factors of Polynomial Functions.

In this one you are looking for the factors and the zeros of polynomial functions. You would divide to find all the zeros. The first polynomial’s exponent tells how many zeros there are. You then find all zeros and tell if they have multiplicity and also the factors. Sometimes the zeros would be imaginary and you would use the quadratic formula to find them. 

After this we looked at Real Zeros of Polynomial Functions.

What would happen if there are no zeros provided to find the rest of the zeros? You use this method which is P/S or Factors of Constant over factors of leading coefficient.

1. You find all the possible zeros from the P/S

2. Test the zeros using remainder thereom

3. Hint- not for test- use graph to see the possible zeros

We then looked at Approximating Zeros.

1. Divide the interval [a,b] in half by finding its midpoint, 

m=a+b/2

2. compute f(m)

3. if f(a) and f(m) have opposite signs, then f has a zero in teh interval [a,m]

if f(m) and f(b) have opposite signs then f has a zero in the interval [m,b] 

if f(m)= o then m is a zero of f. 

You also use the error equation to find out if you are close enough to a zero

error 1/2 (b-a)

the error would need to be  .0005 to be accurate enough to be a zero. 

After this lesson we looked at Rational Functions. 

This lesson was also review from Algebra 2 which consists of vertical asymptotes, horizontal asymptotes and holes. The new thing looked at is slant asymtotes.

To find a vertical asymptote:

set the denominator =0 and find x

To find the horizontal asymptote: 

fx)=anx^n/bmx^m

n=m   y=an/bm

n<m   y=0

n>m   DNE

n= degree of numerator

m= degree of denomator

an= LC of numerator

bm= lc of denominator

also if the HA does not exist (DNE) then there is a slant asymptote. 

To find Slant Asymptote:

Divide the denominator by the numerator and ignore the remainder.

y= answer

Holes:

 Factor the equation and cancel anything possible. What cancels out is a hole, to find the y-int of the hole you plug back in the x to find (x, y)

You would also need to find x-intercepts and y-intercepts:

 Just put 0 for the opposite number to find the answer.

To look more in-depth look at the past posts to understand it better. That is a summary of chapter 3.

See You Next Week!

Week 7

This week we looked again in Chapter 3, Polynomial and Rational Functions.

On Day 1, instead of starting another lesson, we looked again on what we looked at last week which included Polynomial Functions, Division of Polynomial Functions and Synthetic Division, How to Find Zeros of Polynomial Functions and Finding the Real Zeros of Polynomial Functions.

We reviewed on problems to see how they work and if we are doing the problems correctly.

On Day 2, We looked at Approximating Zeros. Some Zeros of Graphs are not Solid Numbers like 1, 2 or 3 but are instead decimals or fractions like -0.481923 or 3/47. In this lesson, we were trying to figure out how to find the zeros to the most approximate number as possible.

The steps to find them is:

1) Divide the interval [a, b] in half by finding its midpoint, 

m=a+b

     2

2) Compute f(m)

3) If f(a) & f(m) have opposite signs, then f has a zero in the interval [a, m]

If f(m) and f(b) have opposite signs then f has a zero in the interval [m,b] 

If f(m)=0 then m is a zero of f and the problem is done.

When solving the problem at the end, you would want the 

error< 0.0005

You find if you are done and find the error with the equation 

1/2 (b-a)

Below is a picture of how you would solve it. 

The problems may take a long time and you would do it until you find a number with 0.0005.

Sometimes you would find it with 1 dec which is < 0.05

2 dec < 0.005

3 dec < 0.0005

4 dec < 0.00005

However the most common one is 3 dec.

That is what we did on day 2.

On day 3, we looked at 3.6 which is Rational Functions. We looked at this in Algebra 2 and it is when you use Vertical and Horizontal Asymptotes. 

Usually the equations would look like 

f(x)=P(x)/Q(x)

To find the Vertical Asymptotes of a problem you would set the denominator=0 and the answer would be x= the answer.

To find Horizontal Asymptotes of a problem, you look at the degree of the first equation:

It would look like

f(x)=anX^n/bmX^m.

n= degree of numerator

m= degree of denominator

an= L.C. of numerator

bm= L.C. of denominator

It would be like n/m

To find the horizontal asymptote though, there are three possibilites

When the degrees are both equal. n=m you would use the coefficient of both answers as the answer. If its 2/2 then it would be y=1

Then n<m the answer would be y=0

n>m then the answer DNE.

However, something we have not seen before is Slant Asymptotes which is only possible if the horizontal asymptote is n>m.

f(x)=m(x)/d(x)

The answer from the vertical asymptote is used to divide numerator of the equation. For example x=2 is the VA. You would divide it by its numerator, for example x^2+4. y= answer.

For all equations, there will be x-intercepts and y-intercepts. You just plug in zero for the opposite to find it. If looking for the x-intercept, make y=0 and solve and vice versa.

To find the holes in a problem. You factor the equation and see if anything can be canceled out. Whatever cancels out is the answer and it would usually look like (x,y). If nothing cancels out, there is no hole. 

This is how Rational Functions work.

On day 4, we took a quiz on 3.4-3.6 to test if we knew the information. We then reviewed quickly on some of the material covered last week and the week before. 

That is what we did on Day Four. Next week we will have a test on Friday and will cover one more lesson before that. 

See You Next Week!

Week 7: Rational Functions

The blog topic for this week is Rational Functions. We looked at this in Algebra 2 and it is when you use Vertical and Horizontal Asymptotes. 

Usually the equations would look like 

f(x)=P(x)/Q(x)

To find the Vertical Asymptotes of a problem you would set the denominator=0 and the answer would be x= the answer.

To find Horizontal Asymptotes of a problem, you look at the degree of the first equation:

It would look like

f(x)=anX^n/bmX^m.

n= degree of numerator

m= degree of denominator

an= L.C. of numerator

bm= L.C. of denominator

It would be like n/m

To find the horizontal asymptote though, there are three possibilites

When the degrees are both equal. n=m you would use the coefficient of both answers as the answer. If its 2/2 then it would be y=1

Then n<m the answer would be y=0

n>m then the answer DNE.

However, something we have not seen before is Slant Asymptotes which is only possible if the horizontal asymptote is n>m.

f(x)=m(x)/d(x)

The answer from the vertical asymptote is used to divide numerator of the equation. For example x=2 is the VA. You would divide it by its numerator, for example x^2+4. y= answer.

For all equations, there will be x-intercepts and y-intercepts. You just plug in zero for the opposite to find it. If looking for the x-intercept, make y=0 and solve and vice versa.

To find the holes in a problem. You factor the equation and see if anything can be canceled out. Whatever cancels out is the answer and it would usually look like (x,y). If nothing cancels out, there is no hole. 

Below is one example of how to solve a rational function.

This is how Rational Functions work.

See You Next Week!

Week 6: Zeros of Functions

This week the topic was Zeros of Functions. On Day three and Day four, we covered zeros of Functions.

On Day three we looked at Zeros and Factors of Polynomial Functions. 

We looked at 

f(x)=d(x)q(x)+r(x)

if r(x)=0, then d(x) is a factor of f(x)

We also looked at

f(x)/x-c × r(x)=f(c)

We tried to look for factors and zeros.

The steps to do so:

1. Complete the factorization

2. List the zeros

The difference between a factor and a zero is

ex:

Factor | Zero

x-7     7

x+9   -9

4x-1   1/4

x-14    14

Also a reminder from Algebra 2 is:

i = √-1

i^2 = -1

a+bi = complex #

a-bi = conjugate

(a+bi)(a-bi) FOIL

Conjugate o’s always occur in conjugate pairs

ex: 1-i <---> 1+i

2i <---> -2i

So we used synthetic division in this type of problem. You would do the exact same thing from the previous lesson but the remainder must be zero. If the remainder is not zero, then the equation does not work. It would not have a zero or a factor.

You would do the same steps with a remainder of zero and then after that. You would turn the number given or the original factor and turn that into a factor: ex (x-2). Then the equation provided from the end. You would try to factor that to get the other factors. Sometimes you would use the Quadratic formula to solve it. Then it would look like for example: (x-2)(x-1)(x+9). that would be the factor and the zeros would be x=-9, 1, 2. That would be the answer. 

HINT: the first degree in the original equation tells you how many zeros there will be in a problem. Sometimes there would be imaginary numbers to make the remaining zeros. For example: x^2-4. There will be 2 zeros in this equation.

Below is two examples of how it would work.

In the second problem, there are imaginary numbers, which appear when you plug it into the quadratic formula to find out how to solve an equation. That is how you find the zeros and factors of a polynomial function.

On day four, we also look at Real Zeros of Polynomial Functions.

In this one we had to so which zeros were real inside a polynomial function. To do this, we had to do 

P= Factors of constant

s   Factors of leading coefficient

The rules on how to find is 

1. Find all possible zeros P/S

2. Test zeros using remainder thereom

HINT: use graph to see possible zeros

This was done in Algebra 2 in the previous year and to begin the problem: a Polynomial will be given to you.

There will be many zeros listed and you are trying to find the real ones and when a factor is not given. Below is an example of how to do it. 

So first you would look at the constant which is the number without x or with a variable. You would find all the possibilities on how to make that number. 

For example from above. -2: to make -2 you could use ±1 and ±2. You would put that on the line for P.

Next you would look at the first coefficient, which is above 3x^4.

You would see how many possibilites would go into 3, which is ±1 and ±3. You would put that on the line for S.

Just like above you see all the it has ±1 and ±2 over ±1 and ±3.

You then times all the possibilites by each other.

For example: 

±1/±1 is one or negative one. Then ±1/±3 is 1/3 or -1/3.

±2/±3 is 2/3 or -2/3

±2/±1  is ±2

So x= -1, 1, 1/3, -1/3, -2, 2, -2/3, 2/3

You then put those into x to figure out After you find one that is equal to 0.

After you find it equal to 0 you put the equation given into Synthetic Division with the zero that works. You then try more zeros and put it into the simplified equation until you have all the zeros needed. Just as this equation needs 4 exact zeros. 

That is how to find the zeros of a function.

See You Next Week!

Week 6

During Week 6, we began a new chapter Polynomial and Rational Functions. 

On day one, we first looked at Polynomial Functions. 

We reviewed on what is a leading coefficient, the degree and the constant.

We then also reviewed on End behavoir which would look like

ex: x--> Infiniti

We then reviwed on even, odd and neither functions.

Lastly we looked at Multiplicity, which is on how many times a point is repeated. 

That is what we did on day one.

On day two, we reviewed again on how to Divide Polynomial Functions. 

It would look like this:

To do long division however with Polynomial functions is similar to the way we would do it as a child.

First you put the right numbers in the right place.

For example:

f(x)= 3x^3 -x^2 -2x +6

d(x)= x^2 +1

Second you would solve it:

Above is an example of how you would solve.

You would sometimes have a remainder and when you do, you would put it over the number you were solving for.

Like if you had a remainder of 7 and the problem was being divided by 4 it would look like 7/4. Just as it is shown above. If there is a remainder it is not a factor.

We then looked at the Remainder Thereom, which is like

f(x)/x-c

the remainder would be f(c)

You would plug c into the equation provided and you would solve it.

for example if the equation was x^2 -1 and the x-c is x-4, you would plug in 4 for x.

You would solve it and the answer is the remainder.

Lastly we looked at Synthetic Division which is the easiest method so far. It is only able to be used when the x is not squared by more than 1. For example x-4.

You would draw a line down and sideways and you would divide. Below are two examples on how it works.

The first number will drop down and you would divide it by the outside number and then added to the next empty space. You would add the numbers on top of each other and you would do the same process over again. At the end the last number will be the remainder unless it is 0.

The other numbers create the equation. 

You would drop down the power by one so if the first number in the original equation was x^3. So if the first number in the answer was 4 it would be 4x^2 not x^3. and you would do it down dropping the square to 0.

That is what we did on Day 2.

On Day three we looked at Zeros and Factors of Polynomial Functions. This is very similar and a continuation from the last lesson. 

We looked at 

f(x)=d(x)q(x)+r(x)

if r(x)=0, then d(x) is a factor of f(x)

We also looked at

f(x)/x-c × r(x)=f(c)

We looked at all of this yesterday, but instead of stopping there, we  try to look for factors and zeros.

The steps to do so:

1. Complete the factorization

2. List the zeros

The difference between a factor and a zero is

ex:

Factor | Zero

x-2     2

x+5   -5

3x-1   1/3

x-4    4

Also a reminder from Algebra 2 is:

i = √-1

i^2 = -1

a+bi = complex #

a-bi = conjugate

(a+bi)(a-bi) FOIL

Conjugate o’s always occur in conjugate pairs

ex: 1-i <---> 1+i

2i <---> -2i

So we used synthetic division in this type of problem. You would do the exact same thing from the previous lesson but the remainder must be zero. If the remainder is not zero, then the equation does not work. It would not have a zero or a factor.

You would do the same steps with a remainder of zero and then after that. You would turn the number given or the original factor and turn that into a factor: ex (x-2). Then the equation provided from the end. You would try to factor that to get the other factors. Sometimes you would use the Quadratic formula to solve it. Then it would look like for example: (x-2)(x-1)(x+9). that would be the factor and the zeros would be x=-9, 1, 2. That would be the answer. 

HINT: the first degree in the original equation tells you how many zeros there will be in a problem. Sometimes there would be imaginary numbers to make the remaining zeros. For example: x^2-4. There will be 2 zeros in this equation.

Below is two examples of how it would work.

In the second problem, there are imaginary numbers. Imaginery numbers appear when you plug it into the quadratic formula to find out how to solve an equation. That is how you find the zeros and factors of a polynomial function.

That is what we did on day 3.

On day four, we looked at Real Zeros of Polynomial Functions. Which is shown below in the post Zeros of Functions.

In this one we had to so which zeros were real inside a polynomial function. To do this, we had to do 

P= Factors of constant

s   Factors of leading coefficient

The rules on how to find is 

1. Find all possible zeros P/S

2. Test zeros using remainder thereom

HINT: use graph to see possible zeros

This was done in Algebra 2 in the previous year and to begin the problem: a Polynomial will be given to you.

There will be many zeros listed and you are trying to find the real ones and when a factor is not given. Below is an example of how to do it. 

So first you would look at the constant which is the number without x or with a variable. You would find all the possibilities on how to make that number. 

For example from above. -2: to make -2 you could use ±1 and ±2. You would put that on the line for P.

Next you would look at the first coefficient, which is above 3x^4.

You would see how many possibilites would go into 3, which is ±1 and ±3. You would put that on the line for S.

Just like above you see all the it has ±1 and ±2 over ±1 and ±3.

You then times all the possibilites by each other.

For example: 

±1/±1 is one or negative one. Then ±1/±3 is 1/3 or -1/3.

±2/±3 is 2/3 or -2/3

±2/±1  is ±2

So x= -1, 1, 1/3, -1/3, -2, 2, -2/3, 2/3

You then put those into x to figure out After you find one that is equal to 0.

After you find it equal to 0 you put the equation given into Synthetic Division with the zero that works. You then try more zeros and put it into the simplified equation until you have all the zeros needed. Just as this equation needs 4 exact zeros. 

That is how to find the zeros of a function.

This is what we did this week. Next week we will have a test on this chapter.

See You Next Week!