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!

Week 5

During Week 5, we went over the Tangent Line, which was the last section of Chapter 2.

We had a review week this week and looked over the material once more to see what we learned and to be prepared for the test.

The test was on friday and it was not as difficult as the last and we will start chapter 3 next week.

See You Next Week!!!

Week 4: Piece Wise Functions

This week we continued on with Chapter 2 on Functions, however, the main priority was Piecewise Functions.

Piecewise functions are functions that are discontinous or continious with graphing.
One example is:

Continuous functions have no holes or gaps while discontinious do.
First you would just graph the first part as you would normally graph it. Like the equation above for x+1. You would graph it like a point-slope function. You then would look at the x is greater than or less then or etc. to see which side you keep and if the equation should have a whole or a dot. Another type of equation we studied in class is the Greatest Integer Function.
It would usually look like: ⟦x⟧

Above are two examples of how it looks like. If you add a two or any other number, it would slop, which is slide opposite.
Lastly then is Absolute Value Function
It is the exact same thing as a regular function but has absolute value signs instead of paranthesis.

If you had y= |-3x+2|
You would ignore the absolute signs and graph it. Once you graph it, you make sure everything that once was negative is postive. So one side would go up making a V shape.
The same would happen with a parabola if it was negative. It would look like a weird W in the end.

That is what we did this week with Piecewise Functions.
See You Next week!

Week 3

During week 3, we continued in the chapter Functions. Since this week was only a four day week due to Labor Day, we only covered three lessons instead of four.

On the first day, we started with the lesson Symmetry and Transformations. We first looked at the power functions of each different type of equation. 

All Power functions start with f(x)=x^n

Below are examples of the ones we looked at. 

We then looked at how to tell which function is an even or an odd function. Even functions are symmetric to the y-axis and to find the symmetry of the y-axis is by turning x into -x. To find the odd functions, you find if it is symmetric to the origin, which is by turning both x and y functions into negatives.

We then looked at vertical and horizontal shifts. Below are the examples of shifting.

y=f(x)+c

y=f(x)-c

y=f(x+c)

y=f(x-c)

We then looked at reflections. To reflect on the x-axis, all you do is by putting the - on the outside of f(x). So it looks like -f(x). To reflect on the y-axis, you put the negative inside like f(-x). Then after, there are vertical streches and compressions. 

y=cf(x)

c>1 stretched or skinny

0<c<1 compressed or fat

This is what we worked on, on day one.

On Day two, we began Linear Functions. The first thing we saw was the constant function, which is y=b. The Linear function is f(x)=ax+b or y=mx+b.

Slope is m= y2-y1/x2-x1 which results in rise over run.

The Point slope equation is: y-y1=m(x-x1)

We then looked at increasing and decreasing. Some graphs have lines that go up like a parabola, yet they also have some that go down, which a parabola does as well and sine or cosine equations.

The slope intercept equation is y=mx+b, which is what we looked at earlier and a horizontal graph is y=b, while a vertical graph is x=a, which is usually an undefined slope. Parallel lines also have the same slope while perpendicular lines turn the slope into the negative reciprical. For example 3 turns into -1/3. 

Lastly, To find the points of intersection, you use substitution or elimination, which goes back to Pre-algebra. You plug in either y or x for the other equation for substition to solve and you find (x,y). 

This is what was learned on day 2.

On day 3, we first had a quiz and then we looked at quadratic functions. Most of the material we have looked at this section is review and this was no exception. 

The quadratic equation is: f(x)= a(x-h)^2 +k

To find the vertex (h, k). Ignore the negative

Axis of symmetry is x=h

to figure out if the equation is up a>0

to figure out if the equation is down a<0

We looked at how to make problems easy to solve by looking at which ones are the vertex or AOS by turning problems back into the original problem look.

Two examples are below:

After we looked at some examples, she left us off with information from Physics, which is Free Falling Objects.

This is what we did this week.
See you next week!

Week 3: Superhero Transformations

This week, we began working on Functions and we worked on a Superhero worksheet to understand it better. Below is the worksheet with my answers on it.

The people below are the people I used to work on the problems. 

This is the worksheet we worked on for this week. 

See you next time!