## Thursday, September 17, 2009

### Period1 - Factoring In!

As a class, I would like you to briefly comment on the findings of your investigation today.

In your response, give me an example of a function (minimum degree 3) with 3 x-intercepts and justify how you know it has 3 x-intercepts.

TThere is an applet on explorelearning.com that can help you play with the functions. Enter "Polynomials & Linear Factors" into the search box and you will have a 5 minute pass to explore.

1. In today's class, I learned that the degree of the function tells you whether the graph has a positve or negative end behaviour. Also the x-intercepts also tells us the number of times it touches the 0. An example is y=(x-1)(x-2).The equation tells us that the number of x-intercepts is 2 and the x-intercepts of the equation y=(x-1)(x-2) would be (1,0) and (2,0).

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3. Today we learned about the degree, x-int's, and the leading coefficient makes a big deal on how your graph is going to end up. If it's a simple quadratic like (x-4)(x+2) the x intercepts would be x-int: (4,-2). degree's were also reviewed today. the law was that if the degree is an even number the end behaviours should be the same. if it's odd then they're opposite. for eg. x^3 would be opposite and x^2 would be the same

4. in class we were working on a sheet were we had to determine the graph of each polynomial function. I learned that the degree in a function equals the total number of variables. I also learned that to figure out the x intercepts of a given function ex
y= (x-2)(x-3) you would have to negative the numbers given so -2 would become 2 while -3 would become 3 and thus the x-intercepts would be (2,0) and (3,0).
And an example of a function with 3 x-intercepts would be f(x)=(x+2)(x-1)(x-3)
i konw this because the total number of variables (in this case x is the variable) is 3.

5. In class today, we learned that by looking at the equation of a line, we can predict many features of that line already. For example, the number of turning points (which is one less than the degree number), the maximum number of x-intercepts (equal to degree number), and the actual x-intercept coordinates themselves (if the equation is in y=(x+2)^2(x-2)^2 format, the x-intercepts would be (2,0) and (-2,0)).

6. y=x(x+1)(x-1)
When expanded, (Expanded: x^3-x) the degree of the function is cubic. Cubic functions have a maximum amount of 3 x-intercepts.
In factored form, we can find the x-intercepts to be: (1,0) (0,0) (-1,0) by subbing in 0 for y and solving for x.

7. In todays lesson, we found out that we are able to find the degree in any given polynomial equation.
For example, in f(x)(x-2)(x+1)(x+3), the degree is 3. You are also able to find the x-intercepts in the equation by solving for zeros. (2,-1,-3)

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10. Today's lesson was pretty cool eh. We learned that factoring an equation allows us to find the x-intercepts of the equation, regardless of what degree it happens to be. That means we can even factor degree 4 equations to find out the intercepts. Pretty cool stuff.
An example of this would be f(x) = (x-3)(x+3)(x-2). The zeroes would be (3,-3, 2).

11. During today's lesson, we discussed how to find the x-intercepts and how many there were by looking at the graph and factoring the equation. By factoring the equation, you are able to solve the x-intercepts no matter the degree.

For example, (x-6)(x+8)(x+1)(x-2) has 4 x-intercepts; 6, -8, -1, and 2.

There are 4 x-intercepts because you can see the number of 'x' variables in this equation, which are 4, thus there are 4 x-intercepts.

12. Today, we learned about the relationship between the x-intercepts and the degree of a polynomial function. I discovered that the maximum number of x-intercepts equals the degree of the polynomial function. The minimum x-intercepts depends on whether the degree is odd or even. If the degree is odd, it will have at least one x-intercept; but if the degree is even, it may not have any x-intercepts.
An example of a polynomial with 3 x-intercepts is: f(x)=(x-2)(x+1)(x+3). I know there are 3 x-intercepts because just by looking at this function in factored form,I noticed that the number of variables in the equation equals the degree number (or vice-versa).I can also calculate that there are 3 x-intercepts algebraically (y=0 and solve for x). Or, you can graph it and there will be 3 x-intercepts.

13. In today's investigate, one would have discovered that the degree of a polynomial function would represent the maximum number of x-intercepts. For example a degree 3 function can at MAXIMUM have 3 x-intercepts. In addition, by looking at the factored form of a function one can easy determine the zeros by setting “y” as 0. Note: x-intercepts have a y value of 0.
One can determine the degree of a function by counting the number of “xs” in the equation. In other words, the degree of a polynomial function is the sum of all the independent variables. For example: f(x) = (x+3)^3 (x+4)(x-4) The example will have a degree of 5 (3+1+1).

An example of a function with 3 x-intercepts can be: f(x)= x(x+3)(x-3).

The 3 x-intercepts will be: (-3.0) (0,0) (3,0)

Algebraically, I know that there are three x-intercepts f(x)= x(x+3)(x-3). As stated above, all x-intercepts have a y value of 0. After letting y = 0, one will just have to solve the roots as one would with a quadratic equation.

14. Today we learned about the relationship in polynomial functions between the degree and the x-intercepts. The degree of a polynomial function would represent the maximum number of x-intercepts it could have. The number of turning points in the function represents the degree of the function minus one. The minimum number of x-intercepts represents whether the degree is odd or even. For odd degrees, there will be at least one x-intercept, and for even degrees it may not even have one.

EXAMPLE:
f(x)=(x-9)(x+3)(x-1): The x-intercepts would be (9,0), (-3,0), (1,0).

You are able to solve this problem algebraically and graphically. Algebraically, you can substitute "y" for "0" you are able to find the x-intercepts.

15. We learned:
- ODD degree polynomials can have a minimum of 1 x-int (due to opposite end behaviours)
- EVEN degree polynomials can have a minimum of 0 x-ints
- The max # of x-ints = n(the degree)
- The max # of "Turning Points" = n-1

Example d: x(x+1)^2
This is a degree 3 polynomial because when the equation's expanded the x-value has a degree of 3.
Because the # of max x-ints = n(the degree), there are 3 x-ints in the above equation.

16. In this morningsclass I learned that the first degree of the polynomial tells you a lot about what the graph will look like.

1. even or odd? an even degree will tell you that the end behavious will be the same. an odd degree will tell you that the end behavious will be opposite.

2. the degree also tells you how many x-intercepts the graph will have.

ie. f(x)= x^3-4x+1
-odd degree: opposite end behaviors
-3 roots (or zeros)

17. Today we learned about the relationships between the degree and the x- intercepts in a polynomial function. In a polynomial, the maximum # of x-intercepts can be indentified by the degree of a function. Even degree polynomials can have a minimum of zero x-intercepts. Maximum number of turning points can also be found by the degree subtracted by one.

An example of a degree three polynomial is:
f(x) = (x+6)(x+1)(x-2)

the zeroes would be -6, -1, and 2

It can be justified as a degree 3 polynomial because when after you expand the equation, the degree would be three.

18. Today's lesson covered things such as: the turning points for an even graph is equal to the total number of local max and min points, and that the min number of x-intercepts for this type of graph is zero. In addition we learned that the max number of x-intercepts is equivalent to the degree of the graph.
For example: f(x)= (x+2)(x-4)(x+3)
the x-intercepts would be -2, 4, and -3. The zeros can be obtained through solving for x.

19. In today's lesson we learned about: local maximun, local minimun points,and turning points. We also learned that the degree in a polynomial function represents how many x-intercept there are going to be on the graph. If the degree was an odd number it means that the end behavious of the graph is going to be opposite. But on the other hand if the degree was an even number than the end behavious of the graph is going to be the same and that graph might not have a zero.
For example: (x+3)(x-8)(x+5) In this function the 3 x-intercept would be -3, +8, and -5. We know that they are zeros because by solving for X.

20. In today's lesson, I learned that the degree (even or odd) not only represents how the function will look but determines the number of maximum x-intercepts.

For example: x^5 - can cross the xaxis 5 times at max, and can have 4 different turning points

Also by factoring out an equation, it gives us access to the x-intercepts

E.g. x^2 + 12x + 32
(x+4)(x+8)
-isolate x to find zeros
therefore, the degree is two, with a max of 2 x-intercepts and we see here that after factoring out the equation the zeros (roots or xintercepts) are -4 and -8

21. this morning we learned about the meaning of the value of the degree on the leading coefficient. if the value if the degree is odd it will have opposite end behaviors. if the value of the degree is even, the end behaviors will be the same. we also learned that depending on the value, there will be a different number of maximum zeros

22. Equations in their factored form can help tell us about the x-intercepts of the graph. Also, the degree of the function tells us the maximum # of x-intercepts there will be.

f(x) = (x+1)(x+2)(x-3)
The x-intercepts can be solved for by letting y = 0.

23. In todays class, we learned about the turning points and how the degree of the fucntion affects how many x intercepts there will be.
For example an equation with 3 x intercept can be:
f(x) = (x-4)(x+5)(x-1)
in this equation the zeroes are 4 , -5 , and 1 which are also the x intercepts

24. Today I learned that I can find the amount of x-intercepts by looking at the degree of the polynomial. I also learned that the equation must be factored first.

An example of a function with 3 x-intercepts is: y=(x+6)(x-4)(x-1) Therefore there are 3 x-intercepts which are -6, 4, and 1. I solved this by isolating x to find the zeroes.

25. After Completing today's assigned investigation I learned that the equation of a polynomial function while in factored form will tell us how many x intercepts there will be. Also I learned that by isolating for x while in factored form will give the co-ordinates of the x intercept
*note that the y value in an x intercept is always 0.

Example:
f(x)=(x-4)(x+2)(x+8)
If you look at this function you immediately know that it is a degree 3 function, so there will be a max number of 3 x intercepts and 2 turning points.
Then you can isolate for X to figure out the 3 x intercept coordinates which are:
(4,0),(-2,0),(-8,0)

26. in today's lesson i learned that the degree and leading co-effecient of a function can tell you about its features:
-degree / 2 is the maximum amount of local mins or maxes
-the degree is the total number of possible x intercepts

an example of a function that is minimum degree three and will have 3 x intercepts is
f(x) = (x-2)(x+1)(x+3)
the zeroes are : 2, -1, -3