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 4) with 4 x-intercepts. Justify how you know it has 4 x-intercepts.
There 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.
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In todays lesson we found that in factored form for any degree the zeros are shown. Also that the number of degrees can be determined by how many x's there are. Finally to determine what degree it is you must add together the number of x's in the form. For example. y=4x(x-2)(x+2)(x+4)
ReplyDeleteThis is an example of a degree 4 polynomial with four different x intercepts at 0, 2, -2, and -4.
The investigation today demonstrated to our class that, in factored form, you can determine the degree of a polynomial based on the number of x's that are being multiplied together. Also, in factored form, you can determine the x-intercepts by substituting a number for x in each bracket that makes that part of the equation equal zero, therefore making the y value equal zero.
ReplyDeleteAn example of a degree four polynomial with 4 x-intercepts is f(x)=(x-2)(x+6)(3x-12)(x+2). I know that it has four x-intercepts because there are 4 individual terms in the polynomial equation. The x-intercepts in my example are 2, -6, 4, and -2.
In this investigation, it was demonstrated that the zeros of a polynomial function can be unveiled by factoring it. Since anything multiplied by zero is also zero, if you make one of the factors of the polynomial equal to zero, then it will make the y-value equal to zero. You can determine the degree of the function by counting all the x values in the polynomial.
ReplyDeleteAn example of this is -3x(x-4)(x+1)^2(5+x)
The degree is 5, and the zeros are 0, 4, -1, and -5.
In today's investigation, it was determined that you can find the degree of the polynomial while it is in factored form by the amount of x's there are. In addition, the x intercepts are easily found while in factored form as well.
ReplyDeleteFor example the polynomial f(x) = 2x(x-4)(x+3)(x-15). This is a degree 4 polynomial with the x intercepts of 0, +4, -3 and 15
In today's investigation, to find the degree of a polynomial with factoring, we should look at the number of x's. To keep it simple, the number of x's is equal to the degree of the polynomial. For example, if a polynomial has 2 x's, then it is a degree 2 graph.
ReplyDeleteAn example of a polynomial with 4 degrees is f(x)=(x-1)(x+1)(x+2)(x-2). The zeros are 1,-1,-2, and 2.
In todays investigation, i found that you can find the degree of the polynomial if you factor the function, and then you can look how many Xs there are.
ReplyDelete5x(x-2)(x+5)(x-2)
degree is 4, x intercepts are 0, 2, -5, 2
Today in class, I learn you can find both the degree of the polynomial and the x-intercepts by factoring the polynomial.
ReplyDeletee.g. 6x(x+2)(x-8)(x-3)
Therefore, the degree we be 4 and the x-intercept would be -2,0,3,8
The investigation, today thought me about how the degree of a polynomial is found by counting all the x's--> for example f(x)=(x+2)(x-1)(x-3)^2, which is a degree 4 polynomial. Also the factors really determine the zeros or the x-intercepts of a polynomial function. In the equation above, if you perform the opposite operation you will be able to solve for x:
ReplyDelete(x+2)=0
x=0-2
First zero is -2
(x-1)=0
x=0+1
Second zero is +1
(x-3)=0
x=0+3
x=3
Last zero is +3
In today's investigation, we discovered that you can determine the degree of a polynomial by counting the number of x variables in the equation. For example, f(x)=-(x-3)(x+2)^5 would be a degree six polynomial. We also discovered that when you find the zeros of the equation, you find the x-intercepts. To find the zeros, you substitute each x value with a number that would make the equation equal zero. For example, the zeros of f(x)=(x-1)(x+3) would be 1 and -3.
ReplyDeleteAn example of a degree four polynomial with four x-intercepts is f(x)=(x+3)(x+2)(x-1)(x+5). The x-intercepts would be -3, -2, 1, and -5.
The results of today's investigation demonstrates that the degree of the polynomial function can be determined by counting the number of 'x's present in the fully factored equation. Note that if an equation looks like this: f(x)=x(x+1)^2 you must take into account that (x+1) is SQUARED. Therefore is actually a degree three polynomial - not a degree two.
ReplyDeleteAn example of a degree 4 polynomial function is: f(x) = (x+5)(x+2)(x-7)(x-4)
The x-intercepts are: x --> (-5,0)(-2,0)(7,0)(4,0)
To derive these x-intercepts - simple plug in y as 0 and solve for x : )
What I found in today's investigation was that the number of the x-values there determine the degree of the function. For example, if the function was f(x) = (x-2)(x+1)(x+3) then there would be 3 x-intercepts because 3 x-values are present.
ReplyDeleteAn example of a function with degree 4 is f(x) = 4x(x+17)(2x+13)(x+7) with the x-intercepts being (0,0), (-17,0), (-13,0), and (-7,0).
In todays investigations, I found that by multiplying all the x's together, you may find the degree of the polynomial. When I first started the investigation, I was counting the turn over points, and found the degree of the polynomial based on that information along with the leading coefficient, however by the end, I learned that factoring was a much easier way.
ReplyDeleteAn example of this would be: f(x)= (3x+1)(x-8)(x-3)^2
In this lesson we have learned to predict the x-intercepts of the graphs at its factorized form any degree the zeros are shown and the degree of the function will be determined by the number of x in the for equation.
ReplyDeleteFor example, a quartic function. In that sense, the formula of finding the x-intercepts will be:
f(x)=(x-a)(x-b)(x-c)(x-d),
which the x-intercepts are a, b, c and d.
Therefore, if there is an equation:
f(x)=x(x-5)(x-1)(x-4).
The x-intercepts will be 0, 5, 1 and 4.
In todays investigation we learned that a Zero can be determind through a factored function, as well as its degree by simply adding the total amount of x's in your function.
ReplyDeleteFor example: F(x)= x(x-2)(x+1)(x+3), so if we count the x's there are four, therfore it is a degree 4 polynomial.If we take another function like as follows:
F(x)= -(x-2)(x+1)(x+3), if we count the x's on this function we see that there are only 3, thefore it is a degree 3 ploynomial.
The zero's in this degree 4 function: F(x)= F(x)= x(x-2)(x+1)(x+3) are:
0,2.-1,-3
In the investigation that we did today, I found that the number of x's is the same as the number of degree of the function. Or in other words, the number of sets of brackets equalls the number of degree of the poynomial. I also found that you can easily tell the roots of the function just by factoring it.
ReplyDeleteAn example: a degree 4 polynomial function is f(x)=(x-5)(x+4)(x-6)(x+3).
In this case, the x-intercepts are: x=(5,0),(-4,0),(6,0),(-3,0).
You can calculate the x-intercepts by taking each set of brackets and setting equal to zero, then solving for "x".
eg:
(x-5)=0
x=5
:)
In the investigation we did today, I found out that you can find the zeros (x-intercepts) when the polynomial function is in factored form. Finding the degree polynomial would be multiplying all the x's together.
ReplyDeleteAn example of a quartic function would be,
f(x)=(x-4)(x-2)(x+1)(x+4)
The x-intercepts would be: 4, 2, -1 and -4
Today in class i learned what can be learned from looking at the factored equation of a power function
ReplyDeleteFor example: x(x+2)(x-4)(x+5) will have x intercepts of 0,-2,4 and -5. Also the number of "x"s in the equation determines the degree of the function. The sign of the first "x" in this case tells us that the graph is opening up
With these few information we can draw a rough sketch of the function and get a better understanding of what it looks like
Today’s investigation helped to find that when a polynomial function is factored, the x-intercepts of the function can be found by solving for y=0 within each factored bracket. The investigation had also helped to see that the amount of x’s in the function determined the degree of the function.
ReplyDeleteAn example of a function with 4x-intercepts, would be something such as f(x)=(x+2)(x-3)(x+4)(x+5). In this equation, when we count up the x’s, we find that the number of x’s is equal to the degree of the function; 4. As we try to make each factored bracket equal to zero, we solve the x-int when y is equal to 0. In this case, the x-intercepts of the polynomial function are -2,3,-4,and -5.
During today’s investigation, I found that based the number of Xs in a function it equals to the same number as the degree of the function. Therefore, if there was a function with 3 Xs, for example x2(x+2), then the degree of the function is 3.
ReplyDeleteAn example of a function with 4 degrees and x-intercepts is 2x(x-9)(x+2)(x+3).
x-intercepts: (-3,0),(-2,0),(0,0), and (0,9)
In class today, we were given an investigation that helped explain through examples what information can be determined from an equation in factored form. The number of x's in the equation determines the degree of the function. For example, in the function f(x)=x(x+2)(x-7), the degree of the function would be 3. This is can be checked because if you were to expand out the function, you would find that the degree is 3. So an easyway to figure out the degree of a polynomial is to count the number of x's in factored form! Also, from the equation, the x-intercepts can be determined by making y=0. For example, in the function f(x)= (x+9)(x+13)(x-3) when y is substituted for 0, the numbers that come out as a result are -9,- 13 and 3, therefore the x-intercepts are at (-9,0), (-13, 0), and (3,0). An easy way to remember this trick is that x-intercepts are the opposite term of the value in each bracket.
ReplyDeleteAn example of a function with 4 x-intercepts is the function f(x)=(x+11)(x-2)(x-17)(x-21) This function can be identified as a degree 4 because it has 4 x's. The x-intercepts can be found by substituting y for 0. From this is can be concluded that the x-intercepts are (-11,0), (2,0), (17,0), and (21,0).
Today's investigation taught us that we are able to count the number of degrees by the number of "x" variables in the equation. For example, the equation f(x)=(x+3)(x+2)(x-1) would be a degree 3 equation. This was a very simple concept to understand.
ReplyDeleteAn example with four x-intercepts would be
f(x)= (x+2)(x-3)(x-5)(x+3). If you count the number of "x's", there would be four. This equation would be described as a degree four equation. The x-intercepts would be (-2), (3), (5), (-3).
The factored form of a polynomial function reveals a lot of information about the graph.
ReplyDeleteAs in the factored form of a quadratic function, the factored form of other polynomial functions reveals the x-intercepts of the graph. The number of x-intercepts is equal to the number of different factors.
All x-intercepts occur at y=0 (or f(x)=0), so to find the x-intercept of a factor, you simply set the factor equal to zero and isolate x.
For example:
The equation is
f(x) = 2x(x-10)(x-3)(x+4)
The x-intercepts are
0 = 2x
0/2 = 2x/2
0 = x
0 = x - 10
0 + 10 = x - 10 + 10
10 = x
0 = x - 3
0 + 3 = x - 3 + 3
3 = x
0 = x + 4
0 - 4 = x + 4 - 4
-4 = x
When there is no coefficient to the x in a factor, reversing the sign on the constant in the factor is a short, easy way to find the x-intercept.
Determining the degree of the function is easy; it only requires counting the x's. When a factor has an exponent, it counts as the same number of factors as the exponent's value; (x-1)^2, for example, counts as two factors, and therefore two x's.
The exponents of the factors also reveal the appearance of the graph as it meets the x-intercept. When the exponent is odd, such as (x-1) which has an invisible exponent of 1, and (x-1)^3, there is no turning point and the function crosses the x-axis. When the exponent is even, such as (x-1)^2, there is a turning point and the function does not cross the x-axis.
Using the first rule, creating a graph with four x-intercepts only requires creating an equation with four unique factors.
f(x) = 2x(x-1)(x+2)[(x-3)^2]
The respective x-intercepts are 0, 1, -2, and 3. Although there are four x-intercepts, this is a degree 5 polynomial function.
Additionally, at (3,0), the graph will have a turning point.
In today's investigation, I learned that the factored equation of a function can be used to calculate the degree of a polynomial function and the x-intercepts. The zeros in the equation are the x-intercepts of the graph. The amount of x's in the equation is equivalent to the degree of the function. If the equation is negative, the leading coefficient of the function is negative, and vice versa. For example, in the equation y=2x(x-7)(x-3)(x+5), the x-intercepts are (7,0), (3,0), and (-5,0). The degree of the function is 4. Since the degree of the function is even, and it is positive, the end behaviours will both be approaching positive infinity.
ReplyDeleteI've learned that a lot of information can be revealed from a polynomial function when you factor it out.
ReplyDelete1. Once you factor out the polynomial you are able to see: how many zeros are present in the graph; and also when looking at how many "x" in the factored form, you can see what the degree of the function is.
2. We have also learned about the min number of x-intercept, max x-intercept, and max number of turning points by looking at the polynomial function. Min #of x-intercept can be determined by looking at the degree of the function. If it's even it would 0 if it is odd then it would be 1. To know the max. #x-intercept it depends on the degree. The max number of turning points is determined by (degree - 1)
An example of an equation.
y = 7x(x+5)(x+3)(x-4)--> this is the factored form of a polynomial function.
1. you can see 4 "x" meaning it's a degree 4 function
2. (-5,0);(-3,0);(4,o) are the zeros of the equation (x-intercept)
Basically in today's investigation we mainly concentrated on determining the x-intercept of an polynomial function (by factoring it). Also, determining what type of degree the function is by looking at the graph.
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ReplyDeleteThrough carrying out this investigation, I discovered that important information regarding the specific function is provided when an equation is in factored form.
ReplyDeleteFirst of all, the degree of the polynomial can be determined by simply calculating the sum of all the variable terms x. An example is as follows:
f(x)= x(x+2)(x-1)(x-3)
When we simply add all the x values, we learn that the degree of this polynomial is 4, making it a quartic (or degree 4) function.
The zeros, otherwise known as the x-intercepts, can also be calculated using this factored equation. If you recall, this is a familiar concept that was thoroughly explored using the quadratic function in grade 11. To determine the zeros of a polynomial function, one simply has to substitute 0 for f(x):
0=(x+2)
0-2=x+2-2
-2=x
This applies to all three brackets. Once all three brackets have been completed, it seems there are only 3 x-intercepts. However, the x preceding the first equation means that there is actually a fourth x-intercept, and that it occurs at the origin or (0,0). In turn, the above equation demonstrates a function with 4 x-intercepts: (-2,0) (1,0) (3,0) and (0,0).
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ReplyDeleteThe investigation demonstrated to us today showed that the number of ‘x’s in an equation revealed the degrees present in a polynomial function.
ReplyDeleteEx: f(x)=(x-1)(x+2)(x-3)
As we can see visually, there are three x values present, and therefore, the degree of the polynomial function is three. In addition to this, because there are three ‘x’s, we are also capable of considering that there are three different x-intercepts in this polynomial function (which would be (1, 0), (-2, 0) and (3, 0)).
Ex: f(x)=-3x(x+7)(x-12)(x+2)
We can determine the degree of this polynomial by looking at the equation and search for the ‘x’s present. Visually, we can see that there are four x values in this equation, and therefore, we can summarize that this polynomial function has a degree of four. To determine the zeros, simply consider what value is required to be inputted so that the value of y becomes ‘0’. Consequently the x-intercepts of this equation is (0,0), (-7,0), (-2,0).
(Note: In polynomial functions that include degrees, the equation f(x)= -2 (x+2)2(x-1) does NOT mean that the equation only has two x-intercepts and is a degree of two—because the equation is an exponent, there are therefore three x values present and the equation has three x-intercepts, shown below:
f(x)=-2(x+2)^2(x-1)
If we factor out the exponent, we see three x’s
f(x) = -2 (x+2) (x+2) (x-1)
Therefore, this polynomial function has a degree of three, and not two.)