Investigating Slopes Assessment

by rocker97 (student)

Investigating Slopes

Alessandro Marazzani

25/11/13

Maths Assessment

Introduction

The aim of our assessment is to find general laws and correlations with the investigation of slopes. In fact, we had been given a sheet, where we had many points to analyse and check and, at the end, come up with a conclusion. We are allowed to use our graphic calculator, mainly because we need to show our results.

I will structure this essay by firstly doing an introduction, and then I will start by answering each point we have been given to analyse and come up with a statement at the end of each point. At the end, I will write a final conclusion that sums up all the results I found in this big investigation of slopes.

Investigation 1

The first investigation of this assessment tells me to find a formula for the gradient when there is the function f(x)= X2. It also tells me to write the answer in the form of f1(x). So, to do this, I will plot the function in the graph, check the gradient, and see if there is a formula for it. This investigation is quick to complete, so I recon, I don’t need to do anything to my results, but only find a connection between them.

The conclusion I can draw out from this investigation is that at the end, I found a correlation with the gradient of my slope. For every value I give to X, every time the gradient is double the value. In fact, I can say that when f(x)= X2, this means that f1(x)= 2X

Investigation 2

This investigation is much more longer than the first investigation. It is also more complex. It tells me to consider two functions, which are f(x)= xn and f(x)=aXn. then I will still need to find an answer for f1(x) for the function f(x)=aXn and then discuss the scope of my analysis. To make my life easier I will divide this investigation in parts. For every part, I will analyse a variety of cases, and come up with a conclusion at the end of each case and part. Consequently, after watching at all my results, I will give a final formula to this investigation.

Part 1

For the first part of the second investigation, I decided to first look at the function f(x)= xn. So, to work with this function, I will give to the input “n” the values of 3,4 and 5. I decided to not give to “n” the value of 2 because I have already analysed the answer of it in the first investigation. So, I will start by giving to “n” the value of 3.

1st Case

I will try, firstly, with “3” for the number that will equal to n. So, f(x)= X3. I have decided to start with number 3 because, as I wrote before, my first investigation already told me the answer if, as my first case, I would have made n=2.

At the conclusion of my first case for my second investigation, I found again one link. As my first investigation, I spotted the correlation after I analysed the gradients. In fact, I arrived at the conclusion of stating that if f(x)= X3, consequently f1(x) will be f1(x)= 3X2

2nd Case

I still continue by increasing the number of one, so that I will not miss any important result. I will try to find a correlation when n=4. Thus f(x)= X4

Again, with these point, I can say by looking at the gradient, that f1(x)= 4X3. Now, I’m starting to have a brief supposition of what my general discussion and results for investigation number 2 will be. However, to be sure, I will try to find a correlation by looking at one more case.

3rd Case

For my last case, I will let n=5. In fact, the formula on which I will investigate the gradient will be f(x)= X5 .

Finally, I have arrived at the conclusion for the first part of Investigation 2. However, before, arriving to this conclusion, I had to analyse my results of the 3rd case. In fact, I can state that when, in this case, f(x)= X5 it means that f1(x)= 5X4

So, as my final conclusion for the first part of investigation number two, I can clearly compare the results I found out with my three cases and come up with a general discussion of the slope of my analysis. In fact, if I look at the cases, every time I increase the number of “n” by 1, every time my power decreased by 1, while “X” was clearly multiplied by the number I chose “n” to be. In a few words, I can say that for the function f(x)= Xn, f1(x) will equal to f1(x)= nXn-1.

Part 2

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This is a preview of the whole essay

Part 2

Then, still for the same investigation, the paper tells me to consider a function of the form f(x)=aXn. Again I will try to find a correlation by using a variety of examples. To help me discover this, I will utilise, again three cases that will give me the possibility to work out the link of this function. Since this time we have two factors that are not constant: “a” and “n”, for every case, I will need to also find out three different answers. For example, for case one I will use n=3. This means that my three other cases for case number 1 will be by substituting “a” with 1,2 and 3. I will do then the same for n=4 and, if necessary, for n=5.

1st Case

The real function I want to study now is f(x)=aX3. For this case I will let a=1 because, the function 1X3, which is X3, I have already found out from previous research. What I found out was that if f(x)=X3 it means that f1(x)= 3X2. I also decided to start with “a” as 1 because it is the first number. This helps me to understand better a correlation, if it exists. To show that I already found out these points, I redrew the table below.

In this case, as I wrote before the answer for f1(x) will be f1(x)=3X2

For this one, I will continue with my sequence of number. Therefore, I will still equal my “n” number 3, however, my “a” will now equal to 2. So, with letter “a” equalling to 2 my function will now be f(x)=2X3

For this one, after looking at the gradient, I realise that the function is similar to the results I found for A, however, this time the gradient is double. So if before, f1(x)=3X2. This time f1(x) is f1(x)=6X2.

For this type of case, now, I will let my “a” number equal to 3. My hypothesis is that, at the end the f1(x) function, will be three times the gradient I found out for f(x)=X3. However, I will do the table normally and check my results.

In fact, my hypothesis, in this case, is correct. The gradient is 3 times bigger than the gradient for the function f(x)=X3. This, thus, means that when a function is f(x)=3X3, f1(x) must be f1(x)=9X2

As a final discussion, for this part, I realised that when “a” is a number, and when n=3, “a” will always be multiplied by 3, and then it will be squared, every time. This can also mean that what I wrote before for the first part of the investigation 2 is correct (f1(x)=nXn-1). However, this time the function is multiplied by the number you give to “a”. Thus, my final hypothesis for this part of the investigation is that for f(x)=aXn, f1(x) must then be f1(x)=a×nXn-1. However, to be sure of this, I will use one more case, to be secure that my hypothesis is correct. But, for now, what I found out from this case is that if f(x)=aX3 then f1(x) is f1(x)=a×3X2.

2nd Case

I will do exactly the same things that I did for the first case of the second part of investigation 2. The function I would like to analyse is f(x)=aX4. So, I will let “a” equal to 1,2 and 3. However, this time there is a change in terms of “n”. In fact I will make it equal to 4. So starting with “a” equalling one and “n” equalling 4, my first function to analyse will be f(x)=1X4, that will become, f(x)=X4. I already found out the results from the preview investigation, however, to be sure it was right I will do it again.

In fact, what I did before was right. So when f(x)=X4, f1(x) must equal to f1(x)=4X3

Now I will still keep my “n” value equal to 4, but this time my “a” number will be 2. Therefore, the function will now become f(x)= 2X4. This is exactly the procedure that I did for the first case.

This time, again, the result of the gradient is exactly double the gradient of the function f(x)=X2. This clearly portrays that when f(x)=2X4, f1(x) is f1(x)=8X3 This can show that it is very likely that my hypothesis is correct since both case 1 and now case two are very similar. To be exactly sure I will now analyse one more function to see if the there is an anomaly of a pretty correct supposition.

Since my hypothesis is very likely to be correct, I just want to study one last function so that I will know that I am certain that my supposition is right. So, lastly, I will keep the value of “n” as 4 and make “a” equal to 3. So now my function will become f(x)=3X4

Exactly as I thought the result is exactly the same procedure as the first case. If the function I analysed now is f(x)=3X4 the result of f1(x) will be f1(x)=12X3. This means two things. Firstly that my hypothesis for the discussion of the scope of my analysis of investigation 2 is correct. Meaning that for when a function is f(x)= aXn my answer for f1(x) will be f1(x)=a×nXn-1. So I believe that investigation 2 is finished. I understand this because, also, at the end of this case I found out the answer for the function f(x)=aX4 will always be f1(x)=a×4X3.

Investigation 3

For this investigation the sheet tells us that I will need to find a conjecture about f1(x) between two functions. The first one will be f(x)=g(x)±h(x) and the second function will be f(x)=g(x)× or ÷h(x). So, also for this investigation, I will need to analyse them separately by dividing them into two parts. Again, for each part I will analyse the cases, and come up with different results that will help me to find the total conclusion for this investigation.

Part 1

1st Case

For the first case of part 1 of investigation 3, I decided to chose the function f(x)=g(x)±h(x). However, In this case, I will allow the function to be f(x)=g(x)+h(x) and then, in the second case, analyse the function f(x)=g(x)-h(x). With the results of both, I will be able to come up with a conclusion for an analysis in part 1. So, as I said before, I will use the function f(x)=g(x)+h(x). I decided to make g(x) equal to X2 and h(x) equal to X3. The function, thus becomes f(x)=X2+X3. I decided to use these two inputs because I already found the answer of f1(x) of both. So, theoretically, I already know the answer without even going to look at the graph. However, to be sure I am correct and, to check my previous investigations, I will do the table anyways. I predict, though, that the answer for the function f(x)=X2+X3 will be f1(x)=2X+3X2.

Exactly what I said in my hypothesis. If we notice the relations, everyone of them have the exact correlation for when f(x)=X2+X3, f1(x) is always f1(x)=2X+3X2. So, for this function the conjecture holds. Now I will do the second case to see if the conjecture holds too.

2nd Case

My second case for this part of the investigation will be to analyse the function f(x)=g(x)-h(x). Like the function before, I will give to g(x) and h(x) two inputs I already analysed to make an hypothesis for the answer of f1(x). In addition, it makes my analysis a lot easier. So, in this case, I will use, for g(x) and h(x), 2X3 and 3X4. Hence my equation will now be f(x)= 2X3-3X4. I predict that my final answer will be f1(x)=6X2-12X3. However, I will do the table to check my hypothesis if it is correct.

Exactly as the first case of part one of investigation 3, from each relation I can clearly see that my conjecture does hold for when f(x)=2X3+3X4, that my answer will be f1(x)=6X2-12X3. This means that as a final discussion for the first part of investigation 3, I can come up with a general formula, for which my conjecture f1(x) holds when the function f(x) is f(x)=g(x)±h(x). The answer for this is f1(x)=g1(x)±h1(x).

Part 2

For part 2 of investigation 3, I will need to check if there is a link about f1(x) between the function f(x)=g(x)× or ÷h(x). Again, I will analyse them separately by dividing this part into two cases. With the two cases, I will then watch my results to conclude with a general formula for this part of my investigation. As, I did for part one, I will give to g(x) and h(x) inputs I already know the answers of, to make a supposition. I will still do a table to then check my results and see if they are correct.

1st Case

For the first case, I decided to look more specifically at the function f(x)=g(x)×h(x). Now I let the two values of g(x) and h(x) equal to X2 and X3. These are exactly the two inputs I used for the first case of part one of this investigation. I decided to use them again because I know they are correct, since they worked out for two investigations. My function I will now analyse, this time, will now be f(x)=X2×X3. From the previous cases I can now exactly say that the conjecture about f1(x) will hold if the final answer will be f1(x)=2X×3X2. This is my prediction. I will however do the table to show my working out.

From the results of my relation here, clearly demonstrates that my conjecture doesn’t hold. I know that my answers were correct from the previous investigations, however, for this one, they don’t seem to work out. This means two things. Firstly, that possibly I made a mistake or secondly, that simply the conjecture doesn’t hold. To check this I will give another value for g(x) and h(x) by still using the function f(x)=g(x)×h(x). This time however I will make g(x) and h(x) be 2X3 and 3X4. So my function now becomes f(x)=2X3×3X4.

Also, for this table the relation don’t seem to work out with the function. This means that I wasn’t incorrect, but that my conjecture doesn’t hold. So, to come up with a formula for this case of part two of the investigation I can say that when f(x) is f(x)=g(x)×h(x), f1(x) will always be f1(x)≠ g1(x)×h1(x). So, my hypothesis I wrote at the start was wrong.

2nd Case

For the second case, I will do exactly the same thing I did for case one of this part of investigation 3. So, since I know that the conjecture about f1(x) for the function f(x)=g(x)×h(x) doesn’t hold. I now predict that also for the function f(x)=g(x)÷h(x) the conjecture won’t hold. However, since for the first case my hypothesis was incorrect, I will now see if the answers of the table will support my prediction. So, for the function f(x)=g(x)÷h(x), I will give the values 2X3 and 3X4 to g(x) and h(x). As I wrote before, I will use these two inputs because I already know the answers of f1(x) of both. So, now, the function I will analyse is f(x)= 2X3÷3X4.

So, also for this function my conjecture doesn’t hold. I know from the previous case that my procedure to find the answer is correct, so, the only possible outcome is that the formula for the first case is exactly the same as the formula for this case. So, when f(x) is f(x)=g(x)÷h(x), f1(x) will equal every time f1(x)≠ g1(x)÷h1(x). So as a conclusion for this investigation I can say that the conjecture about f1(x) holds only for when the function f(x) is f(x)=g(x)±h(x).

Investigation 4

The objective for this investigation is to find maximum and minimum values of the curve y=4X3+21X2-90X+72 by utilising the results obtained in the previous analysis.

By reflecting on the geometrical meaning, the maximum and minimum are points in which a function has horizontal tangents, (y=k), hence of gradient equal to 0.

Therefore, I will operate as it follows:

I will calculate the gradient of the curve by using the formulas I found out in the previous investigations, more specifically I will use:

If f(x) is f(x)=aXn, then f1(x) is f1(x)=a×n×Xn-1
If f(x) is f(x)=g(x)±h(x), then f1(x) is f1(x)=g1(x)±h1(x).

I will impose the gradient equal to 0
I will then resolve the equation, by finding some values to give at the X variable.
I will finally verify from the graphic representation of the curve which values are the minimum and maximum points.

f(x)=4X3+21X2-90X+72

f1(x)=12X2+42X-90

f1(x)=0 12X2+42X-90=0

6(2X2+7X-15)=0

2X2+7X-15=0

From the two values I found, I can clearly notice in my graph that the minimum point will be X=1.5 and, hence, the maximum point will be X=-5.

Investigation 5

This is the last investigation of the assessment. From what it tells me, it is a bit different from the investigations I analysed before. Firstly, this one is a mathematical problem. It tells me that Mr Neuberger sells Mers Bar. Each time, however, for every bar he sells, he needs to spend 15c per bar for the creation of it. It also gives me a table:

It tells me also, that my objective is to find the maximum possible profit by using this table. Firstly I want to calculate, from the table, the best profit he made, therefore, a new data is created.

This table show me that his greatest profit was when he sold 40 bars, making the cost be 30 cents. His second best profit was when he sold 20 bars for 40 cents. This clearly shows me that his possible maximum profit is a number 30 cents and 40 cents. To find the answer to this, I will represent the data of the table graphically.

From this table I have found that there is a linear sequence, so, now I will need to find its formula. So, the first thing to do is to find the gradient. To do this I will use another formula:

= -0.5X

So, now that I know the gradient, I know that the formula for this linear sequence is y=-0.5X+50. I know that the value is 50 because the line passes through that point, for when x=0, therefore y=50. This is the equation of the line of the income. I need now to multiply this, by the number of bars that are required, so, I will give the value of X to the bars. Therefore the equation becomes:

y=X(-0.5X+50)= -0.5X2+50X.

Finally, to work out the maximum profit I will need to subtract form the quadratic line the equation of the cost. This is 15 times the number of bars, which I said before I gave them the value of X. So now everything becomes:

y=-0.5X2+50X-15X=-0.5X2+35X.

This is the equation of the profit; to find the maximum profit I will let y become y1. So, the clear step is this:

y=-0.5X2+35X

y1=-X+35

X=35

Consequently, his maximum profit will be when he will sell bars at 35c. I can also calculate that, if all his sells are linear he will sell exactly 30 bars. I know this because it is the number in between 40 and 20, the number of bars he sold in the best profit of the table. In fact his profit will be of 600c [(35x30)-(30x15)].

With the answer of investigation 5, I can say that my assessment is concluded.