- Level: International Baccalaureate
- Subject: Maths
- Word count: 3641
Investigating Slopes Assessment
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Introduction
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.
f(x)= X2 | ||
X=? | Tangent | Gradient |
X= -2 | Y= -4x+(-4) | -4 |
X= -1 | Y= -2x+(-1) | -2 |
X= 0 | Y= 0 | 0 |
X= 1 | Y= 2x+(-1) | 2 |
X= 2 | Y= 4x+(-4) | 4 |
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
Middle
X=?
Tangent
Gradient
X= -2
Y= 36x+44
36
X= -1
Y= 9x+6
9
X= 0
Y= 0
0
X= 1
Y= 9x+(-6)
9
X= 2
Y= 36x+(-44)
36
- 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.
f(x)= X4 | ||
X=? | Tangent | Gradient |
X= -2 | Y= -32x+(-48) | -32 |
X= -1 | Y= -4x+(-3) | -4 |
X= 0 | Y= 0 | 0 |
X= 1 | Y= 4x+(-3) | 4 |
X= 2 | Y= 32x+(-48) | 32 |
- 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.
f(x)= 2X4 | ||
X=? | Tangent | Gradient |
X= -2 | Y= -64x+(-96) | -64 |
X= -1 | Y= -8x+(-6) | -8 |
X= 0 | Y= 0 | 0 |
X= 1 | Y= 8x+(-6) | 8 |
X= 2 | Y= 64x+(-96) | 64 |
- 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
f(x)= 3X4 | ||
X=? | Tangent | Gradient |
X= -2 | Y= -12x+(-9) | -12 |
X= -1 | Y= -96x+(-144) | -96 |
X= 0 | Y= 0 | 0 |
X= 1 | Y= 12x+(-9) | 12 |
X= 2 | Y= 96x+(-144) | 96 |
- 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.
f(x)=X2+X3 | |||
X=? | Tangent | Gradient | Relation |
X= -2 | Y= 8X+12 | 8 | -4+12 |
X= -1 | Y= X+1 | 1 | -1+1 |
X= 0 | Y= 0 | 0 | 0 |
X= 1 | Y= 5x+(-3) | 5 | 1+1 |
X= 2 | Y= 16x+(-20) | 16 | 4+12 |
Conclusion
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.
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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