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Does anyone need help with Edexcel maths (C1,C2,M1,D1,S1,S2 and FP1)?
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Yeah, C1, I can't do transformations. Like f(x) whatever. I just don't get it.

S2 Bionomial distribution
1st example:
My method was:
X~B(50,0.8)
so...
Y~B(50,0.2)
find P(X<30).
P(X<30) > 0....29 so 30 cases of success, 30...50 so 21 cases of failure
now I have to find a probability in terms of Y where I have a simmilar ratio of fails to succeses but reversed
soo...
P(Y<21) > 0....20 > 21 cases of succes (@p 0.2) and 21...50 > 30 cases of fail (@0.8)
soo...
P(X<30)=1  P(Y<21) So I got it right! It worked!
2nd example
X~B(25,0.8)
so
Y~B(25,0.2)
find P(X>20) > 21...25 > 5 cases of succes (@p=0.8) and 0...20 > 21 cases of fail (@p=0.2)
So:
P(Y<21) > 0..20 > 21 cases of succes (@p=0.2) and 21..25 > 5 cases of fail (@p=0.8)
So
P(X>20) = 1  P(Y<21) = 1  1 = 0 WRONG!!! I should get 0.4027
Please tell me why this method worked in the first example but did not in the second one. And please talk in detail, no one was able to give me detailed answer yet 
YES!!! c1 please! i dont get the whole of chapter 4!!
we haven't been taught it in school and i have my exam in jan!
please help! 
Thanks for the offer, I'll be appearing in this thread with some c1 questions soon.

what is there in chapter 4?

(Original post by GGAMSHING)
YES!!! c1 please! i dont get the whole of chapter 4!!
we haven't been taught it in school and i have my exam in jan!
please help! 
The best way to learn is to write all the transformation rules on a little paper and keep practicing transformation questions. Going back to your little note whenever you are in doubt.
The more you practice, the more likely it will stick. Once it does it is really easy.
f(x) + a => graphs moves up (just add 'a' to ycoordinates and leave x coordinates unchanged). Also know as 'vertical transformation'. Note: 'a' will do the opposite that is graph moves down.
f(x+a) => graphs moves to the left (subtract 'a' for xcoordinates, leave ycoordinates unchanged). Also know as 'horizontal transformation'. If f(xa) graph moves to the right.
f(ax) => multiply x coordinates by 1/a (Compression by a scale factor of 1/a)
af(x) => multiply y coordinates by a (Enlargement by a scale factor of a)
Hope that helps. Good luck 
(Original post by Ihategcse)
what is chap. 4.. i have the book like 10 inches away but cba to get it lol 
Please help with this q:
Find the exact solutions to:
b. e^x + 3e^x = 4 
(Original post by Moa)
S2 Bionomial distribution
1st example:
My method was:
X~B(50,0.8)
so...
Y~B(50,0.2)
find P(X<30).
P(X<30) > 0....29 so 30 cases of success, 30...50 so 21 cases of failure
now I have to find a probability in terms of Y where I have a simmilar ratio of fails to succeses but reversed
soo...
P(Y<21) > 0....20 > 21 cases of succes (@p 0.2) and 21...50 > 30 cases of fail (@0.8)
soo...
P(X<30)=1  P(Y<21) So I got it right! It worked!
2nd example
X~B(25,0.8)
so
Y~B(25,0.2)
find P(X>20) > 21...25 > 5 cases of succes (@p=0.8) and 0...20 > 21 cases of fail (@p=0.2)
So:
P(Y<21) > 0..20 > 21 cases of succes (@p=0.2) and 21..25 > 5 cases of fail (@p=0.8)
So
P(X>20) = 1  P(Y<21) = 1  1 = 0 WRONG!!! I should get 0.4027
Please tell me why this method worked in the first example but did not in the second one. And please talk in detail, no one was able to give me detailed answer yet
X<30 => 29 28 27 26.......... which means X<29
Y will then be 21 22 23 24 ......... which means Y>21 (Because X and Y should add up to 50)
Therefore P(Y>21) = 1  P(Y<20) 'because if u remove '020' you are left with '2150'.
So if we use the same logic on the second example,
P(X>20) = 21 22 23 24 25 i.e X>21
So Y = 4 3 2 1 0 i.e Y<4
using tables P(Y<4)= 0.4207
Hope I explained it properly. 
I'm off now, will be back tommorrow or later tonight. So I'll go through the other questions then. Bye

(Original post by Qk_Sol)
The best way to learn is to write all the transformation rules on a little paper and keep practicing transformation questions. Going back to your little note whenever you are in doubt.
The more you practice, the more likely it will stick. Once it does it is really easy.
f(x) + a => graphs moves up (just add 'a' to ycoordinates and leave x coordinates unchanged). Also know as 'vertical transformation'. Note: 'a' will do the opposite that is graph moves down.
f(x+a) => graphs moves to the left (subtract 'a' for xcoordinates, leave ycoordinates unchanged). Also know as 'horizontal transformation'. If f(xa) graph moves to the right.
f(ax) => multiply x coordinates by 1/a (Compression by a scale factor of 1/a)
af(x) => multiply y coordinates by a (Enlargement by a scale factor of a)
Hope that helps. Good luck 
(Original post by Jingers)
Please help with this q:
Find the exact solutions to:
b. e^x + 3e^x = 4 
(Original post by TheButtylicious)
Yeah, C1, I can't do transformations. Like f(x) whatever. I just don't get it.
I hate chapter 4 
C1 MATHS EDEXCEL
Do we need to be able to prove how we get b+/sqr root b^24ac / 2a from ax^2+bx+c=0 ?
I know some proofs we need to able to show and others we are just shown in books to help with our understanding, so I don't know whether I need to be able to prove how to get this formula?
Thanks 
Hi, could someone help me with this question?
Solve the equation by completing the square: 3x^2  5x = 4
Thank you so much! 
(Original post by TheButtylicious)
Yeah, C1, I can't do transformations. Like f(x) whatever. I just don't get it.(Original post by GGAMSHING)
well, its all about graphs and transformation.
1. [f(x)+1] is a movement of +1 on the y axis.
Think of it like this if the value is outside the bracket as in outside the (x) then it acts on the Y AXIS.
[f(x)+2] would act on the Y axis, as would [f(x)+500]. So anything outside the brackets acts on the y axis. So, if there is a graph of y=x^2, a movement of [f(x)+1] means you move all the y value up by one so the value (0,0) now becomes (0,1).
If it was [f(x)1] it would become (0,1) because it is a movement of one.. but downwards.
Recap: If it is outside the (x) then it acts on the Y axis. If it is () then it moves down, if it is (+) it moves up.
2. [f(x+1)] is a movement of 1 on the X AXIS
If the value is INSIDE like so: (x+1) it acts on the x axis. Just remember if it is inside the bracket with the 'x' it acts on the 'x' because they are 'together.' Going back to the graph of y=x^2. If we were to say transform the graph like so: [f(x+1)] we would move it to the LEFT. Its best remembered if you think about graph sketching, if you have the values (x+1) (x2) (x+5), the points you would draw on the x axis would be: 1, +2, 5. Kind of like the opposite of what is written in the bracket. So, apply the same principal here, it is a movement of in essence 1. So that means the new coordinates of the original point (0,0) would now be (1,0). However, if it said transform as such:
[f(x1)] you move it +1 to the right. So remember, do the opposite when inside the brackets. So if this were to be applied to the graph of y=x^2, the coordinates (0,0) would now be (1,0)
3. If the transformations is: [2f(x)]
If you remember what i previously said, if it is outside the brackets it acts on the Y AXIS. This holds true here as well. This means that we are basically multiplying all the y values by '2', the x values remain that same. So what does that mean? The graph gets longer. The the graph of x^2, when x=2, y also =2. (2,2). If we say transform it by [2f(x)] we would multiply the y value by 2, resulting in: (2,4).
4. If the transformation is [1/2f(x)] Then apply the same principals from (3). However, because it is '1/2' that means you multiply the Y value by '1/2' which is a half. So the coordinate (2,2) now becomes (2,1). As you may have noticed no changes in the 'x' coordinate.
5. If the transformation is [f(2x)]. We are now acting on the X AXIS, because it is inside the brackets. Remember how earlier we said if it was (x+1) we move it 1 opposite of what it suggests? Well, the same thing applies here. if it is [f(2x)] we DIVIDE it by 2. Or basically halve it. So the coordinate (2,2) divide the x value by 2 and you get (1,2).
6. If the transformation is [f(1/2x)] We are gain acting on the x axis. Similar principal to (5) because it says 1/2x we double it/multiply it by 2. SO going back to our coordinate (2,2) it now becomes (4,2) as the x coordinate has been doubled.
7. If the transformation is [f(x)] we flip the graph, so in the x^2 graph now is negative. Instead of going up, it goes down. If the graph was already going down it would flip up. The minimum point (0,0) of the x^2 graph now becomes the maximum point. Or think of it as a multiplication of all the x values by 1. so (2,2) would be (2,2)
8. If the transformation is [f(x)]. We multiply all the y values by 1. So the coordinate (2,2) now would be (2,2)
If you have any questions please feel free to ask. I have tried my best lol :P.

(Original post by zdo0o)
multiply both sides by e^x 
(Original post by Ihategcse)
I'll try an explain the basics.
1. [f(x)+1] is a movement of +1 on the y axis.
Think of it like this if the value is outside the bracket as in outside the (x) then it acts on the Y AXIS.
[f(x)+2] would act on the Y axis, as would [f(x)+500]. So anything outside the brackets acts on the y axis. So, if there is a graph of y=x^2, a movement of [f(x)+1] means you move all the y value up by one so the value (0,0) now becomes (0,1).
If it was [f(x)1] it would become (0,1) because it is a movement of one.. but downwards.
Recap: If it is outside the (x) then it acts on the Y axis. If it is () then it moves down, if it is (+) it moves up.
2. [f(x+1)] is a movement of +1 on the X AXIS
If the value is INSIDE like so: (x+1) it acts on the x axis. Just remember if it is inside the bracket with the 'x' it acts on the 'x' because they are 'together.' Going back to the graph of y=x^2. If we were to say transform the graph like so: [f(x+1)] we would move it to the LEFT. Its best remembered if you think about graph sketching, if you have the values (x+1) (x2) (x+5), the points you would draw on the x axis would be: 1, +2, 5. Kind of like the opposite of what is written in the bracket. So, apply the same principal here, it is a movement of in essence 1. So that means the new coordinates of the original point (0,0) would now be (1,0). However, if it said transform as such:
[f(x1)] you move it +1 to the right. So remember, do the opposite when inside the brackets. So if this were to be applied to the graph of y=x^2, the coordinates (0,0) would now be (1,0)
3. If the transformations is: [2f(x)]
If you remember what i previously said, if it is outside the brackets it acts on the Y AXIS. This holds true here as well. This means that we are basically multiplying all the y values by '2', the x values remain that same. So what does that mean? The graph gets longer. The the graph of x^2, when x=2, y also =2. (2,2). If we say transform it by [2f(x)] we would multiply the y value by 2, resulting in: (2,4).
4. If the transformation is [1/2f(x)] Then apply the same principals from (3). However, because it is '1/2' that means you multiply the Y value by '1/2' which is a half. So the coordinate (2,2) now becomes (2,1). As you may have noticed no changes in the 'x' coordinate.
5. If the transformation is [f(2x)]. We are now acting on the X AXIS, because it is inside the brackets. Remember how earlier we said if it was (x+1) we move it 1 opposite of what it suggests? Well, the same thing applies here. if it is [f(2x)] we DIVIDE it by 2. Or basically halve it. So the coordinate (2,2) divide the x value by 2 and you get (1,2).
6. If the transformation is [f(1/2x)] We are gain acting on the x axis. Similar principal to (5) because it says 1/2x we double it/multiply it by 2. SO going back to our coordinate (2,2) it now becomes (4,2) as the x coordinate has been doubled.
7. If the transformation is [f(x)] we flip the graph, so in the x^2 graph now is negative. Instead of going up, it goes down. If the graph was already going down it would flip up. The minimum point (0,0) of the x^2 graph now becomes the maximum point. Or think of it as a multiplication of all the x values by 1. so (2,2) would be (2,2)
8. If the transformation is [f(x)]. We multiply all the y values by 1. So the coordinate (2,2) now would be (2,2)
If you have any questions please feel free to ask. I have tried my best lol :P.
but can u help me on one of the questions in the mixed exercise on chapter 4?
the question is number 5
 the curve with equation y=f(x) meets the coordinate axes at the points (1,0), (4,0) and (0,3), as shown in the diagram.
using separate diagrams for each, sketch the curve with equation
a) y=f(x1) b) y=f(x)
on each sketch, write in the coordinates of the points at which the curve meets the coordinate axes.
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