x Turn on thread page Beta
 You are Here: Home >< Maths

# help c3 question! no 1 on student room can explain this so far? watch

1. f(x) has coordinates (-1,8) (1,0) (2,-1) (3,0).

y= |f(x+1)|

mark scheme shows that the graph transforms to (8,0) (2,0) (3,1) (4,0). therefore it shifted to the right instead of the left. help why??

y=|f(2x)-1|

for this expression what happens??

thanks
2. (Original post by victor_b1992)
y= modules f(x+1) modules

the graph moves to the right why is this. also what happens if you have

y=modules f(2x) -1 modules

thanks
it f(x+1) is a translation by -1 in the horizontal axis. The modulus of this would mean that this graph would reflect in the x axis (anything below it would go up)

f(2x) is a 'stretch' sf 1/2 in the horizontal axis. the -1 would mean a translation of -1 in the y-axis. The modulus of this would also mean anything below gets reflected above.
3. (Original post by victor_b1992)
y= modules f(x+1) modules

the graph moves to the right why is this. also what happens if you have

y=modules f(2x) -1 modules

thanks
Its transfromations so the first moves right in the x-axis by 1. The second one moves down by 1 in the y-axis and is stretched by 1/2 in the x-axis. Dont for get about the modules part reflects in the x-axis.
4. Modulus? Why not just use the | key? (Shift + backslash)

You should know from C1 that is a negative translation by a units parallel to the x-axis, so the same applies here. It makes sense if you think about it; your value of y takes the value that f(x) would take if x were one more than it is, so y = f(0) at x=-1 and so on, so it moves to the left.

A similar thing happens with (although you'll need to look back at C1 to know what f(2x) - 1 does to f(x)); but bear in mind that since the modulus signs are around the entire RHS, y cannot drop below zero.
5. (Original post by unamed)
it f(x+1) is a translation by -1 in the horizontal axis. The modulus of this would mean that this graph would reflect in the x axis (anything below it would go up)

f(2x) is a 'stretch' sf 1/2 in the horizontal axis. the -1 would mean a translation of -1 in the y-axis. The modulus of this would also mean anything below gets reflected above.

that's what i thought but as it modulus for f(x+1) it translates by +1 in the horizontal axis. and then like normal
6. It Not Like That It Moves To The Right
7. (Original post by nuodai)
Modulus? Why not just use the | key? (Shift + backslash)

You should know from C1 that is a negative translation by a units parallel to the x-axis, so the same applies here. It makes sense if you think about it; your value of y takes the value that f(x) would take if x were one more than it is, so y = f(0) at x=-1 and so on, so it moves to the left.

A similar thing happens with (although you'll need to look back at C1 to know what f(2x) - 1 does to f(x)); but bear in mind that since the modulus signs are around the entire RHS, y cannot drop below zero.
I NO BUT THAT WRONG THAT WHAT I DID BUT AS IT GOT || IT MOVES TO THE RIGHT.
8. (Original post by victor_b1992)
that's what i thought but as it modulus for f(x+1) it translates by +1 in the horizontal axis. and then like normal
There's something wrong with the markscheme then.

(Original post by nuodai)
Modulus? Why not just use the | key? (Shift + backslash)
Because my laptop doesn't have one! {which is why I detest latex}
9. (Original post by victor_b1992)
It Not Like That It Moves To The Right
Nope, that would be rather than .

Take this for example; with (in blue in the diagram in the link above), then (in purple).
10. (Original post by nuodai)
Nope, that would be rather than .

Take this for example; with (in blue in the diagram in the link above), then (in purple).
I think he knows what should happen, but the markscheme says otherwise. He wants to know why.
11. NO THE modules it around f(x) so y=|f(x+1)| therefor it should move to the left by 1 unit however in the mark scheme it moves to the right by 1 unit. and i think this is become of the modules around the function.

do you understand me
12. (Original post by victor_b1992)
NO THE modules it around f(x) so y=|f(x+1)| therefor it move to the right instead of the left. this is what it shows in my book
what is f(x)?
13. (Original post by unamed)
what is f(x)?
it just a curve that goes through points (-1,8) (1,0) (2,-1) (3,0). so the y=|f(x+1)| the graph transforms to (8,0) (2,0) (3,1) (4,0)

WHY?
14. can any1 explain this???
15. surely some1 on this site can explain this
16. Each point is moved leftwards by 1.

Your new coordinates are: (-2,8), (0,0), (1,-1), (2,0)

The y-values are then all positivised (not a word btw), leaving you with:

(-2,8), (0,0), (1,1), (2,0)

That's the answer. Either the mark scheme is wrong, or you've written the question out wrong.
17. I agree with Porkstein. That is correct. You either have the worng question or the mark scheme is wrong.

What board is this from?
18. (Original post by Bhav_Singh)
I agree with Porkstein. That is correct. You either have the worng question or the mark scheme is wrong.

What board is this from?
edexcel. the question is not wrong but the mark scheme might be wrong. however i thought as it was can modules instead of moving it to the left you move it to the right??
19. what month and year? I have a website i can check.
20. (Original post by victor_b1992)
edexcel. the question is not wrong but the mark scheme might be wrong. however i thought as it was can modules instead of moving it to the left you move it to the right??
you DON'T move it to the right with the modulus, you do the transformation WITHOUT the modulus, and then reflect any thing UNDERNEATH the x-axis up. Markscheme=wrong.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: January 19, 2010
Today on TSR

### How do I turn down a guy in a club?

What should I do?

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

## Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE