Hey there! Sign in to join this conversationNew here? Join for free
 You are Here: Home >< Maths

# Quick C1 question Watch

1. This has preceding questions so I'll just fill you in:

Part I)

Describe fully the transformation that transforms the curve to the curve .

I know that's a translation of 2 units in the negative x direction parallel to the x axis.

Then it asks to differentiate the first curve which gives

Now this last part I can't understand how I'm doing it wrong:

Part iii)

Use the preceding parts of the question to find the gradient of the curve at the point where it crosses the y axis.

Well surely when it crosses the y axis, x = 0 - if I differentiate the curve and put in x as zero you get infinity..?
2. Hmm I understand but you get the same result dont you?

Call me an idiot but can it not be written as:

3. Differentiating that gives exactly the same as the first curve - anyone help?
4. (Original post by Dededex)
This has preceding questions so I'll just fill you in:

find the gradient of the curve at the point where it crosses the y axis.

Well surely when it crosses the y axis, x = 0 - if I differentiate the curve and put in x as zero you get infinity..?
I think it's asking for the gradient value of the (x+2)^-1 not the x^-1 graph,

for y = (x+2)^-1
when x = 0, y = 1/2

The y = (x+2)^-1 graph has only been translated to the left by 2 relative to y=x^-1
So the gradient values of the new curve could be thought of as translated to the left by 2 as well.

Sketch the graphs.
5. (Original post by GabsyWabsyWoo)
The y = (x+2)^-1 graph has only been translated to the left by 2 relative to y=x^-1
So the gradient values of the new curve could be thought of as translated to the left by 2 as well.

Sketch the graphs.
Kind of - I've drawn them but still don't understand why you cant get the correct answer with x = 0
6. (Original post by Dededex)
Kind of - I've drawn them but still don't understand why you cant get the correct answer with x = 0
on y=(x+2)^-1 the gradient at x=0 is what we want

on y=x^-1, the gradient at x=2 is -1/4 as found by the gradient function y'=-(x^-2)

if tangents are equal but translated we can assume what the y=(x+2)^-1 gradient is at x=0
7. Sorry for the bad wording. :/ in C1 you "shouldn't know" how to differentiate (x+2)^-1
You learn that in C3; you use what's called the Chain Rule.

Reply
Submit reply
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: December 22, 2010
Today on TSR

### Cambridge interview invitations

Has yours come through yet?

### Official Oxford interview invite list

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll
Useful resources

## Make your revision easier

### 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

Can you help? Study help unanswered threads

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• 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

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.