Join TSR now and get all your revision questions answeredSign up now

Core 1 Graph Question, someone help please? Watch

    • Thread Starter
    Offline

    2
    ReputationRep:
    http://www.mei.org.uk/files/papers/c107ja_3hoza9.pdf


    Can someone kindly give me a step by step solution to question 11 on this paper, it would be greatly appreciated
    Offline

    3
    ReputationRep:
    (Original post by liverpool2044)
    http://www.mei.org.uk/files/papers/c107ja_3hoza9.pdf


    Can someone kindly give me a step by step solution to question 11 on this paper, it would be greatly appreciated
    For part (i) draw suitable lines on the graph to find the points of intersection. Part (ii) is simultaneous equations, and part (iii) you have to consider the graphical plot of the given circle.
    • Thread Starter
    Offline

    2
    ReputationRep:
    Okay thanks i now understand how to do 2 and 3, can you help me with the equations of the graphs i need to draw? like the first one, how do i rearrange it to do it?
    Offline

    3
    ReputationRep:
    (Original post by liverpool2044)
    Okay thanks i now understand how to do 2 and 3, can you help me with the equations of the graphs i need to draw? like the first one, how do i rearrange it to do it?
    Well the graph is y = x + (1/x), and it is asking you to find x + (1/x) = 4. So y = ...?

    EDIT: sorry, it is x + (1/x)
    • Thread Starter
    Offline

    2
    ReputationRep:
    y=4? so the intercept is 4?
    Offline

    3
    ReputationRep:
    (Original post by liverpool2044)
    y=4? so the intercept is 4?
    That's the y coordinate of the intercepts, you need to find the x coordinates because the equation is just in terms of x. How would you do this, knowing that y=4?
    • Thread Starter
    Offline

    2
    ReputationRep:
    ohh look for the x intersection with the graph, thank you!

    What about the second one?
    Offline

    3
    ReputationRep:
    (Original post by liverpool2044)
    ohh look for the x intersection with the graph, thank you!

    What about the second one?
    Exactly what did you get as your points of intersection?

    This one is a little more tricky, you want to end up with x + (1/x) on one side of the equation (so you can use the graph of y = x + (1/x)), so you need to do a bit of rearranging. HINT: consider the 2x term
    • Thread Starter
    Offline

    2
    ReputationRep:
    If its one extra x then surely i just move it over so its =4-x? then draw that line and find the intersections?
    Offline

    3
    ReputationRep:
    (Original post by liverpool2044)
    If its one extra x then surely i just move it over so its =4-x? then draw that line and find the intersections?
    Exactly! Well done.
    • Thread Starter
    Offline

    2
    ReputationRep:
    thanks for your help appreciated, do you reckon you could help me on the other thread im stuck on?
    Offline

    3
    ReputationRep:
    (Original post by liverpool2044)
    thanks for your help appreciated, do you reckon you could help me on the other thread im stuck on?
    Sure, post a link.
    • Thread Starter
    Offline

    2
    ReputationRep:
    http://www.mei.org.uk/files/papers/J..._2013_June.pdf

    12iii? I got to the equation and i know i need to use b^2-4ac but i dont know how to with the long equation?
    Offline

    3
    ReputationRep:
    (Original post by liverpool2044)
    http://www.mei.org.uk/files/papers/J..._2013_June.pdf

    12iii? I got to the equation and i know i need to use b^2-4ac but i dont know how to with the long equation?
    What equation did you get?
    • Thread Starter
    Offline

    2
    ReputationRep:
    x^2 - (k+2)x+2k+1=0
    Offline

    3
    ReputationRep:
    (Original post by liverpool2044)
    x^2 - (k+2)x+2k+1=0
    Looks good to me! Now you have to apply b^2 - 4ac... remember that k is a constant.
    • Thread Starter
    Offline

    2
    ReputationRep:
    This is where im confused! i dont know what to put in where
    Offline

    3
    ReputationRep:
    (Original post by liverpool2044)
    This is where im confused! i dont know what to put in where
    



ax^2 + bx + c = 0

    Just match this up with your equation. I'm guessing you're having a problem with the "c" term?
    • Thread Starter
    Offline

    2
    ReputationRep:
    i think its the bx thats confusing me, so its b^2 - 4 =0?
    Offline

    3
    ReputationRep:
    (Original post by liverpool2044)
    i think its the bx thats confusing me, so its b^2 - 4 =0?
    Nope how many x^2's are there (this is your a). How many x's are there (this is your b). All the other terms which do not have an x in front of them are your "c", just add them all together.
 
 
 
Poll
How are you feeling about your A-level results?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

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

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