Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    21
    (Original post by Zacken)
    Sounds like whatever answers you're looking at are ****ed to hell. :lol:
    So I'm not going crazy :lol: Well I'll get cracking with part (b) I'll update you in a second...(or a more like 5 minutes)
    Offline

    2
    (Original post by homeland.lsw)
    K well this is another one...yuck

    The sketch shows the curve  y = x^2 - 5
    and the normal at  (2 , -6)

    a) Find the equation of the normal at the point
     (2 , -6)

    So I did
    y=x^2 -5
    dy/dx = 2x

    2x=0
    2(2)= 4 (The two coming from the given point)

    so 4 is the gradient of the tangent (?)
    Then -1/4 is the gradient of the normal (negative reciprocal)

    So y= -1/4x + c
    -6= -1/4 (2) + c
    -11/2= c

    Therefore
    y= -1/2x + 11/2 for part a?

    Zacken
    Are you sure you have written the equation for the curve/the coordinates of the point correctly because (2, -6) does not lie on that curve?
    • Thread Starter
    Offline

    21
    (Original post by QT123)
    Are you sure you have written the equation for the curve/the coordinates of the point correctly because (2, -6) does not lie on that curve?
    Yup this is the question direct from the book...
    Name:  image.jpg
Views: 40
Size:  504.7 KB

    Zacken (in case you were curious)

    Edit sorry for the upside downedness...
    Offline

    22
    ReputationRep:
    (Original post by QT123)
    Are you sure you have written the equation for the curve/the coordinates of the point correctly because (2, -6) does not lie on that curve?
    Good point... not sure how I didn't spot that.

    Can you post a picture of the question, homeland?
    Offline

    22
    ReputationRep:
    (Original post by homeland.lsw)
    Yup this is the question direct from the book...
    Name:  image.jpg
Views: 40
Size:  504.7 KB

    Zacken (in case you were curious)

    Edit sorry for the upside downedness...
    This book seems very incompotent, the sketch given doesn't even look like x^2 - 5 at all. :laugh:

    The actual question should be y = x^2 - 5x.
    Offline

    18
    ReputationRep:
    for the benefit of everyone's neck...

    • TSR Support Team
    • Very Important Poster
    • Welcome Squad
    Offline

    20
    ReputationRep:
    TSR Support Team
    Very Important Poster
    Welcome Squad
    (Original post by Student403)
    for the benefit of everyone's neck...
    Thank you .
    • Thread Starter
    Offline

    21
    (Original post by Zacken)
    This book seems very incompotent, the sketch given doesn't even look like x^2 - 5 at all. :laugh:

    The actual question should be y = x^2 - 5x.
    I was thinking that because surely at x=0 , y=-5 so would be no where near the origin...:dontknow:


    (Original post by Student403)
    for the benefit of everyone's neck...
    Like I said, 11 gemmers know what they are doing
    • Thread Starter
    Offline

    21
    Zacken so that means my part (a) is wrong? :getmecoat:
    Offline

    22
    ReputationRep:
    (Original post by homeland.lsw)
    Zacken so that means my part (a) is wrong? :getmecoat:
    According to the way the question is written, it's correct. Because the question is wrong.

    Do it again with y = x^2 - 5x though.
    • Thread Starter
    Offline

    21
    (Original post by Zacken)
    According to the way the question is written, it's correct. Because the question is wrong.

    Do it again with y = x^2 - 5x though.
    dy/dx = 2x - 5
    Then?
    2x-5=0
    2x=5
    x=5/2

    or
    2x-5=0
    2(2)-5
    -1?

    I'm sorry, I'm confused af
    Offline

    22
    ReputationRep:
    (Original post by homeland.lsw)
    2x-5=0
    2(2)-5
    -1?

    I'm sorry, I'm confused af
    Should be this 2(2) - 5, but you shouldn't say =0.

    Just: gradient of tangent is 2(2) - 5 = 4-5 = -1.

    So gradient of normal is 1.
 
 
 
  • 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
    Would you like to hibernate through the winter months?
    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
  • 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

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