Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta
    • Thread Starter
    Offline

    2
    ReputationRep:
    Hi, I'm doing this question for Core 1 revision but I can't find a and b.

    Given that y=ax^2+bx-a^2 has gradient -4 at (-2,-13) find possible values for a and b.

    I've differentiated the equation to get dy/dx=2ax+b-2a
    Then I put the gradient in
    -4=2a (-2)+b-2a
    Now I'm at -4=-6a+b and I don't know what to do next or whether I've done something wrong.
    Offline

    21
    ReputationRep:
    You remove -a^2 , there's no -2a after. You are differentiating relative to x so you move (poor description) powers of x, if a value isn't a coefficient of x or a power of it then its removed.
    Offline

    22
    ReputationRep:
    (Original post by mydearestpotato)
    Hi, I'm doing this question for Core 1 revision but I can't find a and b.

    Given that y=ax^2+bx-a^2 has gradient -4 at (-2,-13) find possible values for a and b.

    I've differentiated the equation to get dy/dx=2ax+b-2a
    Then I put the gradient in
    -4=2a (-2)+b-2a
    Now I'm at -4=-6a+b and I don't know what to do next or whether I've done something wrong.
    Re-check your differentiation, the a^2 is just a constant. Otherwise it's like you saying:

    Well, if I differentiate 4 = 2^2 then I get 2 \times 2^{2-1}...

    The next equation is

    -13 = a(-2) + b(-2) - a^2

    Now solve these two simultaneously (once you've fixed your first question).
    Offline

    18
    ReputationRep:
    I would help you but idk what a gradient is
    Offline

    18
    ReputationRep:
    (Original post by GUMI)
    I would help you but idk what a gradient is
    Seriously
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Vikingninja)
    You remove -a^2 , there's no -2a after. You are differentiating relative to x so you move (poor description) powers of x, if a value isn't a coefficient of x or a power of it then its removed.
    Oh I see what you mean. I've tried again and I got dy/dx=2ax+b
    Offline

    22
    ReputationRep:
    (Original post by mydearestpotato)
    Oh I see what you mean. I've tried again and I got dy/dx=2ax+b
    Yes, that's correct.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    Re-check your differentiation, the a^2 is just a constant. Otherwise it's like you saying:

    Well, if I differentiate 4 = 2^2 then I get 2 \times 2^{2-1}...

    The next equation is

    -13 = a(-2) + b(-2) - a^2

    Now solve these two simultaneously (once you've fixed your first question).

    Okay so dy/dx=2ax+b

    Then to make them equal would it be 2ax+b+4=-13+2a+2b+a^2
    Offline

    22
    ReputationRep:
    (Original post by mydearestpotato)
    Okay so dy/dx=2ax+b

    Then to make them equal would it be 2ax+b+4=-13+2a+2b+a^2
    Huh? Where did you get that from?!

    Do you know what a simultaneous equation is and how to solve one?

    You have two equations:

    2a(-2) + b = -4

    -13 = 2a - 2b - a^2

    Now you can subsitute b = -4 - 2a(-2) into the second equation.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    Yes, that's correct.
    Thank you
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Zacken)
    Huh? Where did you get that from?!

    Do you know what a simultaneous equation is and how to solve one?

    You have two equations:

    2a(-2) + b = -4

    -13 = 2a - 2b - a^2

    Now you can subsitute b = -4 - 2a(-2) into the second equation.
    I've solved it now, took me a while haha
    I got a=-7, b=-32 and a=3,b=8
    Offline

    22
    ReputationRep:
    (Original post by mydearestpotato)
    I've solved it now, took me a while haha
    I got a=-7, b=-32 and a=3,b=8
    Well done.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: May 8, 2016
Poll
Are you going to a festival?
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

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