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

    0
    ReputationRep:
    questionn 4):

    the function f is even and has domain R. For x is greater or equal to 0, f(x) = x^2 - 4ax, where a is a +ve constant.

    a) sketch the curve showing all coordinates at which curve meets the axes.



    i never am able to sketch things with a constants because you obviously cant plot it. can neone help me on how i go about sketching with unknown constants in functions?
    Offline

    2
    ReputationRep:
    The easiest way for this is to complete the square so f(x) = (x - 2a)^2 - 4a^2 so the curve is a parabola shifted 2a to the right and 4a^2 down

    To find the intercepts, first sub in x=0, f(0) = 0 so y-intercept is 0. Then factorise f(x) = x(x - 4a) so curve crosses the x-axis at x = 0 and x = 4a.
    Offline

    2
    ReputationRep:
    (Original post by Jubba)
    ...

    i never am able to sketch things with a constants because you obviously cant plot it. can neone help me on how i go about sketching with unknown constants in functions?
    You can plot it by letting the constant a=1 or any other "easy" value, that looks keeping the function simple-looking and then see what that looks like.
    Offline

    14
    ReputationRep:
    (Original post by Jubba)
    questionn 4):

    the function f is even and has domain R. For x is greater or equal to 0, f(x) = x^2 - 4ax, where a is a +ve constant.

    a) sketch the curve showing all coordinates at which curve meets the axes.



    i never am able to sketch things with a constants because you obviously cant plot it. can neone help me on how i go about sketching with unknown constants in functions?
    this is a stupid question :mad:
    Offline

    14
    ReputationRep:
    (Original post by Jubba)
    questionn 4):

    the function f is even and has domain R. For x is greater or equal to 0, f(x) = x^2 - 4ax, where a is a +ve constant.

    a) sketch the curve showing all coordinates at which curve meets the axes.

    i never am able to sketch things with a constants because you obviously cant plot it. can neone help me on how i go about sketching with unknown constants in functions?
    well it is x^2 parabola, and it is shifted down and right (but still passing through 0) so that the x-intercept is 4a..

    Think about it, when x = 0, y = 0..
    Factorise:

    x^2 - 4ax = x(x - 4a)
    x can equal 4a and give f(x) = 0 then, so the place where it crosses the x axis a second time is at 4a.
 
 
 
Turn on thread page Beta
Updated: June 15, 2004
Poll
Do you like carrot cake?
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.