Results 1 to 8 of 8

Math Help - C3 question

  1. #1
    Member
    Joined
    Sep 2008
    Posts
    239

    C3 question

    f(x) = |2x + 5| for all real values of x

    a) Sketch the graph y = f(x), showing the coordinates of any points where the graph meets the coordinate axes.

    b) evaluate ff(-4)

    g(x) = f(x + k) for all real values of x

    c) state the value of the constant k for which g(x) is symmetrical about the y axis.


    OK for part a i have sketched a line crossing the y axis at +5 and bouncing off the x axis at x=-2.5

    Can anybody confirm this is right?


    For part b i'm not sure what to do. I guess u work out f(-4)= |-8 + 5|

    =|-3| = 3 ....???? then work out f(3)....????

    part c im unsure of...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Jan 2009
    Posts
    36
    Quote Originally Posted by djmccabie View Post
    f(x) = |2x + 5| for all real values of x

    a) Sketch the graph y = f(x), showing the coordinates of any points where the graph meets the coordinate axes.

    b) evaluate ff(-4)

    g(x) = f(x + k) for all real values of x

    c) state the value of the constant k for which g(x) is symmetrical about the y axis.


    OK for part a i have sketched a line crossing the y axis at +5 and bouncing off the x axis at x=-2.5

    Can anybody confirm this is right?


    For part b i'm not sure what to do. I guess u work out f(-4)= |-8 + 5|

    =|-3| = 3 ....???? then work out f(3)....????

    part c im unsure of...

    The points on the graph are correct. It would never cross the x-axis.

     f(x) = |2x + 5|

     ff(-4):

     f(f(-4)) = f( |2(-4) + 5| ) = f( |-8 + 5|) = f(|-3|) = f(3) = |2(3) + 5| = |6 + 5| = 11.

    So yes you are correct here as well.


    g(x) = f(x + k)  for all real values of x

    c) state the value of the constant k for which g(x) is symmetrical about the y axis.

    f(x) = |2x + 5|
    g(x) = f(x+k) = |2(x+k) + 5| = |2x + 2k + 5|
    The answer would be k = -2.5. Try drawing that graph and you'll see it is symmetrical about the y-axis.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2009
    Posts
    36
    fixed it
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Sep 2008
    Posts
    239
    Thanks a lot for your help (quite a lot today haha) just a bit unser about part C

    I understand how you get |2x + 2k + 5|

    But i'm not sure how you derive k=-x
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jan 2009
    Posts
    36
    Sorry i got that bit wrong, but i've fixed it now.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Sep 2008
    Posts
    239
    hi i still dont know how you get this :/ there are 2 unknows in the eaution and it is not equal to anything...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Jan 2009
    Posts
    36
    The only way you will have a symmetric graph of the form  y = |ax + c| is if c = 0. If  c \neq 0 then it won't be symmetrical about the y-axis. So you want to work out what k to put in that will cancel out the +5.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Sep 2008
    Posts
    239
    THANKS! i understand it now
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum