Results 1 to 13 of 13

Math Help - graph

  1. #1
    Member
    Joined
    Aug 2009
    Posts
    211

    graph

    hi, i've got another question on graphs, its part of a worked example, so it actually gives the answer but i don't understand:

    sketch the graph of y = (2x + 3)(x - 1)^2

    the coefficient of x^3 is +2 so the shape of the graph is the same y = x^3 when x is numerically very large (this i understand). When x = 0, y = (3)(-1)^2 = 3 so the graph crosses the y- axis at 3 (this i understand also). When y = 0, (2x + 3)(x - 1)^2 = 0 so x = \frac  {3}{2} - and x = 1 (repeated root) so the graph crosses the x-axis at  x = \frac  {3}{2} - and touches the x-axis at x = 1. When x = -1, y = (1)(-2)^2 = 4 > 3 so the graph must turn in the second quadrant.

    the last two points i don't understand, could someone tell me how you would tell if the graph crosses or only touches the x-axis and how you would tell by using x = -1 or x = + 1 or whatever that it turns in the second quadrant?

    thanks for any help, mark

    ps i tried putting the minus sign on the left for the fractions but it didn't work, i'm going to try to figure out how its done now
    Last edited by mark; September 15th 2009 at 03:19 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,996
    Thanks
    1130
    Quote Originally Posted by mark View Post
    hi, i've got another question on graphs, its part of a worked example, so it actually gives the answer but i don't understand:

    sketch the graph of y = (2x + 3)(x - 1)^2

    the coefficient of x^3 is +2 so the shape of the graph is the same y = x^3 when x is numerically very large (this i understand). When x = 0, y = (3)(-1)^2 = 3 so the graph crosses the y- axis at 3 (this i understand also). When y = 0, (2x + 3)(x - 1)^2 = 0 so x = \frac  {3}{2} - and x = 1 (repeated root) so the graph crosses the x-axis at  x = \frac  {3}{2} - and touches the x-axis at x = 1. When x = 1, y = (1)(-2)^2 = 4 > 3 so the graph must turn in the second quadrant.

    the last two points i don't understand, could someone tell me how you would tell if the graph crosses or only touches the x-axis and how you would tell by using x = -1 or x = + 1 or whatever that it turns in the second quadrant?

    thanks for any help, mark

    ps i tried putting the minus sign on the left for the fractions but it didn't work, i'm going to try to figure out how its done now
    Since the (x-1) term is squared, x= 1 is a double root of the equation. That's why the graph touches rather than crossing at x= 1. Another way to see that is note that near x= 1, 2x+ 3 is always positive so the sign of (2x+3)(x-1)^2 depends only on (x-1)^2. But that is squared so is always postive. It doesn't matter whether x< 1 or x> 2, the function is still positive. Around x=-3/2, (x-1)^2 is always positive so the sign of the function depends on (2x-3) which is negative for x< -3/2 and positive for x> -3/2. So the sign of the function changes from negative to positive: the graph crosses the axis at x= -3/2.

    "When x = 1, y = (1)(-2)^2 = 4 > 3 so the graph must turn in the second quadrant. " is wrong. when x= 1, y= 0. You mean "when x= -1". The value of the function when x= -3/2 is 0, when x= -1 is 4, and when x= 0 is 3. That's what they mean by "turn". The function has gone up from 0 to 4 and then back down to 3 between -3/2 and 0.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Aug 2009
    Posts
    211
    when you say x = 1 is a double root of the equation do you mean the same as "it has two distinct real roots"? because i've learnt that it crosses the x-axis twice when its bigger than 0. i've probably got it mixed up though. i think i understand the stuff about the graph turning now. i'm just confused about this touching the axis crossing the axis business
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Sep 2009
    Posts
    46
    Have you started calculus yet.. like differentiation?

    If the discriminant b2 - 4ac equals zero, the radical in the quadratic formula becomes zero.

    In this case the roots are equal; such roots are sometimes called double roots.

    This applies to your case too.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Aug 2009
    Posts
    211
    no i haven't. one thing i've noticed about this book is that as well as not explaining things well, some things that are really useful to know for certain subjects are just left out till later. why? do you think calculus/differentiation would help?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Sep 2009
    Posts
    46
    Calculus helps enormously when trying to identify turning points on cubic graphs. I don't want to sidetrack the topic too much, but:

    y=2x^3-x^2-4x+3

    This represents the gradient at any point on the graph:
    \frac {dy}{dx}=6x^2-2x-4

    Obviously, the gradient at a turning point is 0:
    0=6x^2-2x-4
    0=2(x-1)(3x+2)

    So clearly, a turning point exists at x=1 and x=-(2/3)

    But nevermind, this is completely meaningless and irrelevant to you! Is there anything else you don't understand from the worked example?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Aug 2009
    Posts
    211
    ha thanks, yeah i don't understand what that means entirely, i noticed that you were factorising though. another thing in this book i don't understand is that one of the questions is sketch the graph of y = x^3 - 1 and another is y = (x - 1)^3. aren't they the same? the graph for the first crosses straight over the x axis whereas the second it lays flat on it then crosses properly, why is this?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Sep 2009
    Posts
    46
    Nooo.. This is a common mistake in mathematics.

    is not the same as

    actually equals:
    y=(x-1)(x-1)(x-1)
    y=x^3-3x^2+3x-1

    That's an entirely different graph to
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Aug 2009
    Posts
    211
    ah yeah, that was a stupid mistake. i still wouldn't know how to draw the graph though.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Sep 2009
    Posts
    46
    Actually, I remember a cute little mnemonic to assist in sketching graphs when calculus hasn't been taught.

    Cubic polynomials, that aren't in the standard form, always tend to have 2 turning points... a hump (maximum point) and a U (minimum point).

    Mneomonic:
    When the x^3 term is positive, the camel's hump is on the left.

    When the x^3 term is negative (eg. -x^3) the camel's hump is on the right.

    Lol, in retrospect it's actually rather pathetic.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Aug 2009
    Posts
    211
    lol sounds pretty good, i understand the turning point stuff now. i just don't get how you can tell whether it crosses or touches the x-axis. no doubt i will in time though because i've managed to understand everything else so far. cheers for the help
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    Sep 2009
    Posts
    46
    Just keep it simple

    If there's a squared it touches, if there's not it crosses.

    EG.


    (2x+3).. is it squared? No, so therefore it crosses this root (aka x-intercept).

    (x-1)^2... is it squared? Yes, so therefore it touches the root (x-intercept).
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Member
    Joined
    Aug 2009
    Posts
    211
    that actually seems to work with some of these questions, i'll let you know how i get on
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Graph Theory / Chromatic Number of a Complete Graph
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: November 15th 2011, 09:59 AM
  2. Replies: 7
    Last Post: August 5th 2011, 01:30 PM
  3. Replies: 0
    Last Post: September 25th 2010, 05:59 AM
  4. Graph Theory - Size of a Line Graph
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: July 25th 2010, 11:15 PM
  5. Replies: 1
    Last Post: February 15th 2009, 05:35 AM

Search Tags


/mathhelpforum @mathhelpforum