1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Solving Rational Functions

Tags:
  1. Jan 10, 2017 #1
    1. The problem statement, all variables and given/known data
    Sketch the graphs of the following functions and show all asymptotes with a dotted line
    y = (2x - 6)/ (x2-5x+4)
    upload_2017-1-10_12-28-44.png

    i) Equation of any vertical asymptote(s)

    ii) State any restrictions or non-permissible value(s)

    iii) Determine coordinates of any intercept(s)

    iv) Describe the behavior of the function as it approaches and leaves vertical asymptotes and/or point of discontinuity

    v) State the horizontal asymptote.


    vi) State the Domain and Range

    2. Relevant equations


    3. The attempt at a solution
    I hope my thread title is correct.
    First off I factored the function
    y = 2(x-3)/(x-4)(x-1)

    i) This gave me my vertical asymptotes: x = 4, x = 1

    ii) Without any points of discontinuity then I don't have any restrictions or non-permissible values (I think)

    iii) X intercept
    0 = (2x-6)/(x2-5x+4)
    (0)(x2-5x+4) = 2x-6
    0 = 2x - 6
    6 = 2x
    x = 3
    (3,0)

    iii) Y intercept
    y = 2(0)-6/(0)2-5(0)+4
    y = -6/4
    y = -3/2
    (0,-3/2)

    iv) I believe I can do this once I figure out my graph

    v) Horizontal Asymptote: y = 2

    vi) D: x ≠ 4, 1

    Now I tried finding my range but substituting my horizontal asymptote into the function
    2 = (2x-6)/(x2-5x+4)
    2x2-10x+8 = 2x-6
    2x2-12x+14 = 0
    Using quadratic equation I get 3±√2

    Now when I put all this onto my graph I don't know where to draw the lines. I think I've placed all my lines where I should but something seems wrong.
     
    Last edited: Jan 10, 2017
  2. jcsd
  3. Jan 10, 2017 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You wrote two functions
    y = 2x - 6/ x2-5x+4 which means ##2x-\frac{6}{x^2}-5x+4##
    and
    y = 2(x-3)/(x-4)(x-1), ##y=2\frac{x-3}{x-4} (x-1)##
    but you seem to work with ##y=\frac{2x-6}{x^2-5x+4}##
    Which function is the real one?
     
    Last edited by a moderator: Jan 10, 2017
  4. Jan 10, 2017 #3
    I see that may have been confusing. I should have written it like this
    y = (2x-6)/(x2-5x+4)
     
  5. Jan 10, 2017 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Yes, it is correct now. And it is 2(x-3)/[(x-4)(x-1)] in factorized form.
    You were right, the vertical asymptotes are at x=1 and x=4. Are these x values permissible for the function? Is the function really continuous?
    How is the horizontal asymptote defined? I do not think y=2 is correct.
     
  6. Jan 10, 2017 #5
    I was under the impression that when you had a function like 2(x-3)/[(x-4)(x-1)] the 2 in the numerator is divided by the denominator and since it is 2/1 then y = 2
     
  7. Jan 10, 2017 #6

    ehild

    User Avatar
    Homework Helper
    Gold Member

    No, you get the horizontal asymptote at the limit x→ -∞ and x→ ∞. If x is very big the constant terms can be ignored, and the function is approximated by 2x/x2=2/x.
     
  8. Jan 10, 2017 #7

    ehild

    User Avatar
    Homework Helper
    Gold Member

    So, what is the horizontal asymptote?
     
  9. Jan 10, 2017 #8
    If its 2/x would you make 2/x = 0?
     
  10. Jan 10, 2017 #9

    ehild

    User Avatar
    Homework Helper
    Gold Member

    yes, y=0 is the horizontal asymptote.
     
  11. Jan 10, 2017 #10

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    No. You have the correct asypmptote, but have expressed it incorrectly.

    You NEVER can have 2/x = 0; however, you do have that 2/x approaches 0 for large |x|, so ##\lim_{x \to \pm \infty} 2/x = 0## would be the right way of saying it.
     
  12. Jan 10, 2017 #11
    So if y = 0 then my range is y ≥ 0 because it doesn't cross the horizontal asymptote. I still don't know how to draw my graph. How do I find out which way to go with it?
     
  13. Jan 10, 2017 #12

    ehild

    User Avatar
    Homework Helper
    Gold Member

    No. Determine the sign of the function when x<1. Remember you obtained y=-3/2 as y intercept.
    Determine the sign of the function in the various domains.
     
  14. Jan 10, 2017 #13
    Ok, so now I start my graph from underneath the horizontal asymptote and go through the y intercept then curve it downwards -∞?
     
  15. Jan 10, 2017 #14
    Also when it says to state non-permissibles. Does that only mean when there are points of discontinuity? Or am I supposed to state the restrictions on x as well? Like x ≠ 4, 1
     
  16. Jan 10, 2017 #15

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Yes. And how will the function look at the other side of the x=1 asymptote?
     
  17. Jan 10, 2017 #16
    If it goes down to -∞ then it will start at the top of the graph coming from ∞ then going through my (3,0) intercept down through -∞ and back to ∞ on the right side of x=4 asymptote and curving towards the right but never crossing y = 0?
     
  18. Jan 10, 2017 #17
    upload_2017-1-10_14-48-9.png Is this how it should look? Because I was looking through my notes and it says that if you have an even exponent then the graph will look different?
     
  19. Jan 10, 2017 #18

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Looks good to me.
    Here's a handy trick: p(x)/q(x) = p(x)q(x)/q(x)2, so, where defined, has the same sign as p(x)q(x). Listing all the roots of pq in order quickly shows the domains in which it is positive and which negative.
     
  20. Jan 10, 2017 #19

    ehild

    User Avatar
    Homework Helper
    Gold Member

    It is correct. Remember, at great x values, both positive and negative, the function looks similar to 2/x (even exponent in the denominator but you have x in the numerator, too.)
    Now: what is the range of the function?
     
  21. Jan 10, 2017 #20
    Range has to be y ∈ R because it crosses the horizontal asymptote, I believe.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Solving Rational Functions
Loading...