Homework Help: Solving Rational Functions

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1. Jan 10, 2017

Schaus

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)

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. Jan 10, 2017

ehild

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
3. Jan 10, 2017

Schaus

I see that may have been confusing. I should have written it like this
y = (2x-6)/(x2-5x+4)

4. Jan 10, 2017

ehild

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.

5. Jan 10, 2017

Schaus

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

6. Jan 10, 2017

ehild

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.

7. Jan 10, 2017

ehild

So, what is the horizontal asymptote?

8. Jan 10, 2017

Schaus

If its 2/x would you make 2/x = 0?

9. Jan 10, 2017

ehild

yes, y=0 is the horizontal asymptote.

10. Jan 10, 2017

Ray Vickson

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.

11. Jan 10, 2017

Schaus

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?

12. Jan 10, 2017

ehild

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.

13. Jan 10, 2017

Schaus

Ok, so now I start my graph from underneath the horizontal asymptote and go through the y intercept then curve it downwards -∞?

14. Jan 10, 2017

Schaus

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

15. Jan 10, 2017

ehild

Yes. And how will the function look at the other side of the x=1 asymptote?

16. Jan 10, 2017

Schaus

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?

17. Jan 10, 2017

Schaus

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?

18. Jan 10, 2017

haruspex

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.

19. Jan 10, 2017

ehild

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?

20. Jan 10, 2017

Schaus

Range has to be y ∈ R because it crosses the horizontal asymptote, I believe.

21. Jan 10, 2017

ehild

Yes y takes all real values. It has y=0 as asymptote, but takes the value y=0 at x=3.

22. Jan 11, 2017