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Getting equation from graph of a rational function with an oblique asymptote.

  • Thread starter Iran11
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Homework Statement


-I have the zero, which is x=-1, however its a squared zero {(x+1)^2)}
-Vertical asymptote is at x=1
-Equation of oblique asymptote is y=x+4


Homework Equations





The Attempt at a Solution


I tried finding the numerator by multiplying the oblique asymptote by the vertical asymptote, but the equation becomes a degree 4 on top and degree 1 on bottom.

Anyone know how to start me off? Any help is appreciated!
 

Answers and Replies

  • #2
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Homework Statement


-I have the zero, which is x=-1, however its a squared zero {(x+1)^2)}
-Vertical asymptote is at x=1
-Equation of oblique asymptote is y=x+4


Homework Equations





The Attempt at a Solution


I tried finding the numerator by multiplying the oblique asymptote by the vertical asymptote
???
, but the equation becomes a degree 4 on top and degree 1 on bottom.
For there to be an oblique asymptote, the degree of the numerator must be 1 more than the degree of the denominator.
Anyone know how to start me off? Any help is appreciated!
From your problem statement,
a zero of multiplicity 2 at x = -1 means that the numerator has to have a factor of (x + 1)2.

For a vertical asymptote at x = 1, there has to be a factor of (x - 1) in the denominator.

For an oblique asymptote of y = x + 4, after doing polynomial long division, the result must be x + 4 + C/(x -1).
 
  • #3
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???
For there to be an oblique asymptote, the degree of the numerator must be 1 more than the degree of the denominator.


From your problem statement,
a zero of multiplicity 2 at x = -1 means that the numerator has to have a factor of (x + 1)2.

For a vertical asymptote at x = 1, there has to be a factor of (x - 1) in the denominator.

For an oblique asymptote of y = x + 4, after doing polynomial long division, the result must be x + 4 + C/(x -1).
So to find C, I have to sub in a point, lets say the y-intercept? What I don't understand is where does the zero at x = -1 of multiplicity 2 go? Thanks for the reply btw
 
  • #4
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I got the eqn. (x^2 + 2x+ 1)/(x-1), all the points work out however when I divide them I don't get the oblique asymptote y=x+4, is this ok?
 
  • #5
SammyS
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So to find C, I have to sub in a point, lets say the y-intercept? What I don't understand is where does the zero at x = -1 of multiplicity 2 go? Thanks for the reply btw
I got the eqn. (x^2 + 2x+ 1)/(x-1), all the points work out however when I divide them I don't get the oblique asymptote y=x+4, is this ok?
What is the y-intercept?

When you say you have a zero of multiplicity 2 (in the denominator, I presume), is that because the graph goes to +∞ on both sides of the vertical asymptote (or -∞ on both sides)?

I suggest that the rational function can be written as [itex]\displaystyle f(x)=x+4+\frac{C}{(x+1)^2}\,.[/itex]
 
  • #6
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No, it's not OK because you don't have the right oblique asymptote.

You want to find polynomials p(x) and q(x) so that p(x)/q(x) = x + 4 + C/q(x).

p(x) will have a factor of (x + 1)2 and q(x) will have a factor of (x -1), but each of these will have other factors.
 
  • #7
i have the similar givens to as well y-int is -1 x-int is -1 oblique asymptote at y=x+4
vertical asymptote x=1
 
  • #8
SammyS
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i have the similar givens to as well y-int is -1 x-int is -1 oblique asymptote at y=x+4
vertical asymptote x=1
madeincanada, welcome to PF.

You should really start your own thread for this.

How would you approach this problem after reading the above posts?
 
  • #9
well the my problem was similar to the topic starters.
well i would use the equation above, and sub in y intercept to solve for c, and multiply the equation x-1 to find the quadratic equation.
 
  • #10
so i got c = 5 y=x+4+[5/(X-1)] (x-1)
(x^2+3x+1)/(x-1) except i don't get all the correct givens.
 
  • #11
SammyS
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I think you need c = 6 .
 

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