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

## 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

## 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!

Mark44
Mentor

## 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

## 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).

???
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

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?

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 $\displaystyle f(x)=x+4+\frac{C}{(x+1)^2}\,.$

Mark44
Mentor
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.

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

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?

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.

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.

SammyS
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I think you need c = 6 .