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

[Iran11]

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!

 

[Mark44]

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).

 

[Iran11]

?
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, let's 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

 

[Iran11]

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]

So to find C, I have to sub in a point, let's 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]

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.

 

[madeincanada]

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]

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?

 

[madeincanada]

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.

 

[madeincanada]

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]

I think you need c = 6 .