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

• Iran11
Wait! How did you get that equation? Did you use polynomial long division?That's not the correct 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

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!

Iran11 said:

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
?
Iran11 said:
, 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.
Iran11 said:
Anyone know how to start me off? Any help is appreciated!

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).

Mark44 said:
?
For there to be an oblique asymptote, the degree of the numerator must be 1 more than the degree of the denominator.

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

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?

Iran11 said:
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 said:
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 $\displaystyle f(x)=x+4+\frac{C}{(x+1)^2}\,.$

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

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

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.

I think you need c = 6 .

1. How do you determine the equation of a rational function with an oblique asymptote from its graph?

The equation of a rational function with an oblique asymptote can be determined by finding the slope of the oblique asymptote and the coordinates of the point where the asymptote intersects the x-axis. These values can then be used to write the equation of the rational function in the form y = mx + b, where m is the slope and b is the y-intercept.

2. What is the significance of an oblique asymptote in a rational function?

An oblique asymptote in a rational function indicates that the function approaches a non-horizontal line as the input values get larger or smaller. This is in contrast to a vertical asymptote, which indicates that the function approaches infinity as the input values approach a certain value.

3. Can a rational function have more than one oblique asymptote?

Yes, a rational function can have more than one oblique asymptote. This can happen when the function has multiple branches or when the numerator and denominator have the same degree.

4. How do you graph a rational function with an oblique asymptote?

To graph a rational function with an oblique asymptote, first plot the oblique asymptote using the equation determined from the graph. Then, plot points on either side of the asymptote and connect them to create the graph of the function. Make sure to also include any vertical asymptotes and points of discontinuity.

5. Can a rational function have an oblique asymptote at a vertical line?

No, a rational function cannot have an oblique asymptote at a vertical line. This is because a vertical line has an undefined slope, and the equation y = mx + b requires a defined slope. A rational function can only have an oblique asymptote at a non-vertical line.

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