Using Information to Sketch Graphs of Rational Functions

[Freddie Neisi]

Let me start but saying that I hope this is the right thread section! I assume it is as it includes trigonometry, which is included in my studies of this course I'm taking currently (MHF4U - Advanced Functions)

Use the information below to sketch the function.
* There is a horizontal asymptote at f(x) = 1
*The denominator of the function is x^2+x-6
*There is a y-intercept at 0
*f(-2) = f(-4) = 0
*f(4) = f(1) = f(-2.5) = 1
*f(x)<0 when -4 < x < -3, -2 < x < 0
*f(x) > 0 when x < -4, -3 < x < -2, 0 < x < 2, x > 2


As far as I can comprehend currently, the horizontal asymptote is located at y=1 on the positive rising scale, which I already have marked, the equation for me right now is just BLANK/x^2+x-6, and the y intercept is 0. That is about all that I was able to comprehend, and I lost myself at the f(x) information, I only need to understand how to form the equation. Any help would be appreciated, and I would like to apologize beforehand if this is the wrong section, feel free to delete the post if it is.

 

[Dr.Peterson]

Let me start but saying that I hope this is the right thread section! I assume it is as it includes trigonometry, which is included in my studies of this course I'm taking currently (MHF4U - Advanced Functions)

Use the information below to sketch the function.
* There is a horizontal asymptote at f(x) = 1
*The denominator of the function is x^2+x-6
*There is a y-intercept at 0
*f(-2) = f(-4) = 0
*f(4) = f(1) = f(-2.5) = 1
*f(x)<0 when -4 < x < -3, -2 < x < 0
*f(x) > 0 when x < -4, -3 < x < -2, 0 < x < 2, x > 2


As far as I can comprehend currently, the horizontal asymptote is located at y=1 on the positive rising scale, which I already have marked, the equation for me right now is just BLANK/x^2+x-6, and the y intercept is 0. That is about all that I was able to comprehend, and I lost myself at the f(x) information, I only need to understand how to form the equation. Any help would be appreciated, and I would like to apologize beforehand if this is the wrong section, feel free to delete the post if it is.

I'd call this advanced or intermediate algebra, but it's close to calculus. That doesn't really matter.

What we need to see is your sketch; that's the goal, not writing the formula for the function, so don't worry at all about the latter.

Have you factored the denominator to find the vertical asymptotes? Have you found other x-intercepts besides 0? Have you noted the direction in which it approaches each asymptote?

And what does "on the positive rising scale" mean?

Show us anything you can do in a sketch, and we'll have more to discuss.

 

[Steven G]

To have a horizontal asymptote of y=1 would mean that the numerator is of degree two with the coefficient of x^2 being 1. Why is that??
But there is a problem! There are three x-intercepts. They are at at x=0, x=-2 and x=-4. This would mean that the degree of f(x) would be at least three.

Even though I fell that the problem is flawed you seem not to know how to convert zeros to factors. There is a theorem that states the following: x=a is a root (or zero) of a function is the same as (x-a) is a factor of the function.

Possibly you did not realize that if f(-2)=0, then x=-2 is a root and hence (x+2) must be a factor.

 

[Dr.Peterson]

To have a horizontal asymptote of y=1 would mean that f(x) is of degree two with the coefficient of x^2 being 1. Why is that??
But there is a problem! There are three x-intercepts. They are at at x=0, x=-2 and x=-4. This would mean that the degree of f(x) would be at least three.

But f(x) is not a polynomial; did you mean that the numerator would have degree 2?

I think you're right that there is a flaw; I think one of the factors of the denominator will turn out to be squared, so that asymptote has multiplicity 2. Then the numerator can have degree 3.

But the question is about sketching the graph, so the error wouldn't be realized until the student goes beyond and thinks about the equation based on the graph.

 

[Steven G]

But f(x) is not a polynomial; did you mean that the numerator would have degree 2?

I think you're right that there is a flaw; I think one of the factors of the denominator will turn out to be squared, so that asymptote has multiplicity 2. Then the numerator can have degree 3.

But the question is about sketching the graph, so the error wouldn't be realized until the student goes beyond and thinks about the equation based on the graph.

Thanks Dr Peterson for finding my error which is now updated. Yes, I meant that the degree of the numerator should be 2.

Here is my thought process to realize that there was a flaw in the problem. If a rational expression has a horizontal asymptote then that means that the degree of the numerator equals the degree of the denominator. The denominator clearly has degree 2 while the numerator has three zeros.

 

[Steven G]

To OP,
Once you understand that the problem is flawed you should change the fact that the y-intercept is at 0 to a more appropriate number.