Rational Function help

Wahamuka

Wahamuka

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The question is how many times does the graph of the rational function g(x)=6x^2 - 5x - 6 / 6x^2 -13x + 6 cross the x-axis?

The answer was given already to be that it crosses the x-axis exactly once but I cant find out why
Ive tried using b^2 -4ac on the numerator but that tells me it crosses twice. So I figured maybe there is an asymptote that prevents the second x-intercept from occurring.

Can somebody explain why this question is so weird?
 
Wahamuka said:
The question is how many times does the graph of the rational function g(x)=6x^2 - 5x - 6 / 6x^2 -13x + 6 cross the x-axis?

The answer was given already to be that it crosses the x-axis exactly once but I cant find out why
Ive tried using b^2 -4ac on the numerator but that tells me it crosses twice. So I figured maybe there is an asymptote that prevents the second x-intercept from occurring.

Can somebody explain why this question is so weird?
Click to expand...
The function you are talking about is (6x^2 - 5x - 6)/(6x^2 - 13x + 6). You need the parentheses in there. (And yes, I know many instructors leave them out. They are wrong to do so.)

I'd do the division, and then we have \(\displaystyle g(x) = 1 + \dfrac{8x - 12}{6x^2 - 13x + 6} = 1 + \dfrac{4(2x - 3)}{(2x - 3)(3x - 2)} = 1 + \dfrac{4}{3x - 2}\)....edited
(For all x except 2x - 3 = 0).

What can you say about how many times this crosses the x-axis?

-Dan
 
Last edited by a moderator:
Wahamuka said:
… tried using b^2-4ac on the numerator but that tells me it crosses twice …
Click to expand...
Hello Wahamuka. The positive discriminant tells us that 6x^2-5x-6 has two roots, but that doesn't necessarily mean the rational function also has two roots. It's possible that a root of the numerator is also a root of the denominator. Such a root is not in the function's domain, so it can't be an x-intercept.

Dan's approach in post #2 is one method. It allows you to see the answer without calculating actual roots or the domain. (It also makes the horizontal asymptote obvious, if you were graphing).

Another approach is to factor both numerator/denominator and then cancel the common factor. You'll get a ratio of linear polynomials, and that means the rational function has one root and one vertical asymptote.

(The cancelled factor's root gives the location of a point discontinuity -- a hole in the given function's graph that the new, simplified function doesn't have. Other than that single difference between their domains, the new function is the same as the given function.)

If you would like more help, please show your relevant work. Cheers

Edited to clarify the difference between given function and simplified version

?
 
I like topsquark's solution best, but here is another way to see the answer that is closer to your original thought process.

[MATH]f(x) = \dfrac{6x^2 - 5x - 6}{6x^2 - 13x + 6x}, \text { provided that } 6x^2 - 13x + 6x \ne 0.[/MATH]
You are quite correct to conclude that

[MATH]f(a) = 0 \implies 6a^2 - 5x - 6 = 0 \implies a = \dfrac{5 \pm \sqrt{25 + 144}}{12} = \dfrac{5 \pm \sqrt{13^2}}{12} = \dfrac{3}{2} \text { or } -\ \dfrac{2}{3}.[/MATH]
But does f(x) exist at both those points? It does not exist if the denominator is zero.

[MATH]6x^2 - 13x + 6x = 0 \implies x = \dfrac{13 \pm \sqrt{169 - 144}}{12} = \dfrac{13 \pm \sqrt{5^2}}{12} = \dfrac{3}{2} \text { or } \dfrac{2}{3}.[/MATH]
[MATH]\dfrac{6 * \left ( \dfrac{3}{2} \right )^2 - 5 * \dfrac{3}{2} - 6}{6 * \left ( \dfrac{3}{2} \right )^2 - 13 * \dfrac{3}{2} + 6} = \dfrac{\dfrac{54 - 30 - 24}{4}}{\dfrac{54 + 24 - 78}{4}} = \dfrac{0}{0} \text { is indeterminate.}[/MATH]
So it can't cross at x = 3/2 because f(x) does not exist there.

Trick question. I think trick questions are stupid.

EDIT: My answer and Otis's mean exactly the same thing. He posted while I was writing.
 
As a supplement to my previous post, Otis suggested to me that I reconsider my previous post. He and I may disagree about the stupidity of trick questions, or perhaps we merely disagree about what constitutes a trick question.

In any case, I have reconsidered. If the question posed did not indicate that the function given is not defined for all reals and if the teaching context in which it was given was focused on finding the zeroes of rational functions, then it is a trick question of the type I strongly dislike. But those suppositions may be false. I do not know the exact wording of the question, nor do I know the context in which the question was posed. Therefore, I should not have been so quick to give a pejorative label to the question.

I leave it up to the OP to clarify.
 
Wahamuka said:
The question is how many times does the graph of the rational function g(x)=(6x^2 - 5x - 6) / (6x^2 -13x + 6) cross the x-axis?

The answer was given already to be that it crosses the x-axis exactly once but I cant find out why.
I've tried using b^2 -4ac on the numerator but that tells me it crosses twice. So I figured maybe there is an asymptote that prevents the second x-intercept from occurring.

Can somebody explain why this question is so weird?
Click to expand...
My impression is that this is a tricky trick question (sort of a double-cross), possibly intended to teach a valuable lesson about caution.

By asking only how many times the graph crosses the axis, it gives the impression that you should be able to answer by a "trick" in the positive sense, without having to do all the work. You could even call it a temptation to do so. And that is what the OP did, using the discriminant to find the number of zeros of the numerator without actually solving or even factoring. That sounds like a smart thing to do, but isn't -- it's too smart for one's own good.

It turns out that the trick fails because one of the two apparent crossings corresponds not to an x-intercept of the rational function, but to a "hole", where both the numerator and the denominator are zero. It's necessary to look at the whole thing in order to be sure what happens.

Now, the kind of trick question that really gets me is one where such a trick (bypassing an actual solution) is what they want you to do, but it turns out that "their" answer is wrong, because the solution turns out not to exist at all. This is similar to those, but since "their" answer is correct, it looks like they know that you have to graph the entire function, or at least think thoroughly about it rather than take the misleading shortcut. That's tricky, but not evil ...

But, yes, I'd like to see the context, to see how much enticement there was for falling into the trap.
 
Yes, I told Jeff that I actually like this exercise, and that I hoped the teacher had covered all the bases first. I agree with Jeff about trick questions of the type where students haven't been prepared. (I'm also not a fan of secret screening or when epiphany is the general expectation.) I also agree with the doc; exercises like this can be used to teach a lesson (eg: be mindful of domain). The OP mentioned asymptotes, so I'd assumed graphing lessons had already covered what they needed to know but they had tripped over a shortcut.

?
 
topsquark said:
\(\displaystyle 1 + \dfrac{4(2x - 3)}{(2x - 3)(3x - 2)} = 1 + \dfrac{4}{2x - 3}\)
Click to expand...
topsquark, your (2x - 3) factors are supposed to cancel with each other. So, the denominator is 3x - 2.

It looks like this is an example of a typo.
 
JeffM said:
I like topsquark's solution best, but here is another way to see the answer that is closer to your original thought process.

[MATH]f(x) = \dfrac{6x^2 - 5x - 6}{6x^2 - 13x + 6x}, \text { provided that } 6x^2 - 13x + 6x \ne 0.[/MATH]
You are quite correct to conclude that

[MATH]f(a) = 0 \implies 6a^2 - 5x - 6 = 0 \implies a = \dfrac{5 \pm \sqrt{25 + 144}}{12} = \dfrac{5 \pm \sqrt{13^2}}{12} = \dfrac{3}{2} \text { or } -\ \dfrac{2}{3}.[/MATH]
But does f(x) exist at both those points? It does not exist if the denominator is zero.

[MATH]6x^2 - 13x + 6x = 0 \implies x = \dfrac{13 \pm \sqrt{169 - 144}}{12} = \dfrac{13 \pm \sqrt{5^2}}{12} = \dfrac{3}{2} \text { or } \dfrac{2}{3}.[/MATH]
[MATH]\dfrac{6 * \left ( \dfrac{3}{2} \right )^2 - 5 * \dfrac{3}{2} - 6}{6 * \left ( \dfrac{3}{2} \right )^2 - 13 * \dfrac{3}{2} + 6} = \dfrac{\dfrac{54 - 30 - 24}{4}}{\dfrac{54 + 24 - 78}{4}} = \dfrac{0}{0} \text { is indeterminate.}[/MATH]
So it can't cross at x = 3/2 because f(x) does not exist there.

Trick question. I think trick questions are stupid.

EDIT: My answer and Otis's mean exactly the same thing. He posted while I was writing.
Click to expand...

This was actually my second solution after realizing why the discriminant method did not work, but I didn’t write it here. It turns out I did some algebra wrong while trying to find the indeterminate solution thats why I posted in the first place. I learned additional info from topsquark’s comment so thank you :)

Also this was taken from a multiple choice exam so for you guys arguing about if it was a trick question, I believe it was because there was a choice there that it crosses the x-axis 2 times. Haha...

Sorry for the late reply but I was busy reviewing for other subjects. Anyway I really appreciate the help guys :))
 
Wahamuka said:
… I did [something wrong] while trying to find the indeterminate solution [and] thats why I posted in the first place …
Click to expand...
Hi. I had thought you posted because you were thinking that maybe an asymptote prevents the second x-intercept from occurring.

? \(\;\) Next time, when your steps do not produce a result you expect, please post your work so we can see what you're doing. Cheers!

\(\;\)
 
Wahamuka said:
Also this was taken from a multiple choice exam so for you guys arguing about if it was a trick question, I believe it was because there was a choice there that it crosses the x-axis 2 times. Haha...
Click to expand...
Another thing worth knowing for the future is that you should always post the entire question, including the choices. They can be important, because they affect what you know, and sometimes how you can solve the problem. You can see how that can affect how we perceive the problem.
 
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