Math Is Fun Forum

FP2Student

Guest

Graphs of Rational Functions

I'm stuck on this question.

Find a condition on p such that

take all values as x varies.

I think that you are meant to change it to the quadratic equation:

And then solve it by saying

But I end up getting

Which I don't know how to solve.

Thank you in advance for your help.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Graphs of Rational Functions

Hi FP2Student;

I am not sure I understand your question. In the rational form:

It is clear that for p >=0 , y can take all positive values and maybe all negative values too. If I introduce p<0 then I will have a pole in the denominator and y will never be able to take all values. It will always be undefined at x^2 = Abs(p).

Last edited by bobbym (2009-09-28 07:16:26)

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Graphs of Rational Functions

It is clear that for p >=0 , y can take all positive values and maybe all negative values too... It will always be undefined at x^2 = Abs(p).

This is not true.  For a quick example, take p = 0.  The only value this function takes is 1.  Another example is p = 1, where my function now becomes:

This function only takes on values from about -0.2 to 1.2, and further it is defined everywhere.

It should be clear that if p >= 0, then the function will not have a vertical asymptote (even if it isn't defined at a point as is the case when p=0).  But not all p < 0 will work.  Graph the case where p = -1, -1.01, and -0.99.  To figure out what's going on, remember (or alternatively, learn now) what can happen if you have a function in the form of 0/0.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Graphs of Rational Functions

Hi Ricky;

I didn't mean for every p, that y could take all positive values.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

FP2Student

Guest

Re: Graphs of Rational Functions

Hey,

Thank you for your replies, I have trieddrawing graphs and I realised that it has to be p<-1 because when p>0, then there are turning points that restrict y from taking any value.

When p=0, I have found that y will always be 1.

And then p=-1, then there is a horizontal asymptote.

Anything below -1 results in 3 parts of the graph which cover all the y values.

I am wondering if there is any algebraic proof of this?

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Graphs of Rational Functions

What happens for -1 < p < 0?

I am wondering if there is any algebraic proof of this?

What you want to show is that there are two vertical asymptotes, and the inside has one side going up to infinity and the other side goes down to negative infinity.  You can show that this is the case for p < 0, p ≠ 1 by limits.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."