Parabolas: Getting a Parabola that Does Not Cross X or Y Axis

[offtheleft]

now, this isn't required for my class but my professor thought it would be cool if anyone in our class could figure it out.

now, I am armed with a ti-nspire calculator and a macbook pro with this nifty grapher application and (despite my limit of knowledge in this field) no matter what i put in i can't quite seem to get it. haha

well here's the question (ill try an word it as best i can):

we want to get a parabola that doesn't cross the x (or y) axis. when we were going over it in class, he said that wed be going into imaginary numbers. does anyone have any idea what he's talking about?

id also like to find out how to turn a parabola on its side, but that's just for me.

heres a list of stuff i had tried...

first thing i did, i took $$y=x^2$$ and tried to move it around with the calculator. i didnt get to far.

i put in multiple equations into the calc and the grapher application. and couldn't really get anywhere. still, probably because of my lack of knowledge of this.

 

[CompuChip]

Well, x^2 + 1 doesn't cross the x-axis... do you mean you want to make a parabola which crosses neither axis?
Also, turning it sideways you should have x = y^2 instead of y = x^2. Try doing something with that.

 

[offtheleft]

well the sideways thing was beneficial. i want the parabola out there... crossing neither axis.

 

[chaoseverlasting]

What you need to do is rotate the axis. Although I don't think you could have a parabola that crosses neither axis... eventually at some point, it will. Maybe you could choose such a parabola that it touches the x and y axes at infinity.

If you take a general parabola (opening to the right),

$$(y-k)^2=4a(x-h)$$ (what you've written is the case for h=0, k=0, a=1/4)

and you apply the transformation,

x'=lx-my
y'=mx+ly

where l and m are between -1 and 1, you get your rotated parabola. If you know something about limits, there's probably another way to do it using the general 2 degree equation of conics.

 

[Defennder]

Hmm, I can't prove it mathematically but it doesn't seem possible. Even if you orientate the parabola 45 degrees clockwise rotation or some other fancy orientation it always appears to cross at least one axis. This seems true even for the degenerate case.

 

[CompuChip]

If you have any parabola
$$y = ax^2 + bx + c$$
the roots (y = 0) are given by
$$x = \frac{-b \pm \sqrt{b^2 - 4 a c}}{2 a}$$

You probably learned that if the thing inside the root becomes negative, there are no solutions. In more advanced mathematics, one can define a way to also take the square root of a negative numbers (doing this, one introduces imaginary numbers). If you do this, then the equation above will always have two solutions (though they may be imaginary). For example, there will be two solutions to $$x^2 + 1 = 0$$.
So there will always be two points where y = 0.

Since you can no longer really speak of an "x-axis" (it's more like an "x-plane" ) the question is whether this makes the answer trivially "no, such a parabola does not exist" or one should read the assignment more carefully

 

[DaxInvader]

Well The teacher talked about imaginairy numbers? isin't i^2 = (-1)

if it is.. then I am working on it

Quote from CompuChip :

For example, there will be two solutions to x^2 + 1 = 0
So there will always be two points where y = 0.


It is still a quadratic fonction if you look at it in the Euclidian geometry.. but if you put "i" anywhere.. You do not stay in Euclidian geometry... That is why you won't have a parabolic "function" that dosen't touch neither axis.. or plane, at infinity. There are some theorie going on with functions.. i heard that if x has two possible values, it is still a fonction.. but i cannot say more.

 

[Defennder]

The original question was whether we can construct a parabola which does not cut either axis, and where the x and y axes are assumed to be real. The answer, as agreed above was no. Now, you have to ask yourself how is it possible to use complex numbers when we've already agreed that the x-y axes themselves are real and not imaginary numbers. So clearly you have to re-define the question for it to make sense, and I can sense it will involve some complex analysis .

 

[Dick]

This thread is going seriously awry. Defennder, there ARE parabolas that don't cross either axis. Take the parabola y=a+x^2. Now rotate it 45 degrees by replacing x->x*cos(pi/4)-y*sin(pi/4) and y->x*sin(pi/4)+y*cos(pi/4). You'll get a quadratic in x and y that has an x*y term (which is normal for conics at an angle). If a is large enough then you'll find it doesn't cross either axis, by putting x=0 or y=0 and trying to solve for the other you will get no real roots.

 

[Defennder]

Ok, my mistake. Apparently I have a poor imagination. The parabola Dick suggested, y = x^2 + 1 where a = 1 doesn't intercept either the y=-x and y=x lines. Just rotate it 45 degrees and there you have it.

 

[offtheleft]

Ok, my mistake. Apparently I have a poor imagination. The parabola Dick suggested, y = x^2 + 1 where a = 1 doesn't intercept either the y=-x and y=x lines. Just rotate it 45 degrees and there you have it.

how do i rotate it 45degrees?


sorry for my delayed response lol

 

[HallsofIvy]

how do i rotate it 45degrees?


sorry for my delayed response lol

To rotate a point or graph by angle $\theta$, replace x with $x cos(\theta)+ y sin(\theta)$ and replace y by $-x sin(\theta)+ y cos(\theta)$
For example if $\theta$ is 90 degrees, $cos(\theta)= 0$ and $sin(\theta)= 1$ so you will be replacing x by y and y by -x: rotating the figure 90 degrees.


In particular, $sin(45)= cos(45)= \sqrt{2}/2$ so you must replace x by $\sqrt{2}/2(x+ y)$ and replace y by $\sqrt{2}/2(-x+ y)$.

y= x²+ 1 becomes ($\sqrt{2}/2$)(-x+ y)= (\sqrt{2}/2)(-x+ y))^2+ 1
[/tex](\sqrt{2}/2)x+ (\sqrt{2}/2)y= (1/2)x^2+ xy+ (1/2)y^2+ 1[/itex]

 

[Mentallic]

how do i rotate it 45degrees?

Take the parabola y=a+x^2. Now rotate it 45 degrees by replacing x->x*cos(pi/4)-y*sin(pi/4) and y->x*sin(pi/4)+y*cos(pi/4). You'll get a quadratic in x and y that has an x*y term (which is normal for conics at an angle). If a is large enough then you'll find it doesn't cross either axis

I always wondered how a graph can be turned at an angle other than 90^o. Too bad I don't understand how it works.

 

[DaxInvader]

What if you only put a contraint?

y = x^2 + 1 { x != 0

you have a graph that dosen't touch the x-axis (+1) a dosn't exist in x=0.

You just put a hole in the function.. But the trigonometric way is also a good answer... and maybe better.

 

[HallsofIvy]

What if you only put a contraint?

y = x^2 + 1 { x != 0

you have a graph that dosen't touch the x-axis (+1) a dosn't exist in x=0.

You just put a hole in the function.. But the trigonometric way is also a good answer... and maybe better.

But that graph is NOT a parabola now, so would not solve the original problem.

 

[DaxInvader]

But that graph is NOT a parabola now, so would not solve the original problem.

in what way it isn't? You just have a hole in it.. Well in my math courses, it still is.. Iv'e talked to my math teacher about that.

 

[Mentallic]

in what way it isn't? You just have a hole in it.. Well in my math courses, it still is.. Iv'e talked to my math teacher about that.

I've been told that such graphs that are the shape of a parabola, but not expressed in the form: $$ax^{2}+bx+c=y$$ are not parabolas.

For e.g. $$f(x)=\frac{5x^3-4x^2+3x}{x}$$ can be simplified to: $$f(x)=5x^2-4x+3$$ which is now in the form of a parabola, but with the key difference that x=0 is not defined, since the denominator in the original function would be 0. This - I have been told - is not a parabola anymore.

 

[symbolipoint]

I've been told that such graphs that are the shape of a parabola, but not expressed in the form: $$ax^{2}+bx+c=0$$ are not parabolas.

That is actually not a parabola; you need two variables, or you need to express a FUNCTION with one variable to the second power and may sometimes include the same variable to the first power - but you still need the variable to the second power. The equation of a quadratic to zero is just a way to find the "roots" of a quadratic function (or the "zeros" for when you have a relation expressing "x" as a relation of "y" with "y" being to the second power, $$ay^{2}+by+c=0$$ )

For e.g. $$f(x)=\frac{5x^3-4x^2+3x}{x}$$ can be simplified to: $$f(x)=5x^2-4x+3$$ which is now in the form of a parabola, but with the key difference that x=0 is not defined, since the denominator in the original function would be 0. This - I have been told - is not a parabola anymore.

PreCalculus will give you a more sophisticated understanding of those. Those two functions are not the same. The one with "x" in the denominator is not a parabola. The easiest way to see is use a graphing utility, such as Arnab's, or Graphmatica, or a graphing calculator. The simplified form of that rational function is not the function itself and has an asymtote at x=0. The shape becomes much different than parabola as you approach x=0

 

[Mentallic]

That is actually not a parabola; you need two variables

Oh yes I just made a slight error, I intended to write the dependant variable, rather than 0. I will fix it

PreCalculus will give you a more sophisticated understanding of those. Those two functions are not the same. The one with "x" in the denominator is not a parabola. The easiest way to see is use a graphing utility, such as Arnab's, or Graphmatica, or a graphing calculator. The simplified form of that rational function is not the function itself and has an asymtote at x=0. The shape becomes much different than parabola as you approach x=0

No. Both those are exactly the same shape. i.e. a parabola. Except that the one with x in the denominator has no existing value at x=0. There are no asymptotes, the graph simply does not exist there. If you were to graph it, you would graph its simplified parabola, but add an open circle where it cuts the y-axis (x=0).

 

[eliotargy]

The solution to the origiinal question is: y=jx^2+c, where c is greater than 0, and j^2 = -1. Go figure. As to rotating, if you rotate a paraboloa 45 degrees then it will ALWAYS cut both axes at some point, and I mean ALWAYS. Drop the talk about infinity and zero. The answer lies in imaginary numbers as stated above. Cheers.

 

[Dick]

in what way it isn't? You just have a hole in it.. Well in my math courses, it still is.. Iv'e talked to my math teacher about that.

Oh, come on. Euclid says nonparallel lines intersect in a point. Are you saying a line with a hole in it is still a line? Then Euclid was wrong. Don't be silly.

 

[symbolipoint]

Originally Posted by symbolipoint

PreCalculus will give you a more sophisticated understanding of those. Those two functions are not the same. The one with "x" in the denominator is not a parabola. The easiest way to see is use a graphing utility, such as Arnab's, or Graphmatica, or a graphing calculator. The simplified form of that rational function is not the function itself and has an asymtote at x=0. The shape becomes much different than parabola as you approach x=0

Observed and analyzed by Mentallic:

No. Both those are exactly the same shape. i.e. a parabola. Except that the one with x in the denominator has no existing value at x=0. There are no asymptotes, the graph simply does not exist there. If you were to graph it, you would graph its simplified parabola, but add an open circle where it cuts the y-axis (x=0).

That is correct. My statement was confused with another idea involving rational functions.

 

[Dick]

Parabolas are becoming less and less fun by the minute as this debate rages on. What's wrong with you people?

 

[Mentallic]

Parabolas are becoming less and less fun by the minute as this debate rages on. What's wrong with you people?

When were parabolas ever fun?

As to rotating, if you rotate a paraboloa 45 degrees then it will ALWAYS cut both axes at some point, and I mean ALWAYS.

I don't see the parabola(?) $$xsin(\frac{\pi}{4})+ycos(\frac{\pi}{4})=(xcos(\frac{\pi}{4})-ysin(\frac{\pi}{4}))^2+1$$ cutting any axes. This is equal to $$y=x^2+1$$ but rotated 45^o clockwise. Try graphing it for yourself. However, the use of a graphing calculator might save you a little time :)

 

[HallsofIvy]

The solution to the origiinal question is: y=jx^2+c, where c is greater than 0, and j^2 = -1. Go figure. As to rotating, if you rotate a paraboloa 45 degrees then it will ALWAYS cut both axes at some point, and I mean ALWAYS. Drop the talk about infinity and zero. The answer lies in imaginary numbers as stated above. Cheers.

And this is true just because you say so? Go back to the orginal point. Where does the parabola, y= x²+ 1 intersect the lines y=x and y= -x? It will intersect y= x where x= x²+ 1 or x²- x+ 1= 0. The roots of that equation are both imaginary and so the parabola does not intersect that line. Similarly the roots of x²+ 1= -x or x²+ x+ 1= 0 are imaginary so the parabola does not intersect that line either. Since x²+ 1 does not intersect either y= x or y= -x and rotating the parabola 45 degrees, so that it's axis of symmetry is y= x, gives a parabola that does not intersect y= 0 or x= 0.

 

[HallsofIvy]

in what way it isn't? You just have a hole in it.. Well in my math courses, it still is.. Iv'e talked to my math teacher about that.

What is you definition of parabola? I know serveral equivalent definition and none of them are satisfied by "a parabola with a hole in it".