# Parabolas are fun

1. Sep 24, 2008

### offtheleft

now, this isnt required for my class but my professor thought it would be cool if anyone in our class could figure it out.

now, im 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 cant quite seem to get it. haha

well heres 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 hes talking about?

id also like to find out how to turn a parabola on its side, but thats 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 couldnt really get anywhere. still, probably because of my lack of knowledge of this.

2. Sep 24, 2008

### 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.

3. Sep 24, 2008

### offtheleft

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

4. Sep 24, 2008

### chaoseverlasting

What you need to do is rotate the axis. Although I dont 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.

5. Sep 24, 2008

### 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.

6. Sep 24, 2008

### 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

7. Sep 24, 2008

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

if it is.. then im 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 wont 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.

Last edited: Sep 24, 2008
8. Sep 24, 2008

### 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 .

9. Sep 24, 2008

### 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.

10. Sep 25, 2008

### 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.

11. Sep 25, 2008

### offtheleft

how do i rotate it 45degrees?

sorry for my delayed response lol

12. Sep 25, 2008

### HallsofIvy

Staff Emeritus
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]

13. Sep 25, 2008

### Mentallic

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

14. Nov 21, 2008

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.

15. Nov 21, 2008

### HallsofIvy

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

16. Nov 21, 2008

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.

17. Nov 21, 2008

### Mentallic

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.

Last edited: Nov 21, 2008
18. Nov 21, 2008

### symbolipoint

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, $$ay^{2}+by+c=0$$ )

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

19. Nov 21, 2008

### Mentallic

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

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).

20. Nov 21, 2008

### 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.