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 [quote="MatthewV"]Yes, but I didn't take a picture of it... [spoiler](Hey, I said that I only had to change one number to re-graph it)[/spoiler][/quote]
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Zag
Posted: Sun Dec 11, 2011 8:04 pm    Post subject: 1

Yeah, yeah.
BraveHat
Posted: Sun Dec 11, 2011 6:36 pm    Post subject: 0

Zag wrote:
y = ± sqrt( - x^2 + 8) : [ -sqrt(8) < x < sqrt(8) ]

wouldn't it be
y = ± sqrt( - x^2 + 8) : [ -sqrt(8) <= x <= sqrt(8) ]?
MatthewV
Posted: Mon Nov 21, 2011 7:06 am    Post subject: -1

Yes, but I didn't take a picture of it... (Hey, I said that I only had to change one number to re-graph it)
Zag
Posted: Mon Nov 21, 2011 1:34 am    Post subject: -2

Not until after I'd posted, correct. So, did you graph it again?
MatthewV
Posted: Mon Nov 21, 2011 1:20 am    Post subject: -3

Hah! You didn't see post #12.
Zag
Posted: Mon Nov 21, 2011 12:27 am    Post subject: -4

Zag wrote:

Plot them for the range x from -10 to 10
and y from 0 to 6
MatthewV
Posted: Sun Nov 20, 2011 11:35 pm    Post subject: -5

Or maybe I just needed to read the part about x going from -10 to 10! (Fortunately I was able to change the entire plot by changing one number)
MatthewV
Posted: Sun Nov 20, 2011 11:25 pm    Post subject: -6

I feel like I am missing something. But your first equation is always going to be more than 530 when x is above zero.

And MATLAB is much nicer than Excel for this
Zag
Posted: Sun Nov 20, 2011 3:25 pm    Post subject: -7

Did you plot them? Can you figure out how I did that?
Amb
Posted: Sun Nov 20, 2011 6:50 am    Post subject: -8

Im using POWER(x,y) because X and Y are cell references. That way I can change it to X1*X1*X1*X1 if I want.
Zag
Posted: Sun Nov 20, 2011 4:12 am    Post subject: -9

Can you plot four equations on top of each other? If so, plot all of these:

y= x^4 + 20 x^3 + 146 x^2 + 460x + 530
y= x^4 - 20 x^3 + 146 x^2 - 460x + 530
y= sqrt(4-x^2) +3
y= - sqrt(4-x^2) +3

Plot them for the range x from -10 to 10
and y from 0 to 6

By the way, rather than writing out POWER(X1,2), you can just write X1*X1
Amb
Posted: Sun Nov 20, 2011 2:59 am    Post subject: -10

I read in a book that if you change the Exponent to higher numbers (even only) the graph changes from a circle to a rounded rectangle. The closer you get to infinity, the more rectangular the shape is.

Well I can tell you if you do this:

=SIN(SQRT((-POWER(A2,\$D\$2)) +\$E\$2))
(D2 = 2, E2 = 8)

You get a sponge shape

Im sure you all really care, but this is fun. Any suggestions for weird things (eg LOG) to throw into equation?
Amb
Posted: Fri Nov 18, 2011 3:21 am    Post subject: -11

Yeah but thats easy enough to overcome. Now that I see the answer, Im surprised I didnt get it earlier.
Zag
Posted: Fri Nov 18, 2011 2:59 am    Post subject: -12

Right. I assume that's only getting you half the circle, though.
Amb
Posted: Fri Nov 18, 2011 2:15 am    Post subject: -13

Oh might be =SQRT((-POWER(A10,2))+8)
Amb
Posted: Fri Nov 18, 2011 2:08 am    Post subject: -14

Actually I used to be really good at algebra back in my school days, but I havent used it since... I was mucking around making cartesian graphs in excel, and wanted a formula to graph a circle

The formula
=SQRT(POWER(-A2,2)+8) doesnt really work though, because A2=9 (aka X=9) produces 9.43
Zag
Posted: Fri Nov 18, 2011 1:43 am    Post subject: -15

Are you taking a class, or something?

(x^2)+(y^2) = 8 ------------------------ Add -(x^2) to both sides of the equation
(x^2)-(x^2) + (y^2) = -(x^2) + 8 ------------ [(x^2)-(x^2)] is just zero, so
y^2 = -(x^2) + 8 ----------------------------- Now take the square root of both sides. Note that y^2 and (-y)^2 equal the same thing, so we have to account for the other side being positive or negative. The symbol ± is said aloud as "plus or minus"
sqrt( y^2 ) = ± sqrt( - x^2 + 8 ) --------------- sqrt( y^2 ) is just y so

y = ± sqrt( - x^2 + 8)

If x^2 gets any larger than 8, then you'll be taking a square root of a negative number, which isn't allowed until calculus. So you might write this as

y = ± sqrt( - x^2 + 8) : [ -sqrt(8) < x < sqrt(8) ]

that is, it only has a value when x is greater than -sqrt(8) and less than sqrt(8).

If you graph this, it makes a circle with a radius of sqrt(8)

Edit: I found an online graphing tool. -- it won't let me put a URL with arguments in here, so you'll have to copy&paste

Code:
http://www.quickmath.com/webMathematica3/quickmath/graphs/equations/basic.jsp#v1=x%5E2%2By%5E2%3D8&v2=-3&v3=3&v4=-3&v5=3
Amb
Posted: Fri Nov 18, 2011 12:48 am    Post subject: -16

If (x^2)+(y^2) = 8

How do you reorganise the equation so that y = .....

I have a whole bunch of workings that are probably wrong. I just want a formula that will calculate Y for any given X, noting that there are values of X that wont work.

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