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Part A: Graph my profile picture on Desmos using one equation. (It doesn't have to be in y=x form.)

 

Part B: Find the other equation that makes this work. (Does not have to be in y=x form.)

 

Part C: Find the point where the 2 parts intersect.

 

Part D: Explain why the 2 parts intersect each other at the point that they do.

 

Part E: Find a combination of 2 equations that gives the exact same graph if possible.

 

Part F: Explain why the nonlinear equation had the equation that it does.

 

I've done up to part C, but now I need some help.

helperid1839321  Jun 1, 2017
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13+0 Answers

 #1
avatar+445 
0

Can I use multiple element equations?

ZZZZZZ  Jun 1, 2017
 #2
avatar+445 
0

I got pretty close, does it have to be exact?

ZZZZZZ  Jun 1, 2017
 #3
avatar+91235 
0

Helperid bought my attention back to this question.

He has not even read my response to him, from a couple of days ago, but I am curious myself.

 

How can I get a graph to look like that curved graph in his profile pic?

(I hope this is not a doh moment - but too bad if it is)

I think it goes through the points (2.75,2.75) (1.75,5) and (5,1.75)

I assumed it was half a hyperbola and got to xy=1.75*5=8.75 but that does not go through (2.75,2.75)

So it is not correct.     sad

 

Can  someone offer a better attack method and answer?

Melody  Jun 24, 2017
 #4
avatar+18777 
+1

*VERY* Hard Math Question for my profile picture

 

I assume two points of the hyperbola are: \(P_1(4,2) \text{ and } P_2 (2,4)\)

 

Then the hyperbola equation is \( xy = 8\), because the Area under the hyperbola is 4*2 = 2*4 = 8 and \(y_{\text{hyperbola}} = \frac{A}{x}\)

The other equation is a line \( y = x\)

 

The intersect Point:

\(\begin{array}{|rcll|} \hline xy&=&8 \quad & | \quad y=x \\ x\cdot x &=&8 \\ x^2 &=&8 \\ x^2 &=&2\cdot 4 \\ x_{\text{intersection} } &=& 2\sqrt{2} \\\\ y_{\text{intersection} } &=& x_{\text{intersection} } \\ y_{\text{intersection} } &=& 2 \sqrt{2} \\ \hline \end{array}\)

 

The intersect Point is: \((2\sqrt{2},\ 2 \sqrt{2} )\)

 

The picture is:

 

 

laugh

heureka  Jun 26, 2017
 #13
avatar+505 
0

Just so you know, the 2 other coordinates were right, heureka. They are (2, 4) and (4, 2) because 4^2 = 2^4 = 8. However, the intersection is (e,e) to the limits of Desmos's calculations.

helperid1839321  Oct 23, 2017
 #5
avatar+91235 
0

Thanks Heureka,

But the curved graph does not go through those 2 points.

Here is a bigger version of the graph.

 

 

I think it goes through the points (2.75,2.75) (1.75,5) and (5,1.75)

 

How do I find a graph that will go through these points ?? 

Melody  Jun 26, 2017
 #6
avatar+18777 
+3

I think it goes through the points (2.75,2.75) (1.75,5) and (5,1.75)

How do I find a graph that will go through these points ?? 

 

The formula of a hyperbola is given in general form: \(y = \frac{a}{x+c} + d \)

 

Set \(P_1(x_1,y_1) = (2.75,2.75)\)
Set \(P_2(x_2,y_2) = (1.75,5)\)
Set \(P_3(x_3,y_3) = (5,1.75)\)

 

\(\begin{array}{|lrcll|} \hline (1) & y_1 &=& \frac{a}{x_1+c} + d \\ & (y_1-d)*(x_1+c) &=& a \\ & y_1x_1+cy_1-dx_1&=& a+cd \\\\ (2) & y_2 &=& \frac{a}{x_2+c} + d \\ & (y_2-d)*(x_2+c) &=& a \\ & y_2x_2+cy_2-dx_2&=& a+cd \\\\ (3) & y_3 &=& \frac{a}{x_3+c} + d \\ & (y_3-d)*(x_3+c) &=& a \\ & y_3x_3+cy_3-dx_3&=& a+cd \\\\ \hline \end{array} \)

 

\(\begin{array}{|lrcll|} \hline (1) & y_1x_1+cy_1-dx_1&=& a+cd \\ (2) & y_2x_2+cy_2-dx_2&=& a+cd \\ (3) & y_3x_3+cy_3-dx_3&=& a+cd \\ \hline (1)-(2) & y_1x_1+cy_1-dx_1 -(y_2x_2+cy_2-dx_2) &=& 0 \\ & c(y_1-y_2)+d(x_2-x_1) &=& y_2x_2-y_1x_1 \\\\ (1)-(3) & y_1x_1+cy_1-dx_1 -(y_3x_3+cy_3-dx_3) &=& 0 \\ & c(y_1-y_3)+d(x_3-x_1) &=& y_3x_3-y_1x_1 \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline (1) & c(y_1-y_2)+d(x_2-x_1) &=& y_2x_2-y_1x_1 \\ (2) & c(y_1-y_3)+d(x_3-x_1) &=& y_3x_3-y_1x_1 \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline c &=& \frac{ \begin{vmatrix} y_2x_2-y_1x_1 & x_2-x_1 \\ y_3x_3-y_1x_1 & x_3-x_1 \end{vmatrix} } { \begin{vmatrix} y_1-y_2 & x_2-x_1 \\ y_1-y_3 & x_3-x_1 \end{vmatrix} } \\ c &=& \frac{ \begin{vmatrix} 5\cdot1.75-2.75\cdot2.75 & 1.75-2.75 \\ 1.75\cdot5-2.75\cdot2.75 & 5-2.75 \end{vmatrix} } { \begin{vmatrix} 2.75-5 & 1.75-2.75 \\ 2.75-1.75 & 5-2.75 \end{vmatrix} } \\ \mathbf{c} &\mathbf{=}& \mathbf{-0.95} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline d &=& \frac{ \begin{vmatrix} y_1-y_2 & y_2x_2-y_1x_1 \\ y_1-y_3 & y_3x_3-y_1x_1 \end{vmatrix} } { \begin{vmatrix} y_1-y_2 & x_2-x_1 \\ y_1-y_3 & x_3-x_1 \end{vmatrix} } \\ d &=& \frac{ \begin{vmatrix} 2.75-5 & 5\cdot 1.75-2.75\cdot 2.75 \\ 2.75-1.75 & 1.75\cdot 5-2.75\cdot 2.75 \end{vmatrix} } { \begin{vmatrix} 2.75-5 & 1.75-2.75 \\ 2.75-1.75 & 5-2.75 \end{vmatrix} } \\ \mathbf{d} &\mathbf{=}& \mathbf{0.95} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline a &=& y_1x_1+cy_1-dx_1-cd \\ a &=& 2.75\cdot 2.75-0.95\cdot 2.75-0.95\cdot 2.75 - (-0.95)\cdot 0.95 \\ \mathbf{a} &\mathbf{=}& \mathbf{3.24} \\ \hline \end{array}\)

 

The formula of a hyperbola is given through  \(P_1, P_2, P_3:\ y = \frac{3.24}{x-0.95} + 0.95\)

 

The graph is:

 

laugh

heureka  Jun 27, 2017
 #7
avatar+91235 
+2

Thank you Heureka,

Now I am very happy :))

Melody  Jun 27, 2017
 #8
avatar+91235 
+1

Helper Id you hae reposted thit same question in all these future spots

https://web2.0calc.com/questions/very-hard-math-question-for-my-profile-picture-repeat

https://web2.0calc.com/questions/very-hard-math-question-for-my-profile-picture-rerepost

https://web2.0calc.com/questions/very-hard-question-for-my-profile-picture

 

Thanks for reposting in the correct way. Each time you have bought us back to your original question very nicely and I appreciate that.

I do not understand however why you are still asking this.

 

You did give an explanation here:

https://web2.0calc.com/questions/very-hard-math-question-for-my-profile-picture-repeat#r1

Heureka did originally put the hyperbola through the wrong points, that is true, but when I pointed out his error he redid it through what lookes to be the correct points. You have not told us what the points are so we did have to estimate them from the pic.

 

This is where heureka re-answered. 

https://web2.0calc.com/questions/very-hard-math-question-for-my-profile-picture#r6

I do not understand why you are not accepting this answer. What is wrong with it?

It goes through the points (2.75,2.75) (1.75,5) and (5,1.75)

 

Why are you not happy with this #6  answer ?????

Melody  Oct 20, 2017
 #9
avatar+91235 
0

 

Arr, I just foung this:

 

In the above-linked question, I told you I had got to part C. I meant up until part B, but then I made a situational approximation. I graphed 

and x log base y = y log base x, and I estimated the interception point to be (e,e) to a measure of precision at least 1/10,000.

Your graph looks like it is probably correct.  Yes you already know that.   :)

 

 

Ok I am finally understanding why helperid is not happy.

This is NOT a hyperbola.

I overlaid this with Heureka's graph and it is not exact, it cannot be because heureka did not use exact points but it was VERY close. This graph sure looks like half a hyperbola with the line y=x crossing it. (Positive quadrant only)

 

SO now what is the question again????

 

Part C: Find the point where the 2 parts intersect.  

 

Part D: Explain why the 2 parts intersect each other at the point that they do.

 

Part E: Find a combination of 2 equations that gives the exact same graph if possible.

 

Part F: Explain why the nonlinear equation had the equation that it does.

Oh there they are, I do not know the answers but I'm interested too :)

 

This is the graph

I'm interested in some discussion on this too    frown

I don't remember seeing a graph like  this before.    

 

 

Melody  Oct 20, 2017
 #10
avatar+505 
0

Small note - I used y^x = x^y, but either way, it comes out the same. I tested both of them, and they are (to the limits of Desmos's calculations) identical.

helperid1839321  Oct 21, 2017
 #11
avatar+91235 
+1

Yes they are identical :)

 

\(y^x=x^y\\ log(y^x)=log(x^y)\\ xlogy=ylogx\)

 

I just realized that the wording in your description is erroneous.

It says

x log base y = y log base x

You need another number or variable after the word 'base'.

Or you can assume they are the same base, which is what I did and leave the word base out altogether.

Melody  Oct 22, 2017
 #12
avatar+505 
+1

Thanks for the grammar note! I'd definitely rather learn that the easy way than the hard way.

helperid1839321  Oct 23, 2017

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