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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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7+0 Answers

 #1
avatar+328 
0

Can I use multiple element equations?

ZZZZZZ  Jun 1, 2017
 #2
avatar+328 
0

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

ZZZZZZ  Jun 1, 2017
 #3
avatar+90581 
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+18616 
+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
 #5
avatar+90581 
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+18616 
+2

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+90581 
+2

Thank you Heureka,

Now I am very happy :))

Melody  Jun 27, 2017

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