How to create a hard graph easily?

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Homework Help Overview

The discussion revolves around understanding how to create a specific type of graph, with participants analyzing a problem related to the representation of linear equations and their slopes. The context involves interpreting a graph and determining the correct answer from multiple choices.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the form of the equation representing the graph, with some attempting to derive expressions for y in terms of x. There are differing opinions on the correct answer, with some questioning the assumptions made about the graph's slope and its implications.

Discussion Status

The conversation is ongoing, with various interpretations being explored. Some participants have offered guidance on how to manipulate the equations, while others express confusion and seek simpler explanations. There is no explicit consensus on the correct answer yet.

Contextual Notes

Participants are working with a graph that has specific characteristics, such as passing through the origin and having a slope that suggests certain constraints on the variables involved. The discussion includes attempts to clarify the setup and the implications of the graph's slope.

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http://im16.gulfup.com/2012-03-07/1331125411751.png



I know the answer is (E) , but HOW ?
 
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Are you sure it's meant to be E) because I'm getting D) :frown:

If you were given a linear graph with the vertical axis being Y, and the horizontal being X, then you could say that Y=mX for some constant m, right?

You can similarly substitute each axis representation into this equation. The vertical is now y-x and the horizontal x+y so let Y=y-x and X=x+y and then rearrange the equation to make y the subject.
 
Sorry,I'm not sure that the correct answer is (E) :(
 
No worries. On the other hand, I'm quite sure the answer is D, but it doesn't even matter, we should just let the math do the talking :wink:
 
The first graph is a straight line passing through the origin so it must be of the form y- x= k(x+ y). That is the same as y- x= kx+ ky or y- ky= x+ kx. Factoring, (1-k)y= (1+k)x or y= (1+k)/(1-k) x. From the picture it appears that k is larger than 1 (the angle is greater than 45 degrees) so 1-k is negative. That, in turn, means that the fraction (1+k)/(1-k) is negative so the graph goes downward to the right. It also follows that k< -1.
 
I didn't understand !
:(


Can you explain it more simply ?
 
Can you follow the procedure I outlined for you in post #2?
 
Since the line passes through the origin, and the slope builds an angle between 45 and 90 degrees, I can say that:
<br /> y - x = a ( y + x), \ a &gt; 1<br />

Solve this for y. What do you get?
 
Dickfore said:
Since the line passes through the origin, and the slope builds an angle between 45 and 90 degrees, I can say that:
<br /> y - x = a ( y + x), \ a &gt; 1<br />

Solve this for y. What do you get?


y(1-a)= x(1+a)

so .. ?

 
  • #10
KKK_123 said:
y(1-a)= x(1+a)

so .. ?


Solve for y.

Plug-in some positive numbers greater than 1, for a.

BTW: Mentallic is right about what's the correct answer .
 
  • #11
KKK_123 said:
Can you explain it more simply ?
That graph in the picture appears to have a slope of about 3, for each mm horizontally it rises about 3mm, roughly (and assuming the scales are the same).

Usually we would express that graph as being the plot: y = 3x[/color]

But on this particular graph, the vertical quantity is given as y-x
and the variable on the horizontal is given as y+x
So substitute these expressions for the y and x in the blue equation, respectively.

Finally, rearrange the result to get y on one side by itself.
 
  • #12
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