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 January 6th, 2015, 04:40 AM #1 Member   Joined: Apr 2014 From: Florida Posts: 69 Thanks: 4 Linear Equations, graphs I need a source where I can learn to do linear equations on graphs and such. Are equations linear and such; Y=X+2 and things like that. Most of the text I have seen give a brief explanation that really isnt sufficient and then produces problems to solve. I need an easy to understand destination that breaks it down if possible.
 January 7th, 2015, 08:16 AM #2 Senior Member   Joined: Aug 2012 Posts: 2,409 Thanks: 753 Think of it as a machine. You put in a value for x, and a value for y comes out the other end. So if the machine is y = x + 2, you put in 2 and 4 comes out. So you graph the point (2,4) on the x-y plane. If you put in 3, 5 comes out, so you graph (3,5). If you put in -5, then -3 comes out; so you graph (-5, -3). That's all there is to it. You put in some random values for x, see what y comes out, and graph (x,y).
January 7th, 2015, 09:52 AM   #3
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Quote:
 Originally Posted by Maschke Think of it as a machine. You put in a value for x, and a value for y comes out the other end. So if the machine is y = x + 2, you put in 2 and 4 comes out. So you graph the point (2,4) on the x-y plane. If you put in 3, 5 comes out, so you graph (3,5). If you put in -5, then -3 comes out; so you graph (-5, -3). That's all there is to it. You put in some random values for x, see what y comes out, and graph (x,y).
Ok to make sure im understanding, substitute a number for X, that would be +2 and result in Y. This would be for any number + X = Y

Example:
y=x+6
x=4
y would be 10

(4,10) would be marked on the graph, correct?

 January 9th, 2015, 05:00 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 You should know, from geometry, that a straight line is determined by two points. So to graph a straight line you just have to calculate two values. Choose what ever values of x you want and use the given formula to find the corresponding y. For example, to graph y= x+ 2, if you choose x= -3, then y= -3+ 2= -1. So mark the point (-3, -1) on your graph. If you choose x= 4, then y= 4+ 2= 6. So mark the point (4, 6) on your graph. Now draw the line through those two points. You can choose any two values of x. In practice you actual draw line will be more accurate if the points are reasonably distant. ("Reasonably" meaning within the length of your straight edge.)
 January 15th, 2015, 09:42 PM #5 Newbie   Joined: Jan 2015 From: Pakistan Posts: 1 Thanks: 0 I'm having trouble with these set of equations 1. Find the arc length function s(t) for the curve r(t) = < t, cosh(t)> with t between 0 and 1 2. Reparameterize the curve r(t) = < (2/(t^2+1)) -1, 2t/(t^2+1)> with respect to arc length from (1,0) in the direction of increasing t. After reparemeterizing, do you recognize the curve? 3. Compute the unit tangent vector T and the unit normal vector N for the curve in question 1. 4. Find T, N and B for the curve <3t^2, 2t^3, 3t> at the point given by t=1.
January 16th, 2015, 04:46 AM   #6
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Quote:
 Originally Posted by Atifraza I'm having trouble with these set of equations 1. Find the arc length function s(t) for the curve r(t) = < t, cosh(t)> with t between 0 and 1 2. Reparameterize the curve r(t) = < (2/(t^2+1)) -1, 2t/(t^2+1)> with respect to arc length from (1,0) in the direction of increasing t. After reparemeterizing, do you recognize the curve? 3. Compute the unit tangent vector T and the unit normal vector N for the curve in question 1. 4. Find T, N and B for the curve <3t^2, 2t^3, 3t> at the point given by t=1.
Please do NOT "hi-jack" someone else's thread to ask a completely different question! And when you do, tell us what you have tried and what you do understand about a problem so we will know where you need help.

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