Shadow Posted December 21, 2008 Posted December 21, 2008 Hey all, as some may know, I've been attempting to learn calculus lately, and despite the fact I though I had already mastered the part of the tutorial I was reading, I did encounter a slight problem for which I cannot find a logical explanation. Let us have a [math]f(t)=2t + 1, t\in N[/math]. Its graph would then be a line going starting at [math]A=[0, 1][/math] and going through [math]B=[1, 3][/math]. If we subtract two from the output, we will get [math]f(t)-2=2t-1[/math]. The graph will be a line starting at [math]A=[0, -1][/math] and continue on through [math]B=[\frac{1}{2},0][/math]. In other words, the graph will be moved down by two. It's minus, so it's down. That makes sense. But if we subtract two from the input, we get [math]f(t-2)=2t-3[/math]. The graph will be a line starting at [math]A=[2, 1][/math] and continue on through [math]B=[3, 3][/math]. In other words, the graph will be moved to the right. It's minus, so logically it should be moved to the left. Why is this so? Cheers, Gabe
Cap'n Refsmmat Posted December 21, 2008 Posted December 21, 2008 Let's suppose t = 5. You put t into f(t-2) and it gives you the same answer you'd get if you did f(3) -- so at t = 5 it's graphing what was originally at t=3. Hence it's to the right.
Petanquell Posted December 22, 2008 Posted December 22, 2008 if [math] t\in N [/math] it won't be a line but dots (on points, don't know how to say it)... The second thing, we talked about it three days ago. Visit math tutor, paragraph about "Transformations". There is quite nice metaphore (I know you like them ) .... See you in half an hour at the teaplace, pq
Shadow Posted December 23, 2008 Author Posted December 23, 2008 I know, I noticed, but had already passed the time limit for editing. I meant [math]Z^+_0[/math]. But thanks for the site ;-)
Petanquell Posted December 23, 2008 Posted December 23, 2008 [math] Z^+_0 [/math] is not enough for line
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