$\displaystyle sin(x+30deg)$

Why graph move to the left?

If this qn is already asked, pls direct me to the thread. Thanks.

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- Sep 9th 2010, 08:22 AMstupidguysin(x+30deg), why graph move to the left?
$\displaystyle sin(x+30deg)$

Why graph move to the left?

If this qn is already asked, pls direct me to the thread. Thanks. - Sep 9th 2010, 08:40 AMyeKciM
do you understand why y=(x+2)^2 is shifted left by 2 ?

same thing meaning that for value that normal y=x^2 would have for x=1 is y=1 , but when shifted in x=1 function y= 1^2 +2 = 3

now if you look at the y=(x-2)^2 function you realize that is the function y=x^2 shifted to the right by 2 :D:D:D - Sep 9th 2010, 08:51 AMmathaddict
- Sep 9th 2010, 09:00 AMstupidguy
Aren't you comparing apples with pears, yeah they are both crunchy fruits but they ain't the same.....LOL?

BTW, anyone have any good ways for me to remember that MINUS is to the RIGHT, PLUS is to the LEFT. It is quite random to me. I can only accept it as graphmatica says so! :( - Sep 9th 2010, 09:09 AMyeKciM
okay :D

look at the function y=x :D:D:D

it's line trough first and 3rd quadrant :D

x=1 -> y=1

x=-1 -> y =-1

now look at the same function but shifted to the left y=x+1

x=0 -> y=1

x=-1 -> y=0

do you see in your head that function y=x is just moved to the left ?

same thing with y=x-1 :D:D:D:D - Sep 9th 2010, 09:10 AMstupidguy
- Sep 9th 2010, 09:18 AMstupidguy
I could hardly understand what u are talking! nvm, thx for ur effort.

I now understand why some ppl are MHF Contributors and some are just members. Try harder! :D - Sep 9th 2010, 09:33 AMundefined
I think once you get 1000 posts you get the title MHF Contributor, perhaps after review by a mod, not sure. Also sometimes there are issues like English not being a first language; overall I'd say your post comes off a bit ungracious towards someone giving their time without compensation to help you.

Anyway normally you associate (+) with right, and (-) with left, as with the real number line.

In terms of graph shifts, though, consider the function**f(x)**versus the function**g(x)=f(x-2)**. How do you get f(0)? In the first one, you plug in**x=0**. In the second one, you plug in**x=2**. Thus, the point (0, f(0)) on the graph of f has been mapped to (2, f(0)) on the graph of g. That is, there has been a shift to the**right**.

So there's a certain amount of "reversal" that you have to keep straight in your head. But if you work with graph transformations a lot, you can get comfortable with this and recognize it in other places. For example, the equation of a unit circle ("unit" means radius=1) with center at $\displaystyle \,(h,k)$ is $\displaystyle (x-h)^2+(y-k)^2=1$. Notice the minus signs. - Sep 9th 2010, 10:01 AMstupidguyapologyQuote:

overall I'd say your post comes off a bit ungracious towards someone giving their time without compensation to help you.