New equation? y=3tan(2x) shifted down 5 and 60 left?

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The discussion revolves around determining the correct equation for shifting the function y=3tan(2x) down by 5 units and left by 60 units. Two potential equations are proposed: y=3tan(2(x-60))-5 and y=3tan(2x-60)-5, with the first being favored. A key point raised is the understanding of horizontal shifts, where shifting right involves f(x-a) rather than f(x+a). The correct transformation for shifting left by 60 units is confirmed to be y=3tan(2(x-60))-5. The conversation emphasizes the importance of correctly applying transformations to functions.
aisha
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Im not sure I think it is either

y=3tan (2(x-60))-5

OR

y=3tan(2x-60)-5

which will become y=3 tan(2 (x-30))-5 but I think this one is wrong

if the first one is right then should the 60 have a negative or plus sign before it?

HELP! QUICK! :cry: :cry:
 
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Think about this. Suppose you have the function f(x) and you want to shift the graph to the right by an amount a. Should the new shifted function be f(x-a) or f(x+a)?
You can answer this by looking at the value of a single point, for example f(0). The shifted function f(x-a) has that value not at x=0, but at x=a, since f(a-a)=f(0), so is that function shifted to the left or to the right?

Also, if the original function is f(x)=3tan(2x) and you want to shift the thing over 60 units, which do you think is correct? 3tan(2(x-60)) or 3tan(2(x-30))?
Which of these is of the form f(x-60)?
 
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