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Hi, I have a question that I don't really understand.
For f(x)=2^x, sketch the graph of the following, labeling asymptotes where appropriate:
a) y=f(x+1)
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Hi Marisca;
Start here:
http://www.mathsisfun.com/algebra/asymptote.html
Graph your function over here:
http://www.mathsisfun.com/graph/functio … ymax=8.932
Can you visually identify a possible horizontal asymptote?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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So something like y=0?
Also, would I need to sub f(x)=2^x into y=f(x+1)?
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Hi Marisca;
Exactly y = 0.
Also, would I need to sub f(x)=2^x into y=f(x+1)
If f(x) = 2^x then f(x+1) = 2^(x+1) wherever you see an x you substitute x + 1.
The equation y = f(x+1) is y = 2^(x+1)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Thank you! I understand it now.
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Hi;
Wunderbar! Glad to help!
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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