I have several problems that look like this. Could someone walk me through the first one: The graph of the function y=f(x) contains the point (2,5) What point is guaranteed to be on the graph of y=f(x+3)-5 ?
Last edited by sy416; Jan 31st 2013 at 01:36 PM. Reason: [solved]
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Originally Posted by sy416 The graph of the function y=f(x) contains the point (2,5) What point is guaranteed to be on the graph of y=f(x+3)-5 ? From the given you know that . WHY? What value of will introduce into ? Then what point do you have?
Originally Posted by Plato From the given you know that . WHY? What value of will introduce into ? Then what point do you have? I don't know. It confuses me with the y's and the f(x)'s
Originally Posted by sy416 I don't know. It confuses me with the y's and the f(x)'s OK. Forget the y's and the f(x)'s You can surely answer this question. What value of will introduce into ?
Originally Posted by Plato OK. Forget the y's and the f(x)'s You can surely answer this question. What value of will introduce into ? 2?
Originally Posted by sy416 2? NO! f(2+3)=f(5) f(2). Try again!
Originally Posted by Plato NO! f(2+3)=f(5) f(2). Try again! BUt that's the thing I don't know how to find the x value that will introduce f(2). How do I do it? I need help understanding the problem.
Wait I think I have it...
Ohh! I have to choose an x value that will bring it back to f(2) .. soo... x= -1 y= f(-1+3)-5 y= f(2) - 5 y= 5-5 y=0 (-1,0) Thank you!
Originally Posted by sy416 BUt that's the thing I don't know how to find the x value that will introduce f(2). How do I do it? I need help understanding the problem. I don't think that you are ready to do this question. If you look at the answer in the 'back-of-the-book' it would be .
Originally Posted by Plato I don't think that you are ready to do this question. If you look at the answer in the 'back-of-the-book' it would be . Nah I got it (scroll up). Thanks.
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