DropBoite Posted May 2, 2012 Report Share Posted May 2, 2012 Hey guys please help me out with this function trouble? The question and the data is in the image attached. How does one answer question a in that screenshot? Reply Link to post Share on other sites More sharing options...
The Economist Posted May 2, 2012 Report Share Posted May 2, 2012 It asks you for the roots of the function - where f(x) crosses the x-axis. The solutions are (0,0) and (4,0). 1 Reply Link to post Share on other sites More sharing options...
DropBoite Posted May 2, 2012 Author Report Share Posted May 2, 2012 (edited) A second question is attached, I genuinely hope you can help me, thank you! Regarding this one , how does one find p and q? what are they symbolic of?! Edited May 2, 2012 by DropBoite Reply Link to post Share on other sites More sharing options...
timtamboy63 Posted May 2, 2012 Report Share Posted May 2, 2012 a) x=0, 4b) x=2c) I'm not too sure, but I'd say x=-2EDIT: That was for the first Q. Reply Link to post Share on other sites More sharing options...
DropBoite Posted May 2, 2012 Author Report Share Posted May 2, 2012 It asks you for the roots of the function - where f(x) crosses the x-axis. The solutions are (0,0) and (4,0). Okay so If I get you that's so simple and I'm an idiot... Thanks !! I don't know how that slipped me, could you be a darling and help me with the other question too? Thanks! Reply Link to post Share on other sites More sharing options...
The Economist Posted May 2, 2012 Report Share Posted May 2, 2012 It asks you for the roots of the function - where f(x) crosses the x-axis. The solutions are (0,0) and (4,0). Okay so If I get you that's so simple and I'm an idiot... Thanks !! I don't know how that slipped me, could you be a darling and help me with the other question too? Thanks! For a) take f(x) and substitute the values x=0 and y=6 (you'll get a relation with only q and p). Then do the same for x=1 and y=4. For b) solve the system of these two equations: p+q=6 and 0.5p+q=4 p=4 and q=2 For c) the limit as x-->+oo of 4(1/2)^x+2=2 so the horizontal asymptote is y=2 1 Reply Link to post Share on other sites More sharing options...
DropBoite Posted May 2, 2012 Author Report Share Posted May 2, 2012 a) x=0, 4b) x=2c) I'm not too sure, but I'd say x=-2EDIT: That was for the first Q.Thank you! Could you elaborate on how I should go about calculating the answer to question c? Thank you! Will rep that post Reply Link to post Share on other sites More sharing options...
DropBoite Posted May 2, 2012 Author Report Share Posted May 2, 2012 It asks you for the roots of the function - where f(x) crosses the x-axis. The solutions are (0,0) and (4,0). Okay so If I get you that's so simple and I'm an idiot... Thanks !! I don't know how that slipped me, could you be a darling and help me with the other question too? Thanks! For a) take f(x) and substitute the values x=0 and y=6 (you'll get a relation with only q and p). Then do the same for x=1 and y=4. For b) solve the system of these two equations: p+q=6 and 0.5p+q=4 p=4 and q=2 For c) the limit as x-->+oo of 4(1/2)^x+2=2 so the horizontal asymptote is y=2 What are those two o's for? Okay but my math teacher didn't do this chapter well can you please tell me what p and q are? Thanks! Reply Link to post Share on other sites More sharing options...
The Economist Posted May 2, 2012 Report Share Posted May 2, 2012 Look at the graph of (1/2)^x as x tends to plus infinity. The limit of that function for high values of x is 0 so in your f(x) that part becomes zero and you only have the value of q left which is 2 so f(x) tends to y=2 as x tends to plus infinity which is why y=2 is the horizontal asymptote. Reply Link to post Share on other sites More sharing options...
DropBoite Posted May 2, 2012 Author Report Share Posted May 2, 2012 (edited) Look at the graph of (1/2)^x as x tends to plus infinity. The limit of that function for high values of x is 0 so in your f(x) that part becomes zero and you only have the value of q left which is 2 so f(x) tends to y=2 as x tends to plus infinity which is why y=2 is the horizontal asymptote. Apologies but you lost me, that's more jargon than I can handle! Okay I'm looking at the guide and it says the equation of the horizontal asymp is y=c so if I have any exponential equation I can simply assume that the number c is the horizontal asymptote? Does that ALWAYS hold true? Additionally an innocent assumption tells me that when there is no c the equation of the asymp is y=0, true? Edited May 2, 2012 by DropBoite Reply Link to post Share on other sites More sharing options...
The Economist Posted May 2, 2012 Report Share Posted May 2, 2012 Look at the graph of (1/2)^x as x tends to plus infinity. The limit of that function for high values of x is 0 so in your f(x) that part becomes zero and you only have the value of q left which is 2 so f(x) tends to y=2 as x tends to plus infinity which is why y=2 is the horizontal asymptote. Apologies but you lost me, that's more jargon than I can handle! Okay I'm looking at the guide and it says the equation of the horizontal asymp is y=c so if I have any exponential equation I can simply assume that the number c is the horizontal asymptote? Does that ALWAYS hold true? Additionally an innocent assumption tells me that when there is no c the equation of the asymp is y=0, true? Hmm no you can't assume that because any exponential function a^x does not always tend to 0 for high values of x. This is the case only when 0<a<1. Anyway, the guide says that the horizontal asymptote is y=c because the horizontal asymptote is always a straight line parallel to the x-axis I think in your exercise you can see the asymptote simply by looking at the graph - especially since the command term is "write down". See how f(x) is almost parallel to the x-axis as y almost becomes 2 but never actually does? 1 Reply Link to post Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.