gib65 Posted October 18, 2005 Share Posted October 18, 2005 I have a couple questions about plotting a curve with assymptotic limits. 1) Suppose you had the equation y = x^2. The domain for x is [-infinity, +infinity]. Therefore, there is no vertical assymptote. How would you have to modify the equation such that there is an vertical asymptote? Would it have to be a totally different equation? Obviously, it needs to be an exponentially increasing/decreasing curve (let's stick with increasing though). 2) Supposing you did modify the equation such that there was a vertical asymptote. What would you have to do to adjust where along the x-axis this asymptote intersected? Link to comment Share on other sites More sharing options...
aommaster Posted October 18, 2005 Share Posted October 18, 2005 1.I would think that assymptotes are formed where there are no values of y which can be produced when x is a certain number. A very common assymptote is 1/x where y cannot be found when x=0. This is, probably, how you would need to modify the equation. 2.As for your second question, I really don't understand what you mean by "What would you have to do to adjust where along the x-axis this asymptote intersected?", especially the 'intersected' part. Hope this helps! Link to comment Share on other sites More sharing options...
gib65 Posted October 18, 2005 Author Share Posted October 18, 2005 What I mean is... take the equation y = 1/x that you mentioned. The asymptote for this is at x=0. What would the equation look like if the asymptote was at x=1? At this point, I can answer my own question: in order for the denominator to be 0 and x be 1, the equation would have to be y = 1/(x-1). Thank you. Link to comment Share on other sites More sharing options...
TD Posted October 18, 2005 Share Posted October 18, 2005 You kind of solved the problem yourself. Vertical asymptotes will always appear when you have a fraction and certain values of x for which the denominator becomes 0 while the nominator doesn't. So a (simple) function of the form y = 1/(x-a) will always yield a vertical asymptote at x = a. Just substitute a to get a vertical asymptote at any desired x-value. Link to comment Share on other sites More sharing options...
aommaster Posted October 18, 2005 Share Posted October 18, 2005 Please allow me to add to that: f(x)= the normal function f(x+a)= is a shift in the left direction (ie the graph moves towards the leftof the graph) It works the opposite of what you might think f(x)+a= the graph moves upwards, exactly what you might think Link to comment Share on other sites More sharing options...
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