gib65 Posted October 18, 2005 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?
aommaster Posted October 18, 2005 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!
gib65 Posted October 18, 2005 Author 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.
TD Posted October 18, 2005 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.
aommaster Posted October 18, 2005 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
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