I the curve (1-2x)/(3x+5). I have been asked to find the verticle and horizontal asymptotes. Can anyone help me with a strategy?
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I dunno but that answer looks a little too complicated...horizontal asymptote: take limit of f(x) as x tends to infinity. What is the relation between the limit value you find and the horizontal asymptote?
That's interesting. I never thought of it like that. That suggests to me that the horizontal asymptote of a function should be equivalent to the vertical asymptote of its inverse (if the inverse exists). Let's see. That might be too general.Originally posted by FZ+
I dunno but that answer looks a little too complicated...
y = (1-2x)/(3x+5)
3xy + 5y = 1 - 2x
x = (1 - 5y )/(3y + 2)
Now, find the value y can't have....
but beware, I have been wrong before (intentionally of course! )