<p>how to find asymptotes of hyperbola? ( For example f(x)=(x+3) /(x+1)
I know that x asymptote is x=-1, but I can't find the theory to find y asymptote. Can someone explain this to me?</p>
<p>If I remember correctly, if the degree of the numerator x is greater than the degree of denominator x, then there is no horizontal asymptote. If the degree of the 2 Xs are equal, then you divide the coefficients to get the asympotote. If the degree of the denominator x is greater than the degree of the numerator x, the horizontal asymptote is x=0. Hope this makes sense.</p>
<p>Basically what LightSource said. So in your example, the horizontal asymptote would be 1, because the coefficient of x in the numerator is 1 and the coefficient of x in the denominator is also 1. 1/1 yields 1.</p>
<p>I dont think that function is a hyperbola…but anyway, for finding the y-asymptote:
You first simplify the function. In this case it’s already simplified.
you then look at the x values of the numerator and the denominator. in here its x and x. Note that as x becomes bigger and bigger, the +3 and +1 on the fractions become negligible. For example, if you plug in 100000 as x, the +3 and +1 are too small to even consider. In other words, as x becomes too big, the fraction is basically (x)/(x) which equals 1. That’s the horizontal asymptote. (This is all basically what LightSource said.)
You can also check this by plugging in this function into your graphing calculator and go to 2nd WINDOW and plug in something like 500000. Then you do 2nd GRAPH to see the horizontal asymptote.</p>
<p>find the limit as x approaches infinity, if it approaches a finite value then that is your horizontal asymptote</p>
<p>eg. in your example, we should try find the limit as x goes to infinity of (x+3)/(x+1) = 1</p>
<p>use your Ti</p>