<p>Every Maths Level 2 Subject test paper asks at least 2-3 questions on asymptotes. I am finding that many people here at CC are having difficulties in working with them. Here is a small guide to help you crack every asymptote problem quite easily.</p>
<p>When are vertical asymptotes formed ?</p>
<p>Vertical asymptotes are formed for values of x such that the denominator of the equation becomes equal to 0. But beware, if the equation contains factorable terms, first factor them out and cancel them if divisible and then check to see when the denominator becomes 0.</p>
<p>For Example: In the equation: y=1/(x-5), the denominator becomes equal to 0 when x = 5. therefore the equation has a vertical asymptote at x = 5.</p>
<p>However in y=(x-1)/(x2-4x+3), it may seem that the denominator becomes 0 for x=1 and x=3.
But this is not so. First factorize the denominator to get:<br>
y=(x-1)/(x-3)(x-1) and Therefore y=1/(x-3)
So the vertical asymptote is formed only at x=3.</p>
<p>When are horizontal asymptotes formed ?</p>
<p>Horizontal Asymptotes are formed according to these three rules. Remember these and find horizontal asymptotes of any equation will be very easy.</p>
<p>If the degree of the numerator is greater than the degree of the denominator there is no horizontal asymptote. </p>
<p>If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is the fraction made by the leading coefficients. </p>
<p>If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is the line y = 0. </p>
<p>Hope that helped. Please feel free to ask any queries related to this on this thread.</p>
<p>Follow the procedure for vertical asymptote: set the denominator equal to zero and find the value of x. Then, substitute that x value into the numerator. If the numerator equals 0, then there is a hole in the graph. If the numerator equals a nonzero constant, it is a vertical asymptote.</p>
<p>Let me give three examples of finding horizontal asymptotes. First read and understand these and then criticize my method.</p>
<ol>
<li><p>y = (x2 + x)/(x+1). Here the degree of x in numerator(2) is greater than that in the denominator(1). Therefore there is no horizontal asymtote.</p></li>
<li><p>y = (3x2 -2)/(4x2 +3). Here the degree of x in numerator(2) is equal to that in the denominator(2). So the horizontal asymptote is formed by the lead coeffecients: 3/4. Therefore y = 3/4 is the horizontal asymptote here.</p></li>
<li><p>y = x/(1+x2). Here the degree of x in numerator(1) is lesser than that in the denominator(2). Therefore the horizontal asymtote is y=0.</p></li>
</ol>
<p>It is that simple.</p>
<p>There is no need to find any limits or perform any substitution at all. You get the asymptotes simply by looking at the equation.</p>
<p>Well, that method simply comes from limits, using the fact that x^a is asymptotically smaller than x^b if a < b (that is, x^a = o(x^b)).</p>
<p>Also, indeterminate forms can lead to pitfalls, and just because a rational function evaluates to 0/0 doesn’t always mean that there is a “hole” there.</p>
<p>For example, let f(x) = (x^3 - x^2 + x - 1)/(x^2 - 2x + 1). f(1) is indeterminate (0/0). However there is a vertical asymptote there. The point is, indeterminate is indeterminate for a reason.</p>