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Hi there. Sometimes the graph of a rational function may have missing points or holes, with or without vertical asymptotes. In this example, we will be working with rational functions whose graphs have a hole. We are asked to find all vertical asymptotes of the graph of each rational function, as seen in parts a and b. In part a, we have h of x equal to, in the numerator, x squared minus 9 over, in the denominator, x minus 3. In part B, we have g of x equal to x plus 2 in the numerator over x squared minus 4 in the denominator. The vertical asymptotes of a rational function are defined by the values of x that make the reduced form of the original function undefined. The reduced form of a rational function is obtained by completely factoring the numerator and denominator of the function and canceling out factors common to the numerator and denominator. For the function in part A, we start by factoring the numerator and denominator. We have that h of x is equal to-- in the numerator, we can factor that using the difference of squares formula. That's factored as x plus 3 times x minus 3. And in the denominator, that factor is already prime, so we leave it alone as x minus 3. At this point, we see that the factors x minus 3 in the numerator and denominator can be cancelled, and so h of x gets reduced to x plus 3 if x is not equal to 3. We see that there is no value for x that makes this reduced form undefined. Therefore, there is no vertical asymptote for the function h of x. We do note that the domain of x excludes the value x equals 3, referred to as a hole. This is because the function is undefined at that value in its original form prior to reducing it. However, x equals 3 is not a vertical asymptote. For the function in part B, we start by factoring the numerator and denominator, as we did for the function in part a. We have that g of x is equal to x plus 2 in the numerator. That factor is already prime. And then we have the difference of squares formula coming in again for x squared minus 4, which factors as x plus 2 times x minus 2. Again, we see that x plus 2 cancels both in the numerator and denominator, and so we have that g of x reduces to the form 1 over x minus 2 if x is not equal to minus 2. Thus, there is a vertical asymptote for this function at x equal 2. Again, we note, as we did in part A, that the domain of x also excludes the value x equal minus 2, called a hole, at x equal minus 2 because the function is undefined at that value in its original form prior to reducing it. However, x equal 2 is a vertical asymptote for this function.

Video Details

Duration: 3 minutes and 8 seconds
Country: United States
Language: English
License: Dotsub - Standard License
Genre: None
Views: 12
Posted by: 3play on Jul 24, 2017

Please translate to spa_la. Account ID: 585. Notes on format and other things are here: http://s3.amazonaws.com/originp3/app/translation-profiles/profiles/c728d56a6e3afc44c0a63b925c143995.html

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