Domains and Asymptotes

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Question

Domains and Asymptotes

Determine the domain of each function in Exercises 69–72. Then use various limits to find the asymptotes.

y = x³ / (x³ − 8)

Accepted Answer

Hello there. Today we're going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Find the domain of the function Y equals 2 x 3, all divided by X 3 minus 64, then use limits to determine its asymptotes. Awesome. So it appears for this particular problem we're asked to solve for two separate answers. Our first answer is we're asked to figure out what the domain of this function of Y is. That's our first answer. And our second answer is we're asked to use limits to help determine the asymptotes of this function of Y. So now that we know that we're trying to find the domain and the asymptotes of this particular function of Y. Our first step that we need to take is focus our attention on solving for the domain. So once again, let us note that our given function states that Y is equal to 2 x 3, all divided by X 3 minus 64. And in order to determine the domain, as we should recall, we need to check where the function is undefined, which this occurs when the denominator equals 0, so we need to take X to the power of 3 minus 64 and set it equal to 0. So when we rearrange this equation to isolate sulfur X all by itself, we will find that X is going to be equal to the cube root of 64, which will mean that X is going to be equal to simply 4. So since the function is undefined at X equals 4, the domain excludes this value. So that will mean that our domain, without a doubt, will be parenthesis, negative infinity 4 union. Parentheses 4 infinity where parentheses means an open interval. So once again it's parentheses negative infinity 4 parent union 4 infinity. So that is our domain. Hooray, we did it. We solve for our first answer. Now we need to determine the asymptotes using limits. So with that said, let us try to find the vertical asymptote first. So as you should recall and note, the vertical asymptote occurs where the denominator is 0 and the function is undefined. So since X equals 4 makes the denominator 0, let us check the following limits. So let us take the limit as X approaches. Or from the left of 2 X to the power of 3, all divided by X 3 minus 64 and the limit as X approaches 4 from the right of 2 X to the power of 3, all divided by X 3 minus 64. So we need to note that as X approaches 4 from the left, that will mean that And let's put a a call. So that will mean as X approaches 4 from the left, that will mean as a result that X to the power 3 minus 64. We approach 0 from the left, which will be a very small negative number, then a positive constant divided by this very small negative number will result in a negative infinity value. So that will mean, without a doubt, we could go ahead and write that. We can go ahead and write that the limit as X approaches 4 from the left of 2 X to the power of 3, so 2 X to the power of 3, all divided by X 3 minus 64, fantastic. will be equal to When we do direct substitution methods, it'll be 2 multiplied by 4 power of 3, all divided by 43 minus 64. Fantastic. So that will mean that when we evaluate this limit, it will be equal to negative infinity. So now shifting gears to focus our attention on as X approaches 4 from the right. So as X approaches 4 from the right, that will mean that X 3 minus 64 will approach 0 from the right, which is a very small positive number. Then a positive constant divided by a very small positive number will result in a positive infinity value, meaning we can go ahead and write that the limit as X approaches 4 from the right of 2 X 3 divided by X 3 minus 64, which is going to be equal to when we use the direct substitution method. So 2 multiplied by 4 of 3, all divided by 43 minus 64, which will give us a positive infinity value. So, thus there will be a vertical asymptote at X equals 4. So let's make a note of that. So, X equals 4 will be our vertical asymptote. Awesome. So I'll draw an arrow. Fantastic. So that is our second answer. But now we need to determine whether or not there are any horizontal asymptotes now. So now our next step is we need to compare the degrees of the numinator and the denominator. So comparing the degrees of our numerator, which is the top part of our division bar, and the denominator, which is the bottom part of our division bar. So comparing the degrees, we need to note that the numerator 2x of the power of 3 has a degree of 3. And the denominator X 3 minus 64 has a degree of 3, and since the degrees are equal, the function will have a horizontal asymptote given by the ratio of the leading coefficients. So that would mean for the horizontal asymptote, so to find the horizontal asymptote, we will need to take the limit of Y. So we need to take the limit of Y as X, so let's make a note of this, as X approaches plus or minus infinity. So, meaning we need to take the limit as X approaches plus or minus infinity of 2 X to the power of 3. All divided by X to the power of 3 minus 64. Fantastic. So now, our next step. we need to divide the numerator and the denominator by X to the power of 3. So we need to take the limit as X approaches plus or minus infinity. Of 2 X to the power of 3. So we need to take 2 X to the power of 3, divided by X to the power of 3, all divided by X to the power of 3, divided by X to the power of 3 minus 64 divided by X to the power of 3, which will be equal to the limit as X approaches plus or minus infinity of 2 divided by 1 minus 64 divided by X to the power of 3. So we need to note that as X approaches plus or minus infinity, we need to note that the term 64 divided by X of the power 3 will approach 0. So, therefore, without a doubt, we can go ahead and write that the limit as X approaches plus or minus infinity of 2 divided by 1 minus 64 divided by X to the power of 3, which will be equal to use the direct substitution method. It will give us 2 divided by 1 minus 0, which will give us an answer of 2. So therefore, without a doubt, the function will have a horizontal ascentote at Y equals 2. Fantastic. So horizontal ascent to HA for short. And that's it. We've officially solved for this problem. So quickly, going back to the top to recap everything that we just solved for together, our final answers without a doubt, have to be the following. So as you can see, we know that our domain is going to be Parents negative infinity 4 union. Our vertical ascentote will be at X equals 4, and our horizontal ascentote will be at Y equals 2. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Domains and Asymptotes

Determine the domain of each function in Exercises 69–72. Then use various limits to find the asymptotes.

y = x³ / (x³ − 8)

Key Concepts

Domain of a Function

Limits

Asymptotes

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