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 = 4 + 3x² / (x² + 1)

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to find the domain of the function Y is equal to 5 + 6 X2 divided by the quantity of X2 + 4 in quantity, then use limits to determine its asymptotes. So the first part of this problem, we need to find the domain of our function Y, and we see that our function Y contains a fraction. So our function Y will be undefined if the denominator of that fraction is equal to 0 at any point. If we look at X2 + 4, which is the denominator of that fraction, we see that that quantity X2 + 4 is always greater than 0. So our denominator is never going to be equal to 0, so that means that Y is defined for all real numbers. So that means that the domain for our function in this case is going to be all real numbers. So that's going to be the open interval from minus infinity. To infinity So that is the first part of our question answered. The next part, we need to use limits to determine its asymptotes. So, if we first look for vertical asymptotes, we noticed that our function is defined for all real numbers. So that means that we do not have any vertical asymptotes, which would correspond to our denominator and our fraction there being equal to 0. So next we'll look for horizontal asymptotes, and we see that our fraction has the same order of polynomial in both the numerator and denominator, so we will have a horizontal asymptotes. So we're going to look at the limit as X approaches infinity. Of the quantity of 5 plus 6 x squad. Divided by the quantity of X2 + 4. So this will give us the equation for our horizontal asymptote. So, in order to take this limit, I'm first going to divide our numerator and denominator for a fraction by X2. So this is going to be equal to the limit as X approaches infinity of the quantity of 5 plus 6 divided by the quantity of 1 + 4 divided by X squared in quantity. So now, when X approaches infinity for this limit, We see that the quantity of 4 divided by X squad is going to approach 0. So this limit is going to be equal to 5 plus 6 divided by the quantity of 1 plus 0, which is equal to 11. We also need to check The limit when X approaches minus infinity. And so we'll take that limit. For the same quantity that we did before and we're going to go ahead and take a shortcut and go ahead and divide both the numerator and denominator of a fraction by x2. So we'll have the limit as x approaches minus affinity of 5 plus 6 divided by the quantity of 1 + 4 divided by x 2. In quantity, which again is equal to 5 plus 6 divided by the quantity of 1 + 0, which is equal to 11. And so we get the same value, whether we're going to the uh as X approaches positive infinity or minus infinity. That means that we have a horizontal asthmatote at Y is equal to 11. So that is the answer to the second half of this problem. So just to recap, we have a domain that is the open interval from minus infinity to infinity, and we have a horizontal asymptote at Y equal to 11. So those are the answers for this problem. So I hope this explanation has been useful, and I'll see you in the next video.

Domains and Asymptotes

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

y = 4 + 3x² / (x² + 1)

Key Concepts

Domain of a Function

Limits

Asymptotes

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