Video: Hyperbolas and Their Asymptotes

By writing the expression (π‘Žπ‘₯ + 𝑏)/(𝑐π‘₯ + 𝑑) in the form ((𝐴)/(𝑃π‘₯ + 𝑄)) + 𝑅, determine the asymptotes of ((5π‘₯ βˆ’ 13)/(π‘₯ βˆ’ 3)) + ((2 + 12π‘₯)/(1 βˆ’ 2π‘₯)).

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Video Transcript

By writing the expression π‘Žπ‘₯ plus 𝑏 divided by 𝑐π‘₯ plus 𝑑 in the form 𝐴 divided by 𝑃π‘₯ plus 𝑄 plus 𝑅, determine the asymptotes of five π‘₯ minus 13 divided by π‘₯ minus three plus two plus 12π‘₯ divided by one minus two π‘₯.

The question is asking us to determine the asymptotes of the sum of two rational functions. And it wants us to do this by writing each term in a different form. And we might ask, why does the question want us to write these terms in this form? This is just a slightly different form for these types of rational functions. However, using this, we can easily find the horizontal and vertical asymptotes.

We know the horizontal asymptotes of our function will be the value of the function as π‘₯ approaches ∞. And we can see in this form as π‘₯ approaches ∞, 𝐴 divided by 𝑃π‘₯ plus 𝑄 approaches zero. So, we can see we’ll have a horizontal asymptote where 𝑦 is equal to 𝑅. And we know that rational functions must have a vertical asymptote for values of π‘₯ where the denominator is equal to zero and the numerator is not equal to zero.

In other words, provided the value of capital 𝐴 is not equal to zero, we just need to solve 𝑃π‘₯ plus 𝑄 equal zero to find our vertical asymptotes. And solving this, we get π‘₯ is equal to negative 𝑄 divided by 𝑃 provided, of course, that 𝑃 is not equal to zero. So, the first thing we need to do is write both of our terms in this form. We’ll do this by using algebraic division. Let’s start with five π‘₯ minus 13 divided by π‘₯ minus three.

First, we know that five π‘₯ divided by π‘₯ is equal to five. Now, we need to multiply π‘₯ minus three by five. And five times π‘₯ minus three is equal to five π‘₯ minus 15. And we want to subtract this from five π‘₯ minus 13 to find our remainder term. First, we have that five π‘₯ minus five π‘₯ is equal to zero. Next, we have negative 13 minus negative 15. That’s negative 13 plus 15 is equal to two. So, we see that π‘₯ minus three goes into five π‘₯ minus 13 five times with remainder two.

In other words, five π‘₯ minus 13 is equal to five times π‘₯ minus three plus two. And to find an expression in the form we want, we’re going to divide through both sides of this equation by π‘₯ minus three. And this gives us that five π‘₯ minus 13 divided by π‘₯ minus three is equal to five plus two divided by π‘₯ minus three. So, using this method, we’ve written the first term given to us in the question in the form the question wants us to. We now want to do the same with our second term.

Again, we’ll do this by algebraic division. We’ll write our terms in increasing powers of π‘₯. We see that 12π‘₯ divided by negative two π‘₯ is equal to negative six. Now, we need to find our remainder term. We’ll do this by subtracting negative six multiplied by negative two π‘₯ plus one from 12π‘₯ plus two. Negative six times negative two π‘₯ is equal to 12π‘₯. And then, negative six times one is just equal to negative six. So, we have 12π‘₯ plus two minus 12π‘₯ plus six, which is of course equal to eight.

So, what we’ve shown is 12π‘₯ plus two is equal to negative six times negative two π‘₯ plus one plus eight. And we’ll divide both sides of this equation through by negative two π‘₯ plus one. Dividing through by negative two π‘₯ plus one gives us the following expression. And, of course, 12π‘₯ plus two divided by negative two π‘₯ plus one is equal to two plus 12π‘₯ divided by one minus two π‘₯. It’s equal to the second term given to us in the question. So, we’ve now rewritten both terms given to us in the question in this form.

So, let’s now combine these expressions. We showed that five π‘₯ minus 13 divided by π‘₯ minus three is equal to five plus two divided by π‘₯ minus three. And two plus 12π‘₯ divided by one minus two π‘₯ is equal to negative six plus eight divided by one minus two π‘₯. And we can simplify this since five minus six is equal to negative one. And we can find the horizontal and vertical asymptotes in this expression by using the exact same logic we did before. We find the horizontal asymptotes by taking limits as π‘₯ is approaching ∞ and π‘₯ is approaching negative ∞.

And in both of these cases, we can see two divided by π‘₯ minus three and eight divided by one minus two π‘₯ are approaching zero. Since their numerators remain constant, however, their denominators are growing without bound. This tells us that we’ll have a horizontal asymptote when 𝑦 is equal to negative one. And for rational functions, we’ll always have a vertical asymptote if the denominator is equal to zero for that value of π‘₯ and the numerator is not equal to zero.

And in this case, our numerators are just constant, so we need to solve π‘₯ minus three equals zero and one minus two π‘₯ equals zero. Solving π‘₯ minus three equals zero gives us π‘₯ is equal to three. And solving one minus two π‘₯ is equal to zero gives us π‘₯ is equal to one-half. Therefore, we’ve shown the asymptotes of five π‘₯ minus 13 divided by π‘₯ minus three plus two plus 12π‘₯ divided by one minus π‘₯ are π‘₯ is equal to one-half, π‘₯ is equal to three, and 𝑦 is equal to negative one.

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