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Question Video: Determining Rational Functions from Their Graphs Mathematics • 10th Grade

Which of the following is the equation of the graphed function 𝑓(π‘₯) whose asymptotes are π‘₯ = 1 and 𝑦 = 2? [A] 𝑓(π‘₯) = (π‘₯ + 1)/(π‘₯ βˆ’ 1) [B] 𝑓(π‘₯) = (π‘₯ + 1)/(π‘₯ + 2) [C] 𝑓(π‘₯) = (2π‘₯ + 1)/(π‘₯ βˆ’ 1) [D] 𝑓(π‘₯) = (2π‘₯ + 1)/(π‘₯ + 1) [E] 𝑓(π‘₯) = (π‘₯ + 1)/(π‘₯ βˆ’ 2)

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Video Transcript

Which of the following is the equation of the graphed function 𝑓 of π‘₯ whose asymptotes are π‘₯ equals one and 𝑦 equals two? We are given five options for 𝑓 of π‘₯. The first option is π‘₯ plus one over π‘₯ minus one. The second is π‘₯ plus one over π‘₯ plus two. The third option is two π‘₯ plus one over π‘₯ minus one. The fourth option is two π‘₯ plus one over π‘₯ plus one. And the fifth option is π‘₯ plus one over π‘₯ minus two.

Let’s begin by highlighting the two asymptotes on the diagram. It’s important to notice that the scaling on both the π‘₯- and the 𝑦-axis is by two. π‘₯ equals one is our vertical asymptote, which we have highlighted in orange. 𝑦 equals two is the horizontal asymptote, which we’ve highlighted in pink. The vertical asymptote in particular tells us at which value of π‘₯ our function is undefined. We recall that a rational function is an algebraic fraction where both the numerator and denominator are polynomials. We also recall that rational functions, just like any fraction, are not defined when the denominator equals zero.

All five options that we were given fit the description of a rational function. In fact, these rational functions are of the form 𝑓 of π‘₯ equal to π‘šπ‘₯ plus 𝑛 over 𝑝π‘₯ plus π‘ž. In other words, the numerator and the denominator each contain a linear polynomial. We recall that for the graph of a rational function of this form, the vertical asymptote is at the root of the denominator. The root of the polynomial in the denominator is found by setting the denominator equal to zero. In general, that means that π‘₯ equals negative π‘ž divided by 𝑝. This will be true as long as 𝑝 does not equal zero and as long as the numerator does not share the same root as the denominator.

Now we are ready to identify the vertical asymptote of each of the five functions that we were given. We will eliminate any option that does not have a vertical asymptote of π‘₯ equals one. To find the root of the denominator of our first option, we set π‘₯ minus one equal to zero. That gives us a vertical asymptote of π‘₯ equals one. So this is a good possibility. However, we cannot be sure that this is the correct answer because we have not yet found the horizontal asymptote of this first option. We can quickly eliminate the second option because the root of the denominator gives us a vertical asymptote of π‘₯ equals negative two. The third option has a vertical asymptote of π‘₯ equals one. So this is another good possibility.

We should notice at this point that a denominator of π‘₯ minus one gives us a vertical asymptote of π‘₯ equals one. Since the last two options do not have a denominator of π‘₯ minus one nor do they give us a vertical asymptote of π‘₯ equals one, we eliminate those possibilities. Therefore, of the five choices we were given, only two remain as possibilities.

Now we recall how to find the horizontal asymptote for the graph of a rational function in this form. To do this, we simply divide the leading coefficient of the numerator by the leading coefficient of the denominator, resulting in the equation 𝑦 equals π‘š divided by 𝑝. We note that π‘₯ is understood to have a leading coefficient of one. We see in our first function that both leading coefficients are one, and one divided by one is one. Therefore, the horizontal asymptote of the first function is 𝑦 equals one.

Then, we will skip to the third option, as we have already eliminated the second option. We see that the numerator has a leading coefficient of two, and the denominator has an understood leading coefficient of one. The quotient of these two leading coefficients gives us our horizontal asymptote of 𝑦 equals two. We can now eliminate the first option because it does not have a horizontal asymptote of 𝑦 equals two. Out of the five options, only option (C) has the correct vertical asymptote and horizontal asymptote.

In other multiple-choice questions, such as this one, we may find that multiple options have the same vertical and horizontal asymptotes. In that case, we will need a third way to differentiate between the two functions with the same asymptotes. To do this, we would simply choose some ordered pairs from the graph and check that those ordered pairs are solutions in the function that we have selected. Ideally, we would check the 𝑦-intercept. The 𝑦-intercept of this graph appears to be the point zero, negative one. We could also use the point two, five. Really, any clear point from the graph would work.

To check the 𝑦-intercept of zero, negative one with our function, we would simply substitute zero for π‘₯. In this case, 𝑓 of zero equals negative one, as expected. However, if we were still deciding between option (A) and option (C), checking the point zero, negative one would not be sufficient because option (A) also has a solution of zero, negative one. So let’s check two, five. Two, five is not a solution in the function given as option (A). But two, five is a solution to the function in option (C). 𝑓 of two equals five.

In conclusion, out of the five options that we were given, the only function that matches the graph and has the asymptotes π‘₯ equals one and 𝑦 equals two is the function 𝑓 of π‘₯ equals two π‘₯ plus one over π‘₯ minus one.

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