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Video: Matching the Rule of a Rational Function with Its Graph

Alex Cutbill

Which of the following graphs represents 𝑓(𝑥) = 1/(𝑥 + 1)? [A] Figure (a) [B] Figure (b) [C] Figure (c) [D] Figure (d)

04:21

Video Transcript

Which of the following graphs represents 𝑓 of π‘₯ equals one over π‘₯ plus one?

We have four options to choose from here. Let’s think about what the graph of 𝑓 of π‘₯ should look like and eliminate the graphs that we can see can’t possibly be right.

Firstly, let’s think where the vertical asymptotes of the graph of 𝑓 of π‘₯ should be. I’ve marked the vertical asymptotes of each of the four options. A rational function has a vertical asymptote where its denominator has a zero. The denominator of our rational function is π‘₯ plus one. And so the graph of this function will have a vertical asymptote when π‘₯ plus one is equal to zero, in other words when π‘₯ is equal to negative one.

Let’s go through our options in turn and check that they have a vertical asymptote in the right place. We can see that option a has a vertical asymptote when π‘₯ is equal to zero, but no vertical asymptote when π‘₯ is equal to negative one. This can’t be the graph of our function. How about option b? This graph has a vertical asymptote when π‘₯ is equal to one. Again, not what we looking for. Option c has a vertical asymptote at the right place at π‘₯ equals negative one. This option is still a possibility. As is option d, it too has a vertical asymptote at π‘₯ equals negative one. Both of these graphs have a horizontal asymptote at 𝑦 equals zero.

Looking at the definition of 𝑓 of π‘₯, this makes sense. As π‘₯ tends to infinity as π‘₯ gets bigger and bigger, 𝑓 of π‘₯ which is one over π‘₯ plus one gets closer and closer to zero. Similarly, as π‘₯ tends to negative infinity as π‘₯ passes minus a 1000, minus 10000, minus a 1000000, and so on, 𝑓 of π‘₯ which is one over π‘₯ plus one gets closer and closer to zero. So how are we going to choose between these two graphs? Graph c, so as 𝑓 of π‘₯ being negative to the left of the vertical asymptote and positive to the right of it. For example from this graph, it appears that 𝑓 of negative two is equal to negative one. It’s negative. And 𝑓 of zero is equal to one.

For option d, the opposite is true. The left of the vertical asymptote, the function appears to be positive. And the right of it, it appears to be negative. For example, it shows 𝑓 of negative two which is to the left of the vertical asymptote being positive equal to one and 𝑓 of zero which is the right of the vertical asymptote be negative, equal to negative one.

So to decide between the two options that we have left, we have to think about what happens to the left of the vertical asymptote when π‘₯ is less than negative one. And what should happen to the right of the vertical asymptote when π‘₯ is greater than negative one? You can prove that when π‘₯ is less than negative one, 𝑓 of π‘₯ which is one over π‘₯ plus one should be less than zero. And when π‘₯ is greater than negative one, 𝑓 of π‘₯ which is one over π‘₯ plus one should be greater than zero. However, there is an easier way. We can simply find 𝑓 of zero.

On option c, it looks like 𝑓 of zero is about one. Maybe we’re not sure exactly what 𝑓 of zero is from the graph, but we know it is at least positive, whereas in option d it looks like 𝑓 of zero is negative. We compute 𝑓 of zero using the definition of the function in the question. Substituting π‘₯ for zero, we get one over zero plus one. And that is of course just one. Based on this fact, we can eliminate option 𝑑 where 𝑓 of zero looks to be negative one and select option c as our correct answer.