Explainer: Graphs of Rational Functions

In this explainer, we will learn how to sketch the graphs of simple rational functions and identify simple rational functions from their graphs.

These are the functions 𝑓(π‘₯)=π‘Žπ‘₯+𝑏𝑐π‘₯+𝑑, where π‘Ž,𝑏,𝑐, and 𝑑 are constants.

In order for 𝑓(π‘₯) to truly represent a rational function, we need to have 𝑐≠0. The easiest such example is 𝑓(π‘₯)=1π‘₯ when π‘Ž=𝑑=0 and 𝑏=𝑐=1, the reciprocal function. The figure shows the curve 𝑦=1π‘₯.

Geometrically, this is a hyperbola. Notice the two asymptotes:

  1. the horizontal asymptote: the line 𝑦=0 which the graph approaches as π‘₯ tends to ∞ and βˆ’βˆž;
  2. the vertical asymptote: the line π‘₯=0 which represents the value 0, missing from the domain of the function 𝑓, and which is such that the curve 𝑦=𝑓(π‘₯) approaches this line as π‘₯ tends to 0 either from above or from below.

Consider the curve

𝑦=3βˆ’5π‘₯βˆ’2(1)

which is the same as

𝑦=3π‘₯βˆ’11π‘₯βˆ’2.(2)

We will use both equations to examine what the graph is like:

  1. The denominator π‘₯βˆ’2 in both forms tells us that the function is not defined at π‘₯=2. This locates the vertical asymptote.
  2. The form in (1) tells us that when π‘₯ is very large (positive or negative), the term 5π‘₯βˆ’2 is very small so that π‘¦β‰ˆ3. This locates the horizontal asymptote as the line 𝑦=3.

The following figure shows the curve 𝑦=3π‘₯βˆ’11π‘₯βˆ’2 together with its two asymptotes.

Notice that this curve is a hyperbola, which looks very much like 𝑦=1π‘₯. If we write (1) as 𝑦=3+βˆ’5π‘₯βˆ’2 rather than as 𝑦=3βˆ’5π‘₯βˆ’2, it is really much more like 𝑦=5π‘₯. Better still, 𝑦=βˆ’5π‘₯. The negative multiple corresponds to the curve lying in the 2nd and 4th quadrants.

Perhaps the easiest way to decide is to ask, β€œAs π‘₯ is going to infinity, is the curve above or below the horizontal asymptote?,” which we can answer with a little algebra, using the form in (2): 𝑦=3π‘₯βˆ’11π‘₯βˆ’2=βˆ’βˆ’π‘₯)=3βˆ’1βˆ’.οŠ©ο—ο—οŠ§οŠ§ο—ο—ο—οŠ¨ο—οŠ§οŠ§ο—οŠ¨ο—(dividethroughby

From this, we see the following.

  1. If π‘₯ is very big and positive (think π‘₯=10,000), then 1βˆ’2π‘₯β‰ˆ1 and is positive, while 3βˆ’11π‘₯ is also positive and close to 3. Therefore, 𝑦 is tending to 3 as π‘₯ is going to infinity, and the horizontal asymptote 𝑦=3.
  2. How does the value of 𝑦 at π‘₯=10,000 compare with 3? Is it greater than or less than 3? Since 3βˆ’11π‘₯<3, it follows that 3βˆ’1βˆ’<31βˆ’β‰ˆ3π‘₯.οŠ§οŠ§ο—οŠ¨ο—οŠ¨ο—forlargepositivevaluesof
    The curve 𝑦=𝑓(π‘₯) lies below the asymptote.

In summary, suppose we are to identify the curve 𝑦=π‘Žπ‘₯+𝑏𝑐π‘₯+𝑑.

Relating the Graph to the Expression π‘Žπ‘₯ + 𝑏/𝑐π‘₯ + 𝑑 Where π‘₯ β‰  0

  1. From 𝑐π‘₯+𝑑=𝑐π‘₯+𝑑𝑐, we see that the vertical asymptote is at π‘₯=βˆ’π‘‘π‘₯.
  2. From π‘Žπ‘₯+𝑏𝑐π‘₯+𝑑=π‘Ž+𝑐+οŒ»ο—οŒ½ο—, we see that the horizontal asymptote is 𝑦=π‘Žπ‘.
  3. By considering the expression above for large positive values of π‘₯, we decide whether the curve is above or below the asymptote 𝑦=π‘Žπ‘ as π‘₯ tends to ∞.

Example 1: Identifying the Graphs of Simple Rational Functions

Which of the following graphs represents 𝑓(π‘₯)=1π‘₯+1?

Answer

From the denominator π‘₯+1, we find the vertical asymptote must be at π‘₯=βˆ’1. Hence, graph (d) can not be the answer. The horizontal comes from dividing through by π‘₯: 1π‘₯+1=1+οŠ§ο—οŠ§ο— which goes to 01=0 when π‘₯ is large. Since these numbers are also positive when π‘₯ is large and positive, the curve must lie above the π‘₯-axis. The graph (c) represents 𝑓(π‘₯)=1π‘₯+1.

The only option is (c).

Suppose that, instead, we wanted to go in the other direction. We are given a graph, and we want to know what rational function it is the graph of. Assuming that it β€œcomes from” the function 𝑓(π‘₯)=1π‘₯, we have two ways to identify the function:

  • by considering transformations of graphs: translations, dilations, and reflections in the axes,
  • by considering the algebraic form 𝑓(π‘₯)=π‘Žπ‘₯+𝑏𝑐π‘₯+𝑑 and determining the coefficients.

As the second method is more applicable generally, let us have a look at it.

  1. Notice that we can always write 𝑓(π‘₯) in the form π‘Žπ‘₯+𝑏π‘₯+𝑑 by dividing the numerator and the denominator by 𝑐, which is not zero since the graph is a hyperbola, not a line. So we assume that 𝑐=1.
  2. We know what 𝑑 is because the denominator is zero when π‘₯=βˆ’π‘‘, so that is where the vertical asymptote is found.
  3. We also know that, in the form above, the horizontal asymptote occurs at 𝑦=π‘Ž since, dividing throughout by π‘₯, we have π‘Žπ‘₯+𝑏π‘₯+𝑑=π‘Ž+1+,οŒ»ο—οŒ½ο— so the graph tells us what π‘Ž must be also.
  4. We are left with one additional piece of data which the graph must give us: the value of 𝑏. One way to get this is by identifying a single point (𝑝,π‘ž) on the graph, be it the intercept with an axis or any other clearly identifiable point. This is because from 𝑦=π‘Žπ‘₯+𝑏π‘₯+𝑑 we get the equation π‘ž=π‘Žπ‘+𝑏𝑝+𝑑 in which the only unknown is 𝑏. Solve for 𝑏 and we are done.

Example 2: Determining the Equation of a Function from Its Graph

What function is represented in the figure below?

Answer

The vertical asymptote is π‘₯=0, so the form of the function is π‘Žπ‘₯+𝑏π‘₯ with constant 𝑑=0.

The horizontal asymptote is the line 𝑦=βˆ’3, which fixes π‘Ž=βˆ’3, and the rational expression is now βˆ’3π‘₯+𝑏π‘₯.

We can see that the point (1,βˆ’4) is on the graph, so we solve for 𝑏 in the equation βˆ’4=βˆ’3(1)+𝑏1βˆ’4=βˆ’3+π‘βˆ’1=𝑏.so

The function graphed is 𝑓(π‘₯)=βˆ’3π‘₯βˆ’1π‘₯𝑓(π‘₯)=βˆ’3βˆ’1π‘₯.

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