Lesson Explainer: Graphs of Rational Functions Mathematics • 10th Grade

In this explainer, we will learn how to graph rational functions whose denominators are linear, determine the types of their asymptotes, and describe their end behaviors.

A rational function is a function defined by an algebraic fraction where both the numerator and denominator of the quotient are polynomials. In particular, a polynomial is a rational function when we consider the denominator to be 1, which is a zero-degree polynomial. However, since the constant function 1 does not have any roots, polynomials are different from other rational functions that have a higher-degree polynomial as the denominator. In particular, the domain of the rational function whose denominator is a linear function has to exclude the number that corresponds to the root of the linear function, while the domain of any polynomial is all real numbers.

Let us examine how this difference impacts the graphs of these functions by considering an example. We know that the graph of a nonconstant polynomial is continuous and tends to either positive or negative infinity at both ends of the graph. We will see that this is not always the case for rational functions by looking at the simplest rational function that is not a polynomial, 𝑓(𝑥)=1𝑥.

In our first example, we will observe a few important characteristics of the graph of this function that are different from the characteristics of polynomials.

Example 1: Evaluating the End Behavior of Rational Functions

Consider the graph of the function 𝑦=1𝑥.

  1. By looking at the graph and substituting a few successively larger values of 𝑥 into the function, what is the end behavior of the graph as 𝑥 increases along the positive 𝑥-axis?
    1. The value of 𝑦 approaches zero as the value of 𝑥 increases.
    2. The value of 𝑦 approaches infinity as 𝑥 increases.
    3. The value of 𝑦 approaches negative infinity as 𝑥 increases.
  2. Similarly, what is the end behavior of the graph as 𝑥 decreases?
    1. The value of 𝑦 approaches ∞.
    2. The value of 𝑦 approaches −∞.
    3. The value of 𝑦 approaches zero.
  3. Finally, by interpreting the graph, what is happening to the function when the value of 𝑥 approaches zero?
    1. The value of 𝑦 approaches positive infinity when 𝑥 gets closer to zero from the negative direction or from the positive direction.
    2. The value of 𝑦 approaches positive infinity when 𝑥 gets closer to zero from the negative direction and approaches negative infinity when 𝑥 gets closer to zero from the positive direction.
    3. The value of 𝑦 approaches negative infinity when 𝑥 gets closer to zero from the negative direction or from the positive direction.
    4. The value of 𝑦 approaches negative infinity when 𝑥 gets closer to zero from the negative direction and approaches positive infinity when 𝑥 gets closer to zero from the positive direction.

Answer

Part 1

We note the 𝑦-values corresponding to 𝑥=1,2,3 on the graph.

From these points, we can observe that the value of 𝑦 approaches zero as the value of 𝑥 increases. This is choice A.

Part 2

Similarly, we note the 𝑦-values corresponding to 𝑥=−1,−2,−3 on the graph.

From these points, we can observe that the value of 𝑦 approaches zero as the value of 𝑥 decreases. This is choice C.

Part 3

We note several points on the graph near the 𝑦-axis.

From these points, we can observe that the value of 𝑦 approaches negative infinity when 𝑥 gets closer to zero from the negative direction and approaches positive infinity when 𝑥 gets closer to zero from the positive direction. This is choice D.

In the previous example, we observed the characteristics of the graph of 𝑦=1𝑥. We can describe these features via asymptotes. Recall that an asymptote is a straight line that a curve, the graph of 𝑦=1𝑥 in this case, approaches. We noted that the 𝑦-values of points on the graph approach zero as 𝑥 tends to ∞ and −∞. This tells us that the curve, or the graph, approaches the horizontal line described by the equation 𝑦=0.

We also noted that as 𝑥-values approach zero, the corresponding 𝑦-values on the graph approach either positive or negative infinity. This means that the graph approaches the vertical line described by the equation 𝑥=0. We can summarize these results.

Property: Graph of the Reciprocal Function

The graph of 𝑦=1𝑥 has horizontal asymptote 𝑦=0 and vertical asymptote at 𝑥=0.

We note that these features of the graph of the rational function 𝑓(𝑥)=1𝑥 are different from general features expected from a nonconstant polynomial. A nonconstant polynomial has neither a vertical nor a horizontal asymptote. This does not mean that all rational functions other than polynomials have vertical and horizontal asymptotes, which is not true. However, it does portray an important distinction between graphs of rational functions and graphs of polynomials.

The curve in the graph of 𝑦=1𝑥 is a hyperbola. We can apply different function transformations (translation, reflection, and dilation) to this function to obtain different hyperbolas as graphs of other rational functions. In the next example, we will identify the graph of a rational function using a function transformation.

Example 2: Matching the Rule of a Rational Function with Its Graph

Which of the following graphs represents 𝑓(𝑥)=1𝑥+1?

Answer

In this example, we need to identify the graph of 𝑦=1𝑥+1 from the given list of graphs. We can obtain this graph from the graph of the parent function 𝑦=1𝑥 by applying the function transformation 𝑥→𝑥+1. Let us begin by recalling the graph of the parent function 𝑦=1𝑥.

In particular, we note that this graph has horizontal asymptote 𝑦=0 and vertical asymptote 𝑥=0.

Recall that the transformation in the 𝑥-variable 𝑥→𝑥+𝑎 graphically represents a horizontal shift to the left by 𝑎 units. Since we are applying the transformation 𝑥→𝑥+1 to obtain the graph of our function from the parent function, we need to shift the graph above to the left by 1 unit.

When we shift the given graph to the left by 1 unit, it will move the vertical asymptote 𝑥=0 of the parent function to the new vertical asymptote 𝑥=−1. The horizontal asymptote 𝑦=0 of the parent function will not move as a result of this transformation. We can see that only choices c and d meet these conditions. However, the graph in choice d is also reflected over the 𝑥-axis, which is not a part of the function transformation to obtain our function 𝑓(𝑥)=1𝑥+1.

Hence, the correct graph of our function is choice c.

In the next example, we will identify a rational function from its graph.

Example 3: Figuring Out the Rule of a Rational Function given Its Graph

What function is represented in the figure below?

Answer

We begin by noting that the given graph resembles the graph of 𝑦=1𝑥. We can obtain this graph from the graph of the parent function 𝑦=1𝑥 by applying a few function transformations. Let us recall the graph of the parent function 𝑦=1𝑥.

In particular, we note that this graph has horizontal asymptote 𝑦=0 and vertical asymptote 𝑥=0. On the other hand, the given graph in this example has horizontal asymptote 𝑦=−3 and vertical asymptote 𝑥=0. This means that a downward shift by 3 units is one of the function transformations used to obtain this graph from the graph of the parent function.

Before we apply the vertical shift, we should examine whether other transformations are involved, since the order of transformations can affect the outcome. There are three different types of transformations to consider: translation, dilation, and reflection. We have already observed the translation involved, and there does not appear to be any sign of stretching or compressing from the graph of the parent function. Hence, we can rule out dilation.

We can notice that the given graph appears to be upside down when compared to the graph of the parent function. More specifically, the value of 𝑦 approaches positive infinity when 𝑥 gets closer to zero from the negative direction and approaches negative infinity when 𝑥 gets closer to zero from the positive direction. This is exactly the opposite behavior of the parent function. This implies that a reflection is another transformation used to obtain our graph. Due to the symmetry of the graph of the parent function, reflecting over the 𝑥-axis leads to the same result as reflecting over the 𝑦-axis. We will say that this is a reflection over the 𝑥-axis.

When combining transformations, we remember that reflections and dilations must be performed before translations. Recall that reflection over the 𝑥-axis is given by the function transformation 𝑓(𝑥)→−𝑓(𝑥), which means that we multiply the function by −1. Hence, the reflected graph of our function is given by the transformation 1𝑥→−1𝑥.

Next, we consider the translation. Recall that a vertical downward shift by 𝑐 units is given by the function transformation 𝑓(𝑥)→𝑓(𝑥)−𝑐. Since we need to shift the graph above downward by 3 units, we need to apply the transformation −1𝑥→−1𝑥−3, which leads to the graph of 𝑦=−1𝑥−3.

This leads to the given graph. Hence, the given figure represents the function 𝑓(𝑥)=−1𝑥−3.

In the previous example, we obtained the expression for a rational function from the given graph by using a translation and a reflection. We noted, in particular, that the reflection over the 𝑥-axis and that over the 𝑦-axis lead to the same outcome for the parent function 1𝑥. This is not true for general functions, but it is true here due to the symmetry of the function 1𝑥. We will address a useful property due to this symmetry after the next example.

In the next example, we will determine missing parameters in a rational function from the given graph.

Example 4: Determining Parameters of Functions from Their Graphs

The graph shows 𝑦=𝑘(𝑥−𝑎)+𝑏. A single point is marked on the graph. What are the values of the constants 𝑎, 𝑏, and 𝑘?

Answer

We begin by noting that the given graph resembles the graph of 𝑦=1𝑥. We can obtain this graph from the graph of the parent function 𝑦=1𝑥 by applying a few function transformations. Let us recall the graph of the parent function 𝑦=1𝑥.

In particular, we note that this graph has horizontal asymptote 𝑦=0 and vertical asymptote 𝑥=0. On the other hand, the given graph in this example has horizontal asymptote 𝑦=−2 and vertical asymptote 𝑥=3. This means that a downward shift of 2 units as well as a shift to the right by 3 units is one of the function transformations used to obtain this graph from the graph of the parent function.

Before we apply the vertical shift, we should examine whether other transformations are involved, since the order of transformations can affect the outcome. There are three different types of transformations to consider: translation, dilation, and reflection. The given graph appear to be oriented the same as the parent function; hence, we can rule out a reflection.

Dilation is a possibility, but it is difficult to detect visually. Since we are given that the point (6,−1) lies on this graph, we can use this information to find out the dilation factor in the end.

Recall that when combining transformations, dilations and reflections must precede translations. There are two types of dilations: horizontal dilation and vertical dilation. Horizontal dilation of a scale factor 𝑑 corresponds to the function transformation 𝑥→𝑥𝑑, while vertical dilation of a scale factor 𝑑 is given by 𝑓(𝑥)→𝑑𝑓(𝑥).

Beginning with a parent function 1𝑥, if we perform a horizontal dilation with scale factor 𝑑 and a vertical dilation with scale factor 𝑑, we can obtain 1𝑥⟶1⟶𝑑=𝑑𝑑𝑥.

Since we do not know the scale factor, the dilated function can be represented as 𝑘𝑥 for some constant 𝑘.

Next, let us consider translations. Recall that a downward shift by 𝑐 units is given by the function transformation 𝑓(𝑥)→𝑓(𝑥)−𝑐, and a shift to the right by 𝑑 units is given by the function transformation 𝑥→𝑥−𝑑. Hence, applying a downward shift of 2 units and then a shift to the right by 3 units to the function 𝑘𝑥 leads to 𝑘𝑥⟶𝑘𝑥−2⟶𝑘(𝑥−3)−2.

Hence, the given graph represents the function 𝑓(𝑥)=𝑘(𝑥−3)−2 for some value 𝑘. This gives us two of the parameters 𝑎=3 and 𝑏=−2. To identify the remaining parameter 𝑘, we can use the fact that the graph of this function passes through the point (6,−1), which means that 𝑓(6)=−1. Substituting 𝑥=6 into the function, 𝑓(6)=𝑘(6−3)−2=𝑘3−2.

Hence, 𝑘3−2=−1𝑘3=1𝑘=3.

Therefore, we have obtained 𝑎=3, 𝑏=−2, and 𝑘=3.

In the previous example, we identified missing parameter values of a rational function representing translations and dilations of the parent function 𝑦=1𝑥. We noted that a dilation is difficult to determine visually; hence, we used an unknown scale factor that we could identify in the end using a point on the graph. Examining this process closer, we can see that whether the dilation is vertical or horizontal does not affect the outcome for the parent function 1𝑥. This is not always the case for general functions, but it is the case here due to the symmetry of the function 1𝑥. Furthermore, we observed previously that the symmetry of this function leads to the same outcome for reflections over the 𝑥- or the 𝑦-axis.

Hence, when starting with the parent function 1𝑥, we can assume that dilation is always done in the vertical direction, 𝑓(𝑥)→𝑓(𝑥)d, and reflection is always over the 𝑥-axis, 𝑓(𝑥)→−𝑓(𝑥). In other words, for some nonzero constant 𝑘, the transformation 1𝑥→𝑘𝑥 can address all reflections and dilations in function transformations.

This leaves only translations, which can be identified by vertical and horizontal asymptotes in the graph as seen in previous examples. Since the parent function has vertical asymptote 𝑥=0 and horizontal asymptote 𝑦=0, a translated reciprocal function with vertical asymptote 𝑥=𝑎 and horizontal asymptote 𝑦=𝑏 is obtained by shifting horizontally by 𝑎 units and shifting vertically by 𝑏 units. These translations are given by the function transformations 𝑥→𝑥−𝑎 and 𝑓(𝑥)→𝑓(𝑥)+𝑏 respectively. Applying these transformations to the reflected and dilated reciprocal function 𝑘𝑥, we obtain 𝑘𝑥⟶𝑘(𝑥−𝑎)⟶𝑘(𝑥−𝑎)+𝑏.

This property is summarized as follows.

Property: Function Transformation of the Reciprocal Function

A hyperbola with vertical asymptote 𝑥=𝑎 and horizontal asymptote 𝑦=𝑏 is the graph of a rational function 𝑓(𝑥)=𝑘(𝑥−𝑎)+𝑏,𝑘≠0.forsome

This property simplifies the work needed to obtain the expression of a rational function from the parent function 1𝑥.

In our next example, we will identify the range of values for an unknown parameter from the given graph of a rational function.

Example 5: Determining Parameters of Functions from Their Graphs

The graph shows 𝑦=𝑘(𝑥−3)−2. We can see that the intersection of its asymptotes is at (3,−2) and that the points (0.5,−1.5) and (1.5,−1) are below and above the graph respectively. Determine the interval in which 𝑘 lies.

Answer

We recall that a hyperbola with vertical asymptote 𝑥=𝑎 and horizontal asymptote 𝑦=𝑏 is the graph of a rational function 𝑓(𝑥)=𝑘(𝑥−𝑎)+𝑏,𝑘≠0.forsome

Since the given graph is a hyperbola with vertical asymptote 𝑥=3 and horizontal asymptote 𝑦=−2, we can obtain the given expression of the function by substituting 𝑎=3 and 𝑏=−2.

To find the range of values for 𝑘, we need to use the given information about the two points (0.5,−1.5) and (1.5,−1). If either of these points were on the graph, we could use the coordinates to find the exact value of 𝑘. But we are only given that they are either above or below the graph.

Let us begin with the first point, (0.5,−1.5). At 𝑥=0.5, we know that the 𝑦-coordinate of the point lying on the graph is given by 𝑓(0.5). Since the point (0.5,−1.5) is below the graph, this means that the 𝑦-coordinate of this point, −1.5, is smaller than the 𝑦-coordinate of the point on the graph, 𝑓(0.5). In other words, −1.5<𝑓(0.5). We have 𝑓(0.5)=𝑘0.5−3−2=𝑘−2.5−2.

Substituting this expression into the inequality and simplifying, we obtain −1.5<𝑘−2.5−20.5<𝑘−2.5.

We can multiply both sides of the inequality by −2.5, which will reverse the direction of the inequality: 0.5×(−2.5)>𝑘, which leads to 𝑘<−1.25.

Next, let us consider the point (1.5,−1). This point lies above the graph, so, using the same logic as before, we can conclude that 𝑓(1.5)<−1. We have 𝑓(1.5)=𝑘1.5−3−2=𝑘−1.5−2.

Hence, 𝑘−1.5−2<−1𝑘−1.5<1𝑘>−1.5.

Using both conditions, the possible values of 𝑘 are −1.5<𝑘<−1.25.

In this explainer, we have discussed rational functions in the form 𝑘(𝑥−𝑎)+𝑏.

After algebraic simplification, this function can be put into the form 𝑚𝑥+𝑛𝑝𝑥+𝑞.

For instance, consider the function −5𝑥−2+3, which can be simplified as follows: −5𝑥−2+3=−5𝑥−2+3(𝑥−2)𝑥−2=−5+3(𝑥−2)𝑥−2=−5+3𝑥−6𝑥−2=3𝑥−11𝑥−2.

Let us see how we can determine the vertical and horizontal asymptotes of a rational function in this form.

We note that the function is not defined at 𝑥=2 due to the denominator 𝑥−2, while the numerator is not equal to zero at 𝑥=2. This indicates that the graph of this function has vertical asymptote 𝑥=2. We remark that if 𝑥=2 is also a zero of the numerator, then both the top and the bottom of the fraction would contain a factor of 𝑥−2, which would cancel out. In this case, there would not be a vertical asymptote. Hence, when we find the root of the denominator, it is important to check whether this root also makes the numerator equal to zero.

Determining the horizontal asymptote requires a little algebra. When we divide the top and the bottom of the fraction by 𝑥, we can write 𝑦=3𝑥−11𝑥−2=−−(𝑥)=3−1−=3−11×1−2×.dividingthroughby

We can see the expression 1𝑥 in the numerator and denominator of the resulting expression. We know that the value of the function 1𝑥 tends to zero as 𝑥 tends to positive or negative infinity. Hence, the value of this fraction will approach 3−11×01−2×0=3.

This tells us that this rational function has horizontal asymptote 𝑦=3. We can repeat this process for general rational functions of the form 𝑦=𝑚𝑥+𝑛𝑝𝑥+𝑞 to obtain the following results.

Property: Asymptotes of the Reciprocal Function

For the graph of a rational function in the form 𝑓(𝑥)=𝑚𝑥+𝑛𝑝𝑥+𝑞 with 𝑝≠0,

  • the vertical asymptote of the graph of this function is at the root of the denominator 𝑝𝑥+𝑞, which is 𝑥=−𝑞𝑝, as long as the numerator does not share the same root,
  • the horizontal asymptote of the graph of this function is 𝑦=𝑚𝑝.

Alternatively, given a rational function in the form 𝑚𝑥+𝑛𝑝𝑥+𝑞, we can convert this to the form 𝑘(𝑥−𝑎)+𝑏 to identify the asymptotes as discussed earlier. Let us demonstrate this process using the earlier example, 3𝑥−11𝑥−2. Starting with this form, we need to create a factor 𝑥−2 on the numerator. We can do this by rewriting the numerator as 3𝑥−11=3𝑥−6−5=3(𝑥−2)−5.

This leads to 3𝑥−11𝑥−2=3(𝑥−2)−5𝑥−2=3(𝑥−2)𝑥−2−5𝑥−2=3−5𝑥−2=−5(𝑥−2)+3.

In this form, we can clearly see that this rational function is obtained from the parent function 1𝑥 by applying function transformations 𝑥→𝑥−2 and 𝑓(𝑥)→𝑓(𝑥)+3, which shift the graph to the right by 2 units and up by 3 units, respectively, as well as reflection and dilation represented by the constant −5 in the numerator. This implies that the graph of this function has vertical asymptote 𝑥=2 and horizontal asymptote 𝑦=3, which agrees with our previous findings.

In our final example, we will identify the rational function in the form 𝑚𝑥+𝑛𝑝𝑥+𝑞 from a given graph.

Example 6: Determining Rational Functions from Their Graphs

Which of the following is the equation of the graphed function 𝑓(𝑥) whose asymptotes are 𝑥=1 and 𝑦=2?

  1. 𝑓(𝑥)=𝑥+1𝑥−1
  2. 𝑓(𝑥)=𝑥+1𝑥+2
  3. 𝑓(𝑥)=2𝑥+1𝑥−1
  4. 𝑓(𝑥)=2𝑥+1𝑥+1
  5. 𝑓(𝑥)=𝑥+1𝑥−2

Answer

Recall that for the graph of a rational function in the form 𝑓(𝑥)=𝑚𝑥+𝑛𝑝𝑥+𝑞 with 𝑝≠0,

  • the vertical asymptote of the graph of this function is at the root of the denominator 𝑝𝑥+𝑞, which is 𝑥=−𝑞𝑝, as long as the numerator does not share the same root,
  • the horizontal asymptote of the graph of this function is 𝑦=𝑚𝑝.

Let us consider the asymptote of the rational function given in each option using this approach.

  1. 𝑓(𝑥)=𝑥+1𝑥−1:
    In this rational function, the denominator is 𝑥−1. Setting this equal to zero leads to the root 𝑥=1. Since the numerator 𝑥+1 is not equal to zero at this value, this leads to the vertical asymptote of the graph 𝑥=1. This matches the given vertical asymptote, so let us check the horizontal asymptote.
    In the given function, we see that 𝑚=1 and 𝑝=1. Hence, the horizontal asymptote of this function is 𝑦=𝑚𝑝=11=1.
    This leads to the horizontal asymptote 𝑦=1. This does not match the given horizontal asymptote; hence, this choice is not the correct answer.
  2. 𝑓(𝑥)=𝑥+1𝑥+2:
    In this rational function, the denominator is 𝑥+2. Setting this equal to zero leads to the root 𝑥=−2. Since the numerator 𝑥+1 is not equal to zero at this value, this leads to the vertical asymptote of the graph 𝑥=−2. This does not match the given vertical asymptote; hence, this choice is not the correct answer.
  3. 𝑓(𝑥)=2𝑥+1𝑥−1:
    In this rational function, the denominator is 𝑥−1. Setting this equal to zero leads to the root 𝑥=1. Since the numerator 2𝑥+1 is not equal to zero at this value, this leads to the vertical asymptote of the graph 𝑥=1. This matches the given vertical asymptote, so let us check the horizontal asymptote.
    In the given function, we see that 𝑚=2 and 𝑝=1. Hence, the horizontal asymptote of this function is 𝑦=𝑚𝑝=21=2.
    This leads to the horizontal asymptote 𝑦=2. This also matches the given horizontal asymptote; hence, this choice is a possible correct answer. Let us see if there are other possible answers listed in the remaining choices.
  4. 𝑓(𝑥)=2𝑥+1𝑥+1:
    In this rational function, the denominator is 𝑥+1. Setting this equal to zero leads to the root 𝑥=−1. Since the numerator 2𝑥+1 is not equal to zero at this value, this leads to the vertical asymptote of the graph 𝑥=−1. This does not match the given vertical asymptote; hence, this choice is not the correct answer.
  5. 𝑓(𝑥)=𝑥+1𝑥−2:
    In this rational function, the denominator is 𝑥−2. Setting this equal to zero leads to the root 𝑥=2. Since the numerator 𝑥+1 is not equal to zero at this value, this leads to the vertical asymptote of the graph 𝑥=2. This does not match the given vertical asymptote; hence, this choice is not the correct answer.

We have only one possible correct choice based on the given asymptotes. We remark here that this is not the only possible rational function that has the given asymptotes. For instance, only considering the asymptotes does not account for distinctions rising from dilations or reflections of the graph. To identify dilations or reflections of the graph, we will need to use a specific point on the graph. For instance, we can see that this graph passes through the point (2,5). Hence, the correct function must satisfy 𝑓(2)=5. Let us verify this with our only candidate, 𝑓(𝑥)=2𝑥+1𝑥−1: 𝑓(2)=2×2+12−1=5.

Hence, the graph of this function passes through the point (2,5). This confirms that this is the correct rational function for this graph.

The correct function is given in answer C.

Let us finish by recapping a few important concepts from this explainer.

Key Points

  • Unlike a graph of a nonconstant polynomial, the graph of a rational function may have vertical and horizontal asymptotes.
  • The graph of 𝑦=1𝑥 is a hyperbola with horizontal asymptote 𝑦=0 and vertical asymptote 𝑥=0.
  • A hyperbola with vertical asymptote 𝑥=𝑎 and horizontal asymptote 𝑦=𝑏 is the graph of a rational function: 𝑓(𝑥)=𝑘(𝑥−𝑎)+𝑏,𝑘≠0.forsome
  • For the graph of a rational function in the form 𝑓(𝑥)=𝑚𝑥+𝑛𝑝𝑥+𝑞 with 𝑝≠0,
    • the vertical asymptote of the graph of this function is at the root of the denominator, 𝑝𝑥+𝑞, which is 𝑥=−𝑞𝑝, as long as the numerator does not share the same root,
    • the horizontal asymptote of the graph of this function is 𝑦=𝑚𝑝.

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