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4.7: Rational Functions - Mathematics


Learning Objectives

  • Use arrow notation.
  • Solve applied problems involving rational functions.
  • Find the domains of rational functions.
  • Identify vertical asymptotes.
  • Identify horizontal asymptotes.
  • Graph rational functions.

Suppose we know that the cost of making a product is dependent on the number of items, (x), produced. This is given by the equation (C (x)=15,000x−0.1x^2+1000.) If we want to know the average cost for producing (x) items, we would divide the cost function by the number of items, (x). The average cost function, which yields the average cost per item for (x) items produced, is

[f(x)=dfrac{15,000x−0.1x^2+1000}{x} onumber]

Many other application problems require finding an average value in a similar way, giving us variables in the denominator. Written without a variable in the denominator, this function will contain a negative integer power.

In the last few sections, we have worked with polynomial functions, which are functions with non-negative integers for exponents. In this section, we explore rational functions, which have variables in the denominator.

Using Arrow Notation

We have seen the graphs of the basic reciprocal function and the squared reciprocal function from our study of toolkit functions. Examine these graphs, as shown in Figure (PageIndex{1}), and notice some of their features.

Several things are apparent if we examine the graph of (f(x)=frac{1}{x}).

  • On the left branch of the graph, the curve approaches the (x)-axis ((y=0)) as (x ightarrow -infty).
  • As the graph approaches (x = 0) from the left, the curve drops, but as we approach zero from the right, the curve rises.
  • Finally, on the right branch of the graph, the curves approaches the (x)-axis ((y=0) ) as (x ightarrow infty).

To summarize, we use arrow notation to show that (x) or (f (x)) is approaching a particular value (Table (PageIndex{1})).

Table (PageIndex{1})
SymbolMeaning
(x ightarrow a^-)(x) approaches a from the left ((x
(x ightarrow a^+)(x) approaches a from the right ((x>a) but close to (a) )
(x ightarrow infty)(x) approaches infinity ((x) increases without bound)
(x ightarrow −infty)(x) approaches negative infinity ((x) decreases without bound)
(f(x) ightarrow infty)the output approaches infinity (the output increases without bound)
(f(x) ightarrow −infty)the output approaches negative infinity (the output decreases without bound)
(f(x) ightarrow a)the output approaches (a)

Local Behavior of (f(x)=frac{1}{x})

Let’s begin by looking at the reciprocal function, (f(x)=frac{1}{x}). We cannot divide by zero, which means the function is undefined at (x=0); so zero is not in the domain. As the input values approach zero from the left side (becoming very small, negative values), the function values decrease without bound (in other words, they approach negative infinity). We can see this behavior in Table (PageIndex{2}).

Table (PageIndex{2})
(x)–0.1–0.01–0.001–0.0001
(f(x)=frac{1}{x})–10–100–1000–10,000

We write in arrow notation

as (x ightarrow 0^−,f(x) ightarrow −infty)

As the input values approach zero from the right side (becoming very small, positive values), the function values increase without bound (approaching infinity). We can see this behavior in Table (PageIndex{3}).

Table (PageIndex{3})
(x)0.10.010.0010.0001
(f(x)=frac{1}{x})10100100010,000

We write in arrow notation

As (x ightarrow 0^+, f(x) ightarrow infty).

See Figure (PageIndex{2}).

This behavior creates a vertical asymptote, which is a vertical line that the graph approaches but never crosses. In this case, the graph is approaching the vertical line (x=0) as the input becomes close to zero (Figure (PageIndex{3})).

Definition: VERTICAL ASYMPTOTE

A vertical asymptote of a graph is a vertical line (x=a) where the graph tends toward positive or negative infinity as the inputs approach (a). We write

As (x ightarrow a), (f(x) ightarrow infty), or as (x ightarrow a), (f(x) ightarrow −infty).

End Behavior of (f(x)=frac{1}{x})

As the values of (x) approach infinity, the function values approach (0). As the values of (x) approach negative infinity, the function values approach (0) (Figure (PageIndex{4})). Symbolically, using arrow notation

As (x ightarrow infty), (f(x) ightarrow 0),and as (x ightarrow −infty), (f(x) ightarrow 0).

Based on this overall behavior and the graph, we can see that the function approaches 0 but never actually reaches 0; it seems to level off as the inputs become large. This behavior creates a horizontal asymptote, a horizontal line that the graph approaches as the input increases or decreases without bound. In this case, the graph is approaching the horizontal line (y=0). See Figure (PageIndex{5}).

Definition: HORIZONTAL ASYMPTOTE

A horizontal asymptote of a graph is a horizontal line (y=b) where the graph approaches the line as the inputs increase or decrease without bound. We write

As (x ightarrow infty ext{ or } x ightarrow −infty), (f(x) ightarrow b).

Example (PageIndex{1}): Using Arrow Notation.

Use arrow notation to describe the end behavior and local behavior of the function graphed in Figure (PageIndex{6}).

Solution

Notice that the graph is showing a vertical asymptote at (x=2), which tells us that the function is undefined at (x=2).

As (x ightarrow 2^−), (f(x) ightarrow −infty,) and as (x ightarrow 2^+), (f(x) ightarrow infty).

And as the inputs decrease without bound, the graph appears to be leveling off at output values of (4), indicating a horizontal asymptote at (y=4). As the inputs increase without bound, the graph levels off at (4).

As (x ightarrow infty), (f(x) ightarrow 4) and as (x ightarrow −infty), (f(x) ightarrow 4).

Exercises (PageIndex{1})

Use arrow notation to describe the end behavior and local behavior for the reciprocal squared function.

Answer

End behavior: as (x ightarrow pm infty), (f(x) ightarrow 0);

Local behavior: as (x ightarrow 0), (f(x) ightarrow infty) (there are no x- or y-intercepts)

Example (PageIndex{2}): Using Transformations to Graph a Rational Function.

Sketch a graph of the reciprocal function shifted two units to the left and up three units. Identify the horizontal and vertical asymptotes of the graph, if any.

Solution

Shifting the graph left 2 and up 3 would result in the function

[f(x)=dfrac{1}{x+2}+3]

or equivalently, by giving the terms a common denominator,

[f(x)=dfrac{3x+7}{x+2}]

The graph of the shifted function is displayed in Figure (PageIndex{7}).

Notice that this function is undefined at (x=−2), and the graph also is showing a vertical asymptote at (x=−2).

As (x ightarrow −2^−), (f(x) ightarrow −infty), and as (x ightarrow −2^+), (f(x) ightarrow infty).

As the inputs increase and decrease without bound, the graph appears to be leveling off at output values of 3, indicating a horizontal asymptote at (y=3).

As (x ightarrow pm infty), (f(x) ightarrow 3).

Analysis

Notice that horizontal and vertical asymptotes are shifted left 2 and up 3 along with the function.

Exercise (PageIndex{2})

Sketch the graph, and find the horizontal and vertical asymptotes of the reciprocal squared function that has been shifted right 3 units and down 4 units.

Solution

The function and the asymptotes are shifted 3 units right and 4 units down. As (x ightarrow 3), (f(x) ightarrow infty), and as (x ightarrow pm infty), (f(x) ightarrow −4).

The function is (f(x)=frac{1}{{(x−3)}^2}−4).

Solving Applied Problems Involving Rational Functions

In Example (PageIndex{2}), we shifted a toolkit function in a way that resulted in the function (f(x)=frac{3x+7}{x+2}). This is an example of a rational function. A rational function is a function that can be written as the quotient of two polynomial functions. Many real-world problems require us to find the ratio of two polynomial functions. Problems involving rates and concentrations often involve rational functions.

Definition: RATIONAL FUNCTION

A rational function is a function that can be written as the quotient of two polynomial functions (P(x)) and (Q(x)).

[f(x)=dfrac{P(x)}{Q(x)}=dfrac{a_px^p+a_{p−1}x^{p−1}+...+a_1x+a_0}{b_qx^q+b_{q−1}x^{q−1}+...+b_1x+b_0},space Q(x)≠0]

Example (PageIndex{3}): Solving an Applied Problem Involving a Rational Function

A large mixing tank currently contains 100 gallons of water into which 5 pounds of sugar have been mixed. A tap will open pouring 10 gallons per minute of water into the tank at the same time sugar is poured into the tank at a rate of 1 pound per minute. Find the concentration (pounds per gallon) of sugar in the tank after 12 minutes. Is that a greater concentration than at the beginning?

Solution

Let t be the number of minutes since the tap opened. Since the water increases at 10 gallons per minute, and the sugar increases at 1 pound per minute, these are constant rates of change. This tells us the amount of water in the tank is changing linearly, as is the amount of sugar in the tank. We can write an equation independently for each:

water: (W(t)=100+10t) in gallons

sugar: (S(t)=5+1t) in pounds

The concentration, (C), will be the ratio of pounds of sugar to gallons of water

[C(t)=dfrac{5+t}{100+10t}]

The concentration after 12 minutes is given by evaluating (C(t)) at (t= 12).

[egin{align} C(12) & =dfrac{5+12}{100+10(12)} &= dfrac{17}{220} end{align} ]

This means the concentration is 17 pounds of sugar to 220 gallons of water.

At the beginning, the concentration is

[egin{align} C(0) & =dfrac{5+0}{100+10(0)} & =dfrac{1}{20} end{align} ]

Since (frac{17}{220}≈0.08>frac{1}{20}=0.05), the concentration is greater after 12 minutes than at the beginning.

Analysis

To find the horizontal asymptote, divide the leading coefficient in the numerator by the leading coefficient in the denominator:

[dfrac{1}{10}=0.1]

Notice the horizontal asymptote is (y= 0.1.) This means the concentration, (C,) the ratio of pounds of sugar to gallons of water, will approach 0.1 in the long term.

Exercise (PageIndex{3})

There are 1,200 freshmen and 1,500 sophomores at a prep rally at noon. After 12 p.m., 20 freshmen arrive at the rally every five minutes while 15 sophomores leave the rally. Find the ratio of freshmen to sophomores at 1 p.m.

Answer

(frac{12}{11})

Finding the Domains of Rational Functions

A vertical asymptote represents a value at which a rational function is undefined, so that value is not in the domain of the function. A reciprocal function cannot have values in its domain that cause the denominator to equal zero. In general, to find the domain of a rational function, we need to determine which inputs would cause division by zero.

Definition: DOMAIN OF A RATIONAL FUNCTION

The domain of a rational function includes all real numbers except those that cause the denominator to equal zero.

How To: Given a rational function, find the domain.

  1. Set the denominator equal to zero.
  2. Solve to find the x-values that cause the denominator to equal zero.
  3. The domain is all real numbers except those found in Step 2.

Example (PageIndex{4}): Finding the Domain of a Rational Function

Find the domain of (f(x)=dfrac{x+3}{x^2−9}).

Solution

Begin by setting the denominator equal to zero and solving.

[x^2-9=0]

[x^2=9][x=pm 3]

The denominator is equal to zero when (x=pm 3). The domain of the function is all real numbers except (x=pm 3).

Analysis

A graph of this function, as shown in Figure (PageIndex{9}), confirms that the function is not defined when (x=pm 3).

There is a vertical asymptote at (x=3) and a hole in the graph at (x=−3). We will discuss these types of holes in greater detail later in this section.

Exercise (PageIndex{4})

Find the domain of (f(x)=frac{4x}{5(x−1)(x−5)}).

Answer

The domain is all real numbers except (x=1) and (x=5).

Identifying Vertical Asymptotes of Rational Functions

By looking at the graph of a rational function, we can investigate its local behavior and easily see whether there are asymptotes. We may even be able to approximate their location. Even without the graph, however, we can still determine whether a given rational function has any asymptotes, and calculate their location.

Vertical Asymptotes

The vertical asymptotes of a rational function may be found by examining the factors of the denominator that are not common to the factors in the numerator. Vertical asymptotes occur at the zeros of such factors.

How To: Given a rational function, identify any vertical asymptotes of its graph

  1. Factor the numerator and denominator.
  2. Note any restrictions in the domain of the function.
  3. Reduce the expression by canceling common factors in the numerator and the denominator.
  4. Note any values that cause the denominator to be zero in this simplified version. These are where the vertical asymptotes occur.
  5. Note any restrictions in the domain where asymptotes do not occur. These are removable discontinuities, or “holes.”

Example (PageIndex{5}): Identifying Vertical Asymptotes

Find the vertical asymptotes of the graph of (k(x)=frac{5+2x^2}{2−x−x^2}).

Solution

First, factor the numerator and denominator.

[k(x)=dfrac{5+2x^2}{2−x−x^2}].

[=dfrac{5+2x^2}{(2+x)(1-x)}]

To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero:

[(2+x)(1−x)=0]

[x=−2,1]

Neither (x=–2) nor (x=1) are zeros of the numerator, so the two values indicate two vertical asymptotes. The graph in Figure (PageIndex{10}) confirms the location of the two vertical asymptotes.

Figure (PageIndex{10}).

Removable Discontinuities

Occasionally, a graph will contain a hole: a single point where the graph is not defined, indicated by an open circle. We call such a hole a removable discontinuity. For example, the function (f(x)=frac{x^2−1}{x^2−2x−3}) may be re-written by factoring the numerator and the denominator.

[f(x)=dfrac{(x+1)(x−1)}{(x+1)(x−3)}]

Notice that (x+1) is a common factor to the numerator and the denominator. The zero of this factor, (x=−1), is the location of the removable discontinuity. Notice also that (x–3) is not a factor in both the numerator and denominator. The zero of this factor, (x=3), is the vertical asymptote. See Figure (PageIndex{11}). [Note that removable discontinuities may not be visible when we use a graphing calculator, depending upon the window selected.]

REMOVABLE DISCONTINUITIES OF RATIONAL FUNCTIONS

A removable discontinuity occurs in the graph of a rational function at (x=a) if (a) is a zero for a factor in the denominator that is common with a factor in the numerator. We factor the numerator and denominator and check for common factors. If we find any, we set the common factor equal to 0 and solve. This is the location of the removable discontinuity. This is true if the multiplicity of this factor is greater than or equal to that in the denominator. If the multiplicity of this factor is greater in the denominator, then there is still an asymptote at that value.

Example (PageIndex{6}): Identifying Vertical Asymptotes and Removable Discontinuities for a Graph

Find the vertical asymptotes and removable discontinuities of the graph of (k(x)=frac{x−2}{x^2−4}).

Solution

Factor the numerator and the denominator.

[k(x)=dfrac{x−2}{(x−2)(x+2)}]

Notice that there is a common factor in the numerator and the denominator, (x–2). The zero for this factor is (x=2). This is the location of the removable discontinuity.

Notice that there is a factor in the denominator that is not in the numerator, (x+2). The zero for this factor is (x=−2). The vertical asymptote is (x=−2). See Figure (PageIndex{12}).

The graph of this function will have the vertical asymptote at (x=−2), but at (x=2) the graph will have a hole.

Exercise (PageIndex{5})

Find the vertical asymptotes and removable discontinuities of the graph of (f(x)=frac{x^2−25}{x^3−6x^2+5x}).

Answer

Removable discontinuity at (x=5).

Vertical asymptotes: (x=0), (x=1).

Identifying Horizontal Asymptotes of Rational Functions

While vertical asymptotes describe the behavior of a graph as the output gets very large or very small, horizontal asymptotes help describe the behavior of a graph as the input gets very large or very small. Recall that a polynomial’s end behavior will mirror that of the leading term. Likewise, a rational function’s end behavior will mirror that of the ratio of the function that is the ratio of the leading terms.

There are three distinct outcomes when checking for horizontal asymptotes:

Case 1: If the degree of the denominator > degree of the numerator, there is a horizontal asymptote at (y=0).

Example: (f(x)=dfrac{4x+2}{x^2+4x−5})

In this case, the end behavior is (f(x)≈frac{4x}{x^2}=frac{4}{x}). This tells us that, as the inputs increase or decrease without bound, this function will behave similarly to the function (g(x)=frac{4}{x}), and the outputs will approach zero, resulting in a horizontal asymptote at (y=0). See Figure (PageIndex{13}). Note that this graph crosses the horizontal asymptote.

Case 2: If the degree of the denominator < degree of the numerator by one, we get a slant asymptote.

Example: (f(x)=dfrac{3x^2−2x+1}{x−1})

In this case, the end behavior is (f(x)≈frac{3x^2}{x}=3x). This tells us that as the inputs increase or decrease without bound, this function will behave similarly to the function (g(x)=3x). As the inputs grow large, the outputs will grow and not level off, so this graph has no horizontal asymptote. However, the graph of (g(x)=3x) looks like a diagonal line, and since (f) will behave similarly to (g), it will approach a line close to (y=3x). This line is a slant asymptote.

To find the equation of the slant asymptote, divide (frac{3x^2−2x+1}{x−1}). The quotient is (3x+1), and the remainder is 2. The slant asymptote is the graph of the line (g(x)=3x+1). See Figure (PageIndex{14}).

Case 3: If the degree of the denominator = degree of the numerator, there is a horizontal asymptote at (y=dfrac{a_n}{b_n}), where (a_n) and (b_n) are respectively the leading coefficients of (p(x)) and (q(x)) for (f(x)=dfrac{p(x)}{q(x)}), (q(x)≠0).

Example: (f(x)=dfrac{3x^2+2}{x^2+4x−5})

In this case, the end behavior is (f(x)≈dfrac{3x^2}{x^2}=3). This tells us that as the inputs grow large, this function will behave like the function (g(x)=3), which is a horizontal line. As (x ightarrow pm infty), (f(x) ightarrow 3), resulting in a horizontal asymptote at (y=3). See Figure (PageIndex{15}). Note that this graph crosses the horizontal asymptote.

Notice that, while the graph of a rational function will never cross a vertical asymptote, the graph may or may not cross a horizontal or slant asymptote. Also, although the graph of a rational function may have many vertical asymptotes, the graph will have at most one horizontal (or slant) asymptote.

It should be noted that, if the degree of the numerator is larger than the degree of the denominator by more than one, the end behavior of the graph will mimic the behavior of the reduced end behavior fraction. For instance, if we had the function

[f(x)=dfrac{3x^5−x^2}{x+3}]

with end behavior

[f(x)≈dfrac{3x^5}{x}=3x^4],

the end behavior of the graph would look similar to that of an even polynomial with a positive leading coefficient.

(x ightarrow pm infty, f(x) ightarrow infty)

HORIZONTAL ASYMPTOTES OF RATIONAL FUNCTIONS

The horizontal asymptote of a rational function can be determined by looking at the degrees of the numerator and denominator.

  • Degree of numerator is less than degree of denominator: horizontal asymptote at (y=0).
  • Degree of numerator is greater than degree of denominator by one: no horizontal asymptote; slant asymptote.
  • Degree of numerator is equal to degree of denominator: horizontal asymptote at ratio of leading coefficients.

Example (PageIndex{7}): Identifying Horizontal and Slant Asymptotes

For the functions listed, identify the horizontal or slant asymptote.

  1. (g(x)=dfrac{6x^3−10x}{2x^3+5x^2})
  2. (h(x)=dfrac{x^2−4x+1}{x+2})
  3. (k(x)=dfrac{x^2+4x}{x^3−8})

Solution

For these solutions, we will use (f(x)=dfrac{p(x)}{q(x)},space q(x)≠0).

  1. (g(x)=frac{6x^3−10x}{2x^3+5x^2}): The degree of (p=)degree of (q=3), so we can find the horizontal asymptote by taking the ratio of the leading terms. There is a horizontal asymptote at (y =frac{6}{2}) or (y=3).
  2. (h(x)=frac{x^2−4x+1}{x+2}): The degree of (p=2) and degree of (q=1). Since (p>q) by 1, there is a slant asymptote found at (dfrac{x^2−4x+1}{x+2}).
  3. (k(x)=frac{x^2+4x}{x^3−8}) : The degree of (p=2) < degree of (q=3), so there is a horizontal asymptote (y=0).

Example (PageIndex{8}) Identifying Horizontal Asymptotes

In the sugar concentration problem earlier, we created the equation (C(t)=frac{5+t}{100+10t}).

Find the horizontal asymptote and interpret it in context of the problem.

Solution

Both the numerator and denominator are linear (degree 1). Because the degrees are equal, there will be a horizontal asymptote at the ratio of the leading coefficients. In the numerator, the leading term is (t), with coefficient 1. In the denominator, the leading term is 10t, with coefficient 10. The horizontal asymptote will be at the ratio of these values:

(t ightarrow infty,space C(t) ightarrow frac{1}{10})

This function will have a horizontal asymptote at (y=frac{1}{10}).

This tells us that as the values of t increase, the values of (C) will approach (frac{1}{10}). In context, this means that, as more time goes by, the concentration of sugar in the tank will approach one-tenth of a pound of sugar per gallon of water or (frac{1}{10}) pounds per gallon.

Example (PageIndex{9}): Identifying Horizontal and Vertical Asymptotes

Find the horizontal and vertical asymptotes of the function

(f(x)=dfrac{(x−2)(x+3)}{(x−1)(x+2)(x−5)})

Solution

First, note that this function has no common factors, so there are no potential removable discontinuities.

The function will have vertical asymptotes when the denominator is zero, causing the function to be undefined. The denominator will be zero at (x=1,–2,)and (5), indicating vertical asymptotes at these values.

The numerator has degree (2), while the denominator has degree 3. Since the degree of the denominator is greater than the degree of the numerator, the denominator will grow faster than the numerator, causing the outputs to tend towards zero as the inputs get large, and so as (x ightarrow pm infty), (f(x) ightarrow 0). This function will have a horizontal asymptote at (y =0.) See Figure (PageIndex{16}).

Exercise (PageIndex{6})

Find the vertical and horizontal asymptotes of the function:

(f(x)=dfrac{(2x−1)(2x+1)}{(x−2)(x+3)})

Answer

Vertical asymptotes at (x=2) and (x=–3)

horizontal asymptote at (y =4).

INTERCEPTS OF RATIONAL FUNCTIONS

A rational function will have a (y)-intercept at (f(0),) if the function is defined at zero. A rational function will not have a (y)-intercept if the function is not defined at zero.

Likewise, a rational function will have (x)-intercepts at the inputs that cause the output to be zero. Since a fraction is only equal to zero when the numerator is zero, x-intercepts can only occur when the numerator of the rational function is equal to zero.

Example (PageIndex{10}): Finding the Intercepts of a Rational Function

Find the intercepts of (f(x)=dfrac{(x−2)(x+3)}{(x−1)(x+2)(x−5)}).

Solution

We can find the y-intercept by evaluating the function at zero

(f(0)=dfrac{(0−2)(0+3)}{(0−1)(0+2)(0−5)})

(=−dfrac{6}{10})

(=−dfrac{3}{5})

(=−0.6)

The x-intercepts will occur when the function is equal to zero:

(0=dfrac{(x−2)(x+3)}{(x−1)(x+2)(x−5)}) This is zero when the numerator is zero.

(0=(x−2)(x+3))

(x=2,−3)

The y-intercept is ((0,–0.6)), the x-intercepts are ((2,0)) and ((–3,0)).See Figure (PageIndex{17}).

Exercise (PageIndex{7})

Given the reciprocal squared function that is shifted right 3 units and down 4 units, write this as a rational function. Then, find the x- and y-intercepts and the horizontal and vertical asymptotes.

Answer

For the transformed reciprocal squared function, we find the rational form.

(f(x)=dfrac{1}{{(x−3)}^2}−4=dfrac{1−4{(x−3)}^2}{{(x−3)}^2}=dfrac{1−4(x^2−6x+9)}{(x−3)(x−3)}=dfrac{−4x^2+24x−35}{x^2−6x+9})

Because the numerator is the same degree as the denominator we know that as (x ightarrow pm infty), (f(x) ightarrow −4); so (y=–4) is the horizontal asymptote. Next, we set the denominator equal to zero, and find that the vertical asymptote is (x=3), because as (x ightarrow 3), (f(x) ightarrow infty). We then set the numerator equal to (0) and find the x-intercepts are at ((2.5,0)) and ((3.5,0)). Finally, we evaluate the function at 0 and find the y-intercept to be at ((0,−frac{35}{9})).

Graphing Rational Functions

In Example(PageIndex{10}), we see that the numerator of a rational function reveals the x-intercepts of the graph, whereas the denominator reveals the vertical asymptotes of the graph. As with polynomials, factors of the numerator may have integer powers greater than one. Fortunately, the effect on the shape of the graph at those intercepts is the same as we saw with polynomials.

The vertical asymptotes associated with the factors of the denominator will mirror one of the two toolkit reciprocal functions. When the degree of the factor in the denominator is odd, the distinguishing characteristic is that on one side of the vertical asymptote the graph heads towards positive infinity, and on the other side the graph heads towards negative infinity. See Figure (PageIndex{18}).

When the degree of the factor in the denominator is even, the distinguishing characteristic is that the graph either heads toward positive infinity on both sides of the vertical asymptote or heads toward negative infinity on both sides. See Figure (PageIndex{19}).

For example, the graph of (f(x)=dfrac{{(x+1)}^2(x−3)}{{(x+3)}^2(x−2)}) is shown in Figure (PageIndex{20}).

  • At the x-intercept (x=−1) corresponding to the ({(x+1)}^2) factor of the numerator, the graph "bounces", consistent with the quadratic nature of the factor.
  • At the x-intercept (x=3) corresponding to the ((x−3)) factor of the numerator, the graph passes through the axis as we would expect from a linear factor.
  • At the vertical asymptote (x=−3) corresponding to the ({(x+3)}^2) factor of the denominator, the graph heads towards positive infinity on both sides of the asymptote, consistent with the behavior of the function (f(x)=frac{1}{x^2}).
  • At the vertical asymptote (x=2), corresponding to the ((x−2)) factor of the denominator, the graph heads towards positive infinity on the left side of the asymptote and towards negative infinity on the right side, consistent with the behavior of the function (f(x)=frac{1}{x}).

Howto: Given a rational function, sketch a graph.

  1. Evaluate the function at 0 to find the y-intercept.
  2. Factor the numerator and denominator.
  3. For factors in the numerator not common to the denominator, determine where each factor of the numerator is zero to find the x-intercepts.
  4. Find the multiplicities of the x-intercepts to determine the behavior of the graph at those points.
  5. For factors in the denominator, note the multiplicities of the zeros to determine the local behavior. For those factors not common to the numerator, find the vertical asymptotes by setting those factors equal to zero and then solve.
  6. For factors in the denominator common to factors in the numerator, find the removable discontinuities by setting those factors equal to 0 and then solve.
  7. Compare the degrees of the numerator and the denominator to determine the horizontal or slant asymptotes.
  8. Sketch the graph.

Example (PageIndex{11}): Graphing a Rational Function

Sketch a graph of (f(x)=frac{(x+2)(x−3)}{{(x+1)}^2(x−2)}).

Solution

We can start by noting that the function is already factored, saving us a step.

Next, we will find the intercepts. Evaluating the function at zero gives the y-intercept:

(f(0)=dfrac{(0+2)(0−3)}{{(0+1)}^2(0−2)})

(=3)

To find the x-intercepts, we determine when the numerator of the function is zero. Setting each factor equal to zero, we find x-intercepts at (x=–2) and (x=3). At each, the behavior will be linear (multiplicity 1), with the graph passing through the intercept.

We have a y-intercept at ((0,3)) and x-intercepts at ((–2,0)) and ((3,0)).

To find the vertical asymptotes, we determine when the denominator is equal to zero. This occurs when (x+1=0) and when (x–2=0), giving us vertical asymptotes at (x=–1) and (x=2).

There are no common factors in the numerator and denominator. This means there are no removable discontinuities.

Finally, the degree of denominator is larger than the degree of the numerator, telling us this graph has a horizontal asymptote at (y =0).

To sketch the graph, we might start by plotting the three intercepts. Since the graph has no x-intercepts between the vertical asymptotes, and the y-intercept is positive, we know the function must remain positive between the asymptotes, letting us fill in the middle portion of the graph as shown in Figure (PageIndex{21}).

The factor associated with the vertical asymptote at (x=−1) was squared, so we know the behavior will be the same on both sides of the asymptote. The graph heads toward positive infinity as the inputs approach the asymptote on the right, so the graph will head toward positive infinity on the left as well.

For the vertical asymptote at (x=2), the factor was not squared, so the graph will have opposite behavior on either side of the asymptote. See Figure (PageIndex{22}). After passing through the x-intercepts, the graph will then level off toward an output of zero, as indicated by the horizontal asymptote.

Exercise (PageIndex{8})

Given the function (f(x)=frac{{(x+2)}^2(x−2)}{2{(x−1)}^2(x−3)}), use the characteristics of polynomials and rational functions to describe its behavior and sketch the function.

Answer

Horizontal asymptote at (y=frac{1}{2}). Vertical asymptotes at (x=1) and (x=3). y-intercept at ((0,frac{4}{3})).

x-intercepts at ((2,0)) and ((–2,0)). ((–2,0)) is a zero with multiplicity (2), and the graph bounces off the x-axis at this point. ((2,0)) is a single zero and the graph crosses the axis at this point.

Writing Rational Functions

Now that we have analyzed the equations for rational functions and how they relate to a graph of the function, we can use information given by a graph to write the function. A rational function written in factored form will have an x-intercept where each factor of the numerator is equal to zero. (An exception occurs in the case of a removable discontinuity.) As a result, we can form a numerator of a function whose graph will pass through a set of x-intercepts by introducing a corresponding set of factors. Likewise, because the function will have a vertical asymptote where each factor of the denominator is equal to zero, we can form a denominator that will produce the vertical asymptotes by introducing a corresponding set of factors.

WRITING RATIONAL FUNCTIONS FROM INTERCEPTS AND ASYMPTOTES

If a rational function has x-intercepts at (x=x_1,x_2,...,x_n), vertical asymptotes at (x=v_1,v_2,…,v_m), and no (x_i=) any (v_j), then the function can be written in the form:

(f(x)=adfrac{ {(x−x_1)}^{p_1} {(x−x_2)}^{p_2}⋯{(x−x_n)}^{p_n} }{ {(x−v_1)}^{q_1} {(x−v_2)}^{q_2}⋯{(x−v_m)}^{q_n}})

where the powers (p_i) or (q_i) on each factor can be determined by the behavior of the graph at the corresponding intercept or asymptote, and the stretch factor (a) can be determined given a value of the function other than the x-intercept or by the horizontal asymptote if it is nonzero.

Given a graph of a rational function, write the function.

  1. Determine the factors of the numerator. Examine the behavior of the graph at the x-intercepts to determine the zeroes and their multiplicities. (This is easy to do when finding the “simplest” function with small multiplicities—such as 1 or 3—but may be difficult for larger multiplicities—such as 5 or 7, for example.)
  2. Determine the factors of the denominator. Examine the behavior on both sides of each vertical asymptote to determine the factors and their powers.
  3. Use any clear point on the graph to find the stretch factor.

Example (PageIndex{12}): Writing a Rational Function from Intercepts and Asymptotes

Write an equation for the rational function shown in Figure (PageIndex{24}).

Solution

The graph appears to have x-intercepts at (x=–2) and (x=3). At both, the graph passes through the intercept, suggesting linear factors. The graph has two vertical asymptotes. The one at (x=–1) seems to exhibit the basic behavior similar to (dfrac{1}{x}), with the graph heading toward positive infinity on one side and heading toward negative infinity on the other. The asymptote at (x=2) is exhibiting a behavior similar to (dfrac{1}{x^2}), with the graph heading toward negative infinity on both sides of the asymptote. See Figure (PageIndex{25}).

We can use this information to write a function of the form

(f(x)=adfrac{(x+2)(x−3)}{(x+1){(x−2)}^2})

To find the stretch factor, we can use another clear point on the graph, such as the y-intercept ((0,–2)).

(−2=adfrac{(0+2)(0−3)}{(0+1){(0−2)}^2})

(-2=adfrac{−6}{4})

(a=dfrac{−8}{−6}=dfrac{4}{3})

This gives us a final function of (f(x)=frac{4(x+2)(x−3)}{3(x+1){(x−2)}^2}).

Key Equations

Rational Function(f(x)=dfrac{P(x)}{Q(x)}=dfrac{a_px^p+a_{p−1}x^{p−1}+...+a_1x+a_0}{b_qx^q+b_{q−1}x^{q−1}+...+b_1x+b_0},space Q(x)≠0)

Key Concepts

  • We can use arrow notation to describe local behavior and end behavior of the toolkit functions (f(x)=frac{1}{x}) and (f(x)=frac{1}{x^2}). See Example (PageIndex{1}).
  • A function that levels off at a horizontal value has a horizontal asymptote. A function can have more than one vertical asymptote. See Example.
  • Application problems involving rates and concentrations often involve rational functions. See Example.
  • The domain of a rational function includes all real numbers except those that cause the denominator to equal zero. See Example.
  • The vertical asymptotes of a rational function will occur where the denominator of the function is equal to zero and the numerator is not zero. See Example.
  • A removable discontinuity might occur in the graph of a rational function if an input causes both numerator and denominator to be zero. See Example.
  • A rational function’s end behavior will mirror that of the ratio of the leading terms of the numerator and denominator functions. See Example, Example, Example, and Example.
  • Graph rational functions by finding the intercepts, behavior at the intercepts and asymptotes, and end behavior. See Example.
  • If a rational function has x-intercepts at (x=x_1,x_2,…,x_n), vertical asymptotes at (x=v_1,v_2,…,v_m), and no (x_i=) any (v_j), then the function can be written in the form

Rational Expressions and Functions: Your Complete Guide

Learning about Rational Expressions and Functions can be tough. Once you feel you mastered one type of problem you get stumped on the next. This course is structured to not leave you behind in the dust. I start off each section with basic definitions and processes you will need to know moving through the course. I then present two types of videos to you for each skill. First is the overview video where I explain the concept as a whole like a typical lecture in a classroom. I then work through multiple examples showing you step by step how to complete different types of problems. We both know watching someone do math is not the best way to learn. You have to practice! Each section you are provided with multiple worksheets to practice your skills as well as the answers to check your work. Revert back to videos if you get stuck and forget how to solve the problems. Once you feel you have a good grasp of your understanding it is time to take your quiz. There are multiple quizzes provided for each section. Take the quizzes as many times as you need to earn 100%.

There is no pressure you are hear to learn. By taking this course you will not only gain a better understanding of Rational Expressions and Functions but you will gain confidence to solve more problems on your own. That is why I created this course. I want students to no longer fear learning math or walking into their math class because they just don't understand. Everyone can learn math. Some it just takes a little longer, some just need a little boast and some need a course like I designed to guide them through the material. Heck once you complete this course, show your teacher! You deserve and A. I am here for you and by joining this course you are now one of my students just as important to me as the 140 students I teach in the classroom during the school year. So please keep in touch, let me know how I am doing and if there is anything extra I can provide to assist you with your learning of Rational Expressions and Equations.


Frequently bought together

Review

Iteration of Rational Functions

Complex Analytic Dynamical Systems

"This book makes available a comprehensive, detailed, and organized treatment of the foundations of the theory of iteration of rational functions of a complex variable. The material covered extends from the original memoirs of Fatou and Julia to the recent and important results and methods of Sullivan and Shishikura. Many of the details of the proofs have not occurred in print before ."―ZENTRALBLATT MATH


Algebra II

Emphasis on functions returns as the functional families learned in Algebra I are studied in more detail. Algebra II expands on quadratic, polynomial, exponential, radical and rational functions and introduces logarithmic functions. Students learn how to use and operate on complex numbers so that complex solutions of quadratic and polynomial functions can be considered. This course focuses on more abstract and theoretical concepts, allowing students to improve their critical thought processes. The primary goal of this course is for students to own a strong understanding of all types of functions and prepare them for Pre-Calculus.

Review of Equations & Exponents

  1. 1.1 Introduction to Equations
  2. 1.2 Multistep & Distributive Part I
  3. 1.3 Multistep & Distributive Part II
  4. 1.4 Equations with Multiple Variables
  5. 1.5 Slope & Rate of Change
  6. 1.6 Calculating Slope Using a Graph
  7. 1.7 Calculate Slope Using dydx & Intercepts
  8. 1.8 Slope-Intercept Form
  9. 1.9 Equation of a Line
  10. 1.10 Introduction to Exponents
  11. 1.11 The Product Property
  12. 1.12 The Quotient Property
  13. 1.13 Zero and Negative Exponents
  14. 1.14 Fractional Exponents
  15. 1.15 Power of a Power Property
  16. 1.16 Power of a Product Property
  17. 1.17 Power of a Fraction
  18. 1.18 Simplifying Algebraic Expressions with Exponents

Systems of Linear Equations & Inequalities

  1. 2.1 Intro to Systems of Linear Equations & Inequalities
  2. 2.2 Solving a System of Equations by Substitution
  3. 2.3 Solving a System of Equations by Elimination
  4. 2.4 Graphing Systems of Linear Equations
  5. 2.5 Graphing Systems of Inequalities
  6. 2.6 Linear Programming

Polynomial Functions

  1. 3.1 Introduction to Polynomials
  2. 3.2 Adding & Subtracting Polynomials
  3. 3.3 Multiplying Polynomials
  4. 3.4 Dividing Polynomials Using Long Division
  5. 3.5 Dividing Polynomials Using Synthetic Division
  6. 3.6 Remainder Theorem
  7. 3.7 Factor Theorem
  8. 3.8 Factoring Polynomials Using the GCF
  9. 3.9 Factoring Using Difference of Squares
  10. 3.10 Factoring Perfect Square Trinomials
  11. 3.11 Factoring Trinomials
  12. 3.12 Factoring Using Sums or Difference of Cubes
  13. 3.13 Solving Equations Using Factoring

Functions

  1. 4.1 Introduction to Functions
  2. 4.2 Function Notation & Evaluation
  3. 4.3 Domain of a Function & Interval Notation
  4. 4.4 Range of a Function & Interval Notation
  5. 4.5 Adding & Subtracting Functions
  6. 4.6 Multiplying & Dividing Functions
  7. 4.7 Composition of Functions
  8. 4.8 Inverse Functions
  9. 4.9 Composition and Inverse
  10. 4.10 Piecewise Functions
  11. 4.11 Step Functions

Complex Numbers

  1. 5.1 Introduction to Imaginary & Complex Numbers
  2. 5.2 Adding & Subtracting Complex Numbers
  3. 5.3 Multiplying Complex Numbers
  4. 5.4 Complex Conjugates
  5. 5.5 Dividing Complex Numbers
  6. 5.6 Absolute Value & Complex Numbers

Polynomial Functions Solving & Graphing

  1. 6.1 Finding a Polynomial Given the Roots
  2. 6.2 Location Principle & Multiplicity of Zeros
  3. 6.3 Rational Root Theorem
  4. 6.4 The Quadratic Formula
  5. 6.5 The Complex Conjugate Root Theorem
  6. 6.6 Fundamental Theorem of Algebra & Descartes Rule of Signs
  7. 6.7 Graphing Polynomials

Quadratic Functions

  1. 7.1 Introduction to Quadratic Functions
  2. 7.2 Graphing Quadratic Functions in Vertex Form
  3. 7.3 Solving Quadratics with Square Roots
  4. 7.4 Solving Quadratics with Completing the Square
  5. 7.5 Converting Quadratic Function to Vertex Form using Completing the Square
  6. 7.6 Graphing Quadratic Inequalities
  7. 7.7 Applications of Quadratics

Exponential & Logarithmic Functions

  1. 8.1 Intro to Exponential & Logarithmic Properties
  2. 8.2 Exponential Growth
  3. 8.3 Exponential Decay
  4. 8.4 Logarithmic Functions
  5. 8.5 Evaluating Logarithmic Functions
  6. 8.6 Product Property of Logarithms
  7. 8.7 Quotient Property of Logarithms
  8. 8.8 Power Property of Logarithms
  9. 8.9 Exponential-Logarithmic Inverse Properties
  10. 8.10 Application of Logarithms
  11. 8.11 The Natural Exponential Function
  12. 8.12 The Natural Logarithm
  13. 8.13 Solving Logarithmic Equations

Exponential & Logarithmic Functions

  1. 9.1 Introduction to Rational Functions
  2. 9.2 Direct Variation
  3. 9.3 Inverse Variation
  4. 9.4 Joint & Combined Variation
  5. 9.5 Simplifying Rational Expressions
  6. 9.6 Adding & Subtracting Rational Expressions
  7. 9.7 Multiplying Rational Expressions
  8. 9.8 Dividing Rational Expressions
  9. 9.9 Complex Fractions
  10. 9.10 Solving Rational Equations
  11. 9.11 Graph of a Rational Function
  12. 9.12 Graph of a Rational Function Continued

Radical Functions

  1. 10.1 Introduction to Radical Functions
  2. 10.2 Simplifying Radicals - Numerical
  3. 10.3 Simplifying Radicals - Algebraic
  4. 10.4 Multiplying Radicals
  5. 10.5 Dividing Radicals
  6. 10.6 Adding & Subtracting Radicals
  7. 10.7 Solving Radical Equations
  8. 10.8 Graphing Radical Functions

Conic Sections

  1. 11.1 Introduction to Conic Sections
  2. 11.2 Distance & Midpoint Formulas
  3. 11.3 Parabolas Part I
  4. 11.4 Parabolas Part II
  5. 11.5 Circles Part I
  6. 11.6 Circles Part II
  7. 11.7 Ellipses Part I
  8. 11.8 Ellipses Part II
  9. 11.9 Hyperbolas Part I
  10. 11.10 Hyperbolas Part II
  11. 11.11 Solving Non Linear Systems

Statistics & Probability

  1. 12.1 Introduction to Statistics
  2. 12.2 Independent & Dependent Events
  3. 12.3 Measures of Central Tendency
  4. 12.4 Hisotgrams & Circle Graphs
  5. 12.5 Stem & Leaf Plots
  6. 12.6 Box & Whisker Plots
  7. 12.7 Scatter Plots
  8. 12.8 Permutations
  9. 12.9 Combinations
  10. 12.10 Measures of Dispersion

Series & Patterns

  1. 13.1 Introduction to Series & Patterns
  2. 13.2 Sequences & Series
  3. 13.3 Arithmetic Sequences
  4. 13.4 Arithmetic Series
  5. 13.5 Geometric Sequences
  6. 13.6 Finite Geometric Series
  7. 13.7 Infinite Geometric Series
  8. 13.8 Pascal's Triangle
  9. 13.9 Binomial Theorem

Trigonometry

  1. 14.1 Introduction to Trigonometric Functions
  2. 14.2 Finding an Unknown Angle
  3. 14.3 Reciprocal Ratios
  4. 14.4 Sine Law
  5. 14.5 Cosine Law
  6. 14.6 Angles in Standard Position
  7. 14.7 Special Triangles & Exact Values

Matrices

  1. 15.1 Introduction to Matrices
  2. 15.2 Basic Matrix Operations
  3. 15.3 Matrix Multiplication
  4. 15.4 Determinant of a Matrix
  5. 15.5 Inverse of a Matrix

Rational Functions Contents : This page corresponds to § 3.5 (p. 289) and § 3.6 (p. 299) of the text. Suggested Problems from Text: Introduction

A rational function is one that can be written as a polynomial divided by a polynomial. Since polynomials are defined everywhere, the domain of a rational function is the set of all numbers except the zeros of the denominator.

f(x) = x / (x - 3). The denominator has only one zero, x = 3. So the domain of f is the set of all numbers other than 3.

Domain of f: (-inf, 3) union (3, inf).

The graph of f is pictured below.

Asymptotes

Look again at the graph of f(x) = x / (x - 3) shown above. Since 3 is not in the domain of f, there is no point on the graph with first coordinate 3, so there has to be a break in the graph. In fact, we can see that the function values become unbounded (go to infinity or negative infinity) as x approaches 3. The following tables of function values illustrate this behavior.

We say that f (x) approaches infinity as x approaches 3 from the right , or

The phrase from the right is important. It means that we are using values for x that are larger than 3 and getting close to 3. The next table shows the behavior of f as x approaches 3 from the left.

We say that f(x) approaches negative infinity as x approaches 3 from the left. Here we will use a superscript - to indicate approaching from the left.

In the last section we pointed out that polynomial graphs either rise to the right or fall to the right. The graph of the rational function f does neither of these. It appears from the picture that the points on the graph of f approach the horizontal line y = 1 as x goes right and as x goes left. The tables below provide further evidence that this is the case.

f(x) -> 1 as x -> inf and f(x) -> 1 as x -> - inf,

The vertical line x = 3 and the horizontal line y = 1 are examples of asymptotes . An asymptote is a line that a graph gets close to as x goes to plus or minus infinity or a particular number.

Definition of Horizontal and Vertical Asymptotes

The line x = a is a vertical asymptote of the graph of f if f(x) -> ± infinity as x -> a from the left or the right.

The line y = b is a horizontal asymptote of the graph of f if f(x) -> b as x -> ± infinity.

In general, the graph of a rational function will have a vertical asymptote at a zero of the denominator. The exception to this rule is the case where the numerator and denominator share a zero.

g(x) = (x 2 - 4) / (x - 2). g is a rational function and g is not defined at 2 because 2 is a zero of the denominator. However, 2 is also a zero of the numerator, and we can simplify the quotient.

(x 2 - 4) / (x - 2) = (x - 2)(x + 2)/(x - 2) = x + 2.

However, the function g is not equal to the function h(x) = x + 2, because h is defined at 2 while g is not! (This point is usually mentioned in a College Algebra class and then promptly forgotten.)

The graph of g does not have a vertical asymptote through 2. The graph of g is the line y = x + 2 with a hole where the point (2, 4) would be.

You can usually see from a graphing utility whether or not the graph of a rational function has a horizontal asymptote. However it is generally not clear from the picture exactly what number the asymptote goes through on the y-axis. Generating a table as we did above gives you a much better idea of where the horizontal asymptote is.

f(x) = x 2 / (x - 3). The graph of f is pictured below.

We see from the graph that f(x) -> -inf as x -> 3 - , and f(x) -> inf as x -> 3 + . It is also clear from the picture that the graph has no horizontal asymptote . The graph rises to the right and falls to the left.

The dotted line is a slant asymptote of the graph of f. Slant asymptotes are discussed on page 302 of the text.

It is possible to tell whether or not a rational function has a horizontal asymptote, and if so, exactly where it is, by analyzing the leading terms of the numerator and denominator. This procedure is described in detail on page 291 of the text.

It is certainly possible to be fooled by a graph. Consider the graph pictured below.

This appears to be a graph with the y-axis as a vertical asymptote. In fact this is the graph of f(x) = 5x / (x 2 + 0.01), as rendered by the Java Grapher. The denominator has no (real) zeros, so the graph has no vertical asymptotes.

Use a graphing utility to graph f(x) = 3x 2 / (x 2 - 16). Find all asymptotes. Answer

Application

We are going to enclose a corral adjacent to a river as in the diagram below. No fence is needed on the river side. The enclosed area needs to be 800 square yards. Find the dimensions x and y that require the least amount of fence.

Since we are given that the area must be 800 sq yds, and the area of a rectangle is the product of the two dimensions, this gives us an equation that x and y must satisfy.

x y = 800.

Solving for y yields

y = 800 / x.

Let F stand for the length of fence used, in yards. Since there is one side of length x and two sides of length y, we have

F = x + 2y.

Substituting for y yields

F = x + 2(800 / x).

F = x + 1600 / x.

F is a rational function. Its graph has the y-axis (x = 0) as a vertical asymptote, because F is not defined at 0. The graph of F has no horizontal asymptote, but the line y = x is a slant asymptote.

F = x + 1600 / x

Note that since the variable x in this problems stands for a length, we are only interested in values x > 0, so we may focus on quadrant I of the graph of F.

We need to find the first coordinate of the relative minimum point in quadrant I, for that is the length x corresponding to the smallest values of F, and our goal is to use the smallest possible amount of fence.

One of the graded assignments for the course will be to complete this problem by using a graphing utility to approximate the coordinates of the relative minimum point and find the x and y dimensions of the corral that require the least amount of fence.


Solved Examples

Find the x-intercepts of the function give below:

Set the numerator of this rational function equal to zero and solve for ( ext ):

[ egin ext x^2+ ext x - 2 &= 0 ( ext x&minus1)( ext x+2) &= 0 end ]

Solutions for this polynomial are -

and plot the graph for the function.

Factoring the numerator, we have

[ egin 3 ext x^3&minus6 ext x &= 0 3 ext x( ext x^2&minus2) &= 0 end ]

Given the factor ( 3 ext x ), the polynomial equals ( 0 ) when ( 3 ext x = 0 ) or ( ext x = 0 ).

Let the second factor equal zero, and solve for ( ext x ) :

Thus, the three roots or x-intercepts are:

( herefore ext< x-intercepts >: 0, &minus&radic2 ext < and >&radic2 )

Sketch the graph of the following function.
[ ext < f(x) >= dfrac < ext x+3 >< ext x&minus1>]

So, we&rsquoll start off with the intercepts.

Now, we need to determine the asymptotes.

Let&rsquos first find the vertical asymptotes.

( egin ext x &minus1 &= 0 ext x &= 1 end )
So, we&rsquove got one vertical asymptote. This means that there are now two regions of ( ext x )&rsquos.

They are ( ext x < 1 ) and ( ext x >1 ).

Now, the largest exponent in the numerator and denominator is ( 1 ) and so by the fact, there will be a horizontal asymptote at the line.

Now, we just need points in each region of ( ext x )&rsquos.

Since the ( ext ) and ( ext < x-intercept >) are already in the left region we won&rsquot need to get any points there.

That means that we&rsquoll just need to get a point in the right region.

It doesn&rsquot really matter what value of ( ext x ) we pick here we just need to keep it fairly small so it will fit onto our graph.

&rArr ( (2,5) )
Okay, putting all this together gives the following graph.

Find the vertical asymptotes of

Notice that, based on the linear factors in the denominator, singularities exists at ( ext x= -4 ) and ( ext x= -1 ).

Also, notice that one linear factor ( ( ext x+4) ) cancels with the numerator.

However, one linear factor ( ( ext x+4)) remains in the denominator because it is squared.

Therefore, a vertical asymptote exists at
[ ext x= -4 ]
The linear factor ( ( ext x+1) ) also does not cancel out thus, a vertical asymptote also exists at

Vertical asymptotes are at

( herefore) ( ext x = -1 ext < and > ext x = &minus4 )

Find any horizontal or oblique asymptote of

Since, the polynomials in the numerator and denominator have the same degree ( (2) ), we can identify that there is one horizontal asymptote and no oblique asymptote.

The coefficient of the highest power term is ( 6 ) in the numerator and ( 3 ) in the denominator.

Hence, horizontal asymptote is given by ( ext y= dfrac63=2 )

( herefore ext : ext y = 2 )


Solved Examples on Rational Functions

For ACT Students
The ACT is a timed exam. $60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no "negative" penalty for any wrong answer.

For JAMB and CMAT Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE-FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For GCSE Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits. separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Review:
Given a Rational Function:
To Find the Vertical Asymptote (VA):
(1.) Simplify the function
(2.) Set the denominator to zero (if applicable after simplifying the function)
(3.) Solve for the value of $x$
(4.) $VA: x = value$

To Find the Horizontal Asymptote (HA):
(1.) Arrange the numerator in standard form
(2.) Arrange the denominator in standard form
(3.) If the degree of the numerator is less than the degree of the denominator, $HA: y = 0$
(4.) If the degree of the numerator is the same as the degree of the denominator,
$HA: y = dfrac$

If the degree of the numerator is greater than the degree of the denominator, then onto the Slant Asymptote (or Oblique Asymptote)

To Find the Slant Asymptote (SA):
If the degree of the numerator is greater than the degree of the denominator,
$SA: y = quotientofdfrac$

To Find the $x-intercept$:
(1.) Set $y = 0$
(2.) Solve for the value of $x$
(3.) $x-intercept = (value, 0)$

To Find the $y-intercept$:
(1.) Set $x = 0$
(2.) Solve for the value of $y$
(3.) $y-intercept = (0, value)$

Solve all questions.
Show all work.

(1.) Determine the $x-intercept$ and the $y-intercept$ for the function.

$ f(x) = dfrac [5ex] $ Show/Hide Answer

(2.) ACT Consider the graph of the equation $y = dfrac<3x - 12><2x - 6>$ in the standard $(x, y)$ coordinate plane.
Which of the following equations represents the vertical asymptote of the graph?

$ F.:: x = 2 [3ex] G.:: x = 3 [3ex] H.:: x = 4 [3ex] J.:: x = 6 [3ex] K.:: x = 12 [3ex] $ Show/Hide Answer

(3.) Determine the intercepts and the vertical asymptote of

$ f(x) = dfrac [5ex] $ Show/Hide Answer

(4.) ACT Which of the following linear equations gives the vertical asymptote for the graph of $y = dfrac<201x + 202><203x + 204>$ in the standard $(x, y)$ coordinate plane?

Only two asymptotes apply to the function: the Vertical Asymptote (VA) and the Slant Asymptote (SA)
So, we shall find both asymptotes.
Then, we shall answer the question (the intersection of both asymptotes)

$ underline [3ex] y = dfrac<2x(x + 2)> [5ex] Nothingcancels [3ex] Set hedenominator o solveforx [3ex] x - 3 = 0 [3ex] x = 0 + 3 [3ex] x = 3 [3ex] VA:x = 3 [3ex] underline [3ex] y = dfrac<2x(x + 2)> [5ex] y = dfrac<2x^2 + 4x> [5ex] y = dfrac<2x^2 + 4x + 0> [5ex] equire egin 2x + 10 [-3pt] x - 3 enclose<2x^2 + 4x + 0>kern-.5ex [-3pt] underline<2x^2 - 6xphantom<000>> [-3pt] 10x + 0phantom <0>[-3pt] underline10x - 30> [-3pt] phantom<0>30 [-3pt] end [3ex] SA: y = 2x + 10 [3ex] underline [3ex] VA:x = 3 [3ex] SA: y = 2x + 10 [3ex] y = 2(3) + 10 [3ex] y = 6 + 10 [3ex] y = 16 [3ex] Intersection:(x, y) = (3, 16) $

(6.) ACT In the standard $(x, y)$ coordinate plane, the graph of which of the following equations has the line $x = 2$ as a vertical asymptote?

To determine the vertical asymptote:
(1.) Simplify the function
(2.) Set the denominator to zero
(3.) Solve for $x$

$ All hefunctionsarealreadysimplified [3ex] Lookingat hedenominators: [3ex] x - 2 = 0 [3ex] x = 2 [3ex] CorrectOption: C $

To determine the vertical asymptote:
(1.) Simplify the function
(2.) Set the denominator to zero
(3.) Solve for $x$

(10.) ACT At what value(s) of x is $dfrac<(x - 3)^2>$ undefined?

F. 0 only
G. 0 and 3 only
H. -3 only
J. -3 and 0 only
K. -3, 0, and 3 only

The function is undefined when the denominator is zero.
In other words, find the vertical asymptote of the function.

$ underline [3ex] f(x) = dfrac<(x - 3)^2> [5ex] Nothingcancels [3ex] Set hedenominator o solveforx [3ex] x^2 = 0 [3ex] x = pm sqrt <0>[3ex] x = pm 0 [3ex] x = 0 [3ex] VA:x = 0 $

(12.) ACT The equation $y = dfrac<2x^2 - 18>$ has 2 vertical asymptotes and 1 horizontal asymptote.
What is the horizontal asymptote?

$ F. x = 0 [3ex] G. x = 3 [3ex] H. x = 9 [3ex] J. y = 0 [3ex] K. y = 2 [3ex] $ Show/Hide Answer

(14.) ACT In the standard $(x, y)$ coordinate plane, when $a e 0$ and $b e 0$, the graph of $f(x) = dfrac<2x + b>$ has a horizontal asymptote at:

$ F. y = 2 [3ex] G. y = a [3ex] H. y = -a [3ex] J. y = -dfrac <2>[5ex] K. y = dfrac [5ex] $ Show/Hide Answer

Use the following information to answer questions 15 - 17

ACT Consider the rational function $f(x) = dfrac$, whose graph is shown in the standard $(x, y)$ coordinate plane below.

(15.) ACT What is the value of $f(x)$ at $x = 4$?

$ F. 4 [3ex] G. 2 [3ex] H. dfrac<5> <3>[5ex] J. -1 [3ex] K. -dfrac<7> <3>[5ex] $ Show/Hide Answer

(16.) ACT What is the domain of $f(x)$?
(Note: The domain of a function is all the $x-values$ for which the function is defined.)

A. All real values of $x$ except $pm 3$

B. All real values if $x$ except $dfrac<9><7>$

C. All real values of $x$ except $7$

D. All real values of $x$ except $pm 3$ and $7$

E. All real values of $x$ where $x le -6$ or $x ge 8$

In this case: (based on this question):
Th denominator cannot be zero
Why?
Because division by zero is undefined
Therefore, to find the domain, set the denominator to zero and exclude the value of $x$ that makes the denominator to be zero

$ x - 7 = 0 [3ex] x = 0 + 7 [3ex] x = 7 [3ex] $ Therefore, the domain is the set of all the real values of $x$ except $7$

(17.) ACT How many horizontal and/or vertical asymptotes are there for the graph of $f(x)$?

$ F. 4 [3ex] G. 3 [3ex] H. 2 [3ex] J. 1 [3ex] K. 0 [3ex] $ Show/Hide Answer

To determine the vertical asymptote:
(1.) Simplify the function
(2.) Set the denominator to zero
(3.) Solve for $x$

$ underline [3ex] f(x) = dfrac [5ex] f(x) = dfrac [5ex] f(x) = dfrac<(x + 3)(x - 3)> [5ex] Nothingcancels [3ex] Set hedenominator o solveforx [3ex] x - 7 = 0 [3ex] x = 7 [3ex] VA:x = 7. only1verticalasymptote [5ex] underline [3ex] numeratorinstandardform . yes [3ex] denominatorinstandardform . yes [3ex] DegreeofNumerator = 2 [3ex] DegreeofDenominator = 1 [3ex] $ Because the Degree of the Numerator is greater than the Degree of the Denominator, there is no horizontal asymptote.
In this case, we have a slant asymptote.
Therefore, we have only 1 vertical asymptote.
The correct option is J.


Properties of Rational Numbers

Here are some properties based on arithmetic operations such as addition and multiplication when performed on the rational irrational numbers.

  • Closure Property: This property states that when any two rational numbers are added, the result is also a rational number.

Example: 1/2 + 1/3 = (3 + 2)/6 = 5/6
So it is closed under addition, the same way for other operations also it remains closed.
Example: 1/2 -1/3=1/6
1/2* 1/3=1/6
1/2/1/3=3/2

  • Commutative Property: The sum of two rational numbers are commutative, as they can be added in any order.

For example:1/2+1/3=5/6 or 1/3+1/2=5/6
The product of two rational numbers is a rational number.
Example: 1/2 x 1/3 = 1/6 or 1/3*1/2=1/6
This property holds true for addition and multiplication , but it does not hold true for Subtraction and division of rational numbers.

  • Associative Property: In this property, when we Take any three rational numbers a, b and c.

(a + b) + c and a + (b + c) is the same .
It states that you can add or multiply numbers regardless of how they are grouped.

For example, given numbers are 5, -6 and 2/3 ( 5 &ndash 6 ) + 2/3= -1+2/3=-1/3 Now, 5 + ( -6 + 2/3)=-1/3.
But again this property does not hold true for subtraction and division.

  • Distributive Property: This property says that when we multiply a sum of integers by another integer it equals to the same result when we multiply each integer by another integer and then add the products together. In other words, it is the distributive property of multiplication by making use of integers.

Distributive property states that for any three integers x, y and z we have
x * ( y + z ) = (x * y) +( x * z)
2*(3+4)=(2*3)+(2*4) = 14


Graph some more complicated functions

Graphs of normal polynomials are always unbroken lines or smooth curves. Rational functions, by contrast, sometimes have breaks which separate the graph into two or more disconnected pieces. A break in the graph of a function is called a discontinuity. Think about it as a gap on the road you are driving.

Polynomials have no breaks or discontinuities because they deal mainly with the process of addition, subtraction and multiplication. Rational functions deal with the process of division. Division makes room for discontinuity to take place in the graph of the function, particularly in places where a division by zero exists. Before we go on, it is vital for you to remember that division by zero is undefined or does not exist. In other words, a fraction cannot have zero in the denominator. Here are three samples of fractions that are undefined: 5/0, 12/0, x/0 and/or any other fraction which has zero in the denominator. The equation (y=frac<5>) is undefined at exactly one place: when (x=1).

Example A:

In this case, x CANNOT = 2 in the denominator because it will create division by zero. If x = 2 in the denominator of the above fraction, there will be a break or discontinuity in the graph or picture of the function.

How to graph the above function:

1) Create an (x, y) table.
2) Select different values for x OTHER THAN x = 2.
3) Use algebra to simplify the fraction.
4) Graph each point on the xy-plane. NOTE: The xy-plane is also called the Coordinate Plane. This idea is taught in geometry.

I will select the following 4 values for x: 0, 3, 4 and 5.

Here are the four points in (x, y) form:

After graphing all four points, the graph will look like this:

NOTE: There are functions with squares, cubes, etc that will NEVER yield division by zero regardless what values we decide to select for x. The squaring and cubing process will do away with division by zero.

Again, follow the steps given in Sample A.

I decided to select the following 4 values for x: 0, -1, 3 and 1/2.

This is what our function looks like:

A break in the graph of the function can happen in two ways:

1) A Missing Point (also called Removable Singularity or Discontinuity).

2) Asymptotes. I will discuss asymptotes in math lesson: Graphing Rational Functions. Part 2.

In the function y = x^2 - x + 12/(x - 4), if x = 4, there is discontinuity or a gap in the graph of the function because when x = 4, division by zero is the result.

Here's what the above graph looks like when x = 4:

When facing division by zero, I can factor the numerator and denominator of the fraction. I will now factor the above fraction. Why factor? After factoring, I'll be able to select ANY value for x in terms of making a graph without removable singularity.

In the equation y = x + 3, I can select ANY value for x to find points in the (x, y) form.

The first type of break is known as REMOVABLE SINGULARITY, which is a missing point on the graph of the function. Removable singularity takes place when a chosen value for x leads to division by zero. If you factor and then simplify the rational function, division by zero can be avoided.

Asymptotes: An Introduction

In the function f(x) = (x + 3)/(x - 2), x CANNOT equal 2 because it would yield division by zero in the denominator, which does not exist. But what happens when x almost equals 2?
For example, in the sample above, if x = 2.3, then this will happen:

We can see here that as the value of x decreases, y increases. The closer x will get to 2, the bigger y will be. Of course, the opposite is also true. As x increases, y decreases.

Geometrically speaking, there will be a near-vertical line that appears as x gets closer to 2. It will approach, but never actually touch, x=2. The vertical line x=2 is called the asymptote. Here is an illustration:

Asymptote on the right upward side of the vertical line looks like this:


The above pictures represent vertical asymptotes. Both asymptotes get closer to 2 but never quite reach the vertical line at point (2, 0 ).

There are many more pictures of asymptotes. Look for them in future lessons.


Review

1. Write a function that fits the following criteria:

2. Write a function that fits the following criteria:

3. Write a function that fits the following criteria:

4. Write a function that fits the following criteria:

5. Write a function that fits the following criteria:

Give the equations of the vertical asymptotes for the following functions.

Identify the holes and equations of the vertical asymptotes of the following rational functions.


Watch the video: Part 1 Graphing Rational Functions (December 2021).