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Algebra 2 View details

3. Graphing Reciprocal Functions

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eCourses /

Algebra 2 /

Chapter 4

Graphing Reciprocal Functions

This lesson zeroes in on the graphing of reciprocal functions, highlighting unique features like asymptotes. Grasping these functions is vital for various disciplines such as engineering, physics, and economics. For instance, engineers may need to understand how certain variables are inversely related for precise calculations. The lesson explains how to spot asymptotes, lines that the graph approaches but never crosses. It also covers the domain and range, which indicate the set of possible input and output values. This foundational knowledge is crucial for tackling real-world challenges where variable behavior is key.

Lesson Settings & Tools

	16 Theory slides
	10 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

Image Credits

Slide of 16

This lesson introduces a new function family and analyze its characteristics. The functions of this family have the form

f (x) = \frac{1}{a ( x )},

where

a (x)

is a nonzero linear function. Furthermore, the transformations of the most basic function in this family

y = \frac{1}{x}

will be studied.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Challenge

Graphing Functions of the Form $f (x) = \frac{a x + b}{c x + d}$

Consider the function shown.

f (x) = \frac{3 x - 1}{x - 2}

a Use long division to rewrite the function in the form

y = \frac{a}{x - h} + k .

b Identify the asymptotes, domain, and range of the function.

c Plot some points around the asymptotes by making a table of values and draw the function.

d Graph the function by transforming the graph of

y = \frac{1}{x} .

Explore

Comparing Graphs of Functions of the Form $f (x) = \frac{a}{x - h} + k$

The graphs of three functions of the form

y = \frac{a}{x - h} + k

are shown in the applet.

Function I: f(x)=6/(x+4)-4, Function II: g(x)= 4/(x-4)+4, Function III:h(x) = 5/(x+6)+2

Identify the asymptotes, domain, and range of each function.

	Asymptotes	Domain	Range
Function I
Function II
Function III

What is the relation between the asymptotes, and domain and range? Is it possible to identify them just by looking at the function rule?

Discussion

Reciprocal Functions

The reciprocal function is a function that pairs each

x -

value with its reciprocal.

The function rule of this reciprocal function is obtained algebraically by writing the rule for the pairings shown in the table.

$f (x) = \frac{1}{x}, x \neq = 0$

The graph of the function $f (x) = \frac{1}{x}$ is a hyperbola, which consists of two symmetrical parts called branches. It has two asymptotes, the $x -$ and $y -$ axes. The domain and range are all nonzero real numbers.

Graph of y = 1/x with horizontal asymptote at y = 0 and vertical asymptote at x = 0. Points (-2, -1/2), (-1, -1), (-0.5, -2), (2, 1/2), (1, 1), and (0.5, 2) are plotted on the graph.

The graph of $y = \frac{1}{x}$ can be used to graph other reciprocal functions. This can be done by applying different transformations.

Name	Equation	Characteristics
Parent Reciprocal Function	$y = \frac{1}{x}$	$Domain : Range : Asymptotes : R - {0} R - {0} x - and y - axes$
Inverse Variation Functions	$y = \frac{a}{x}$
General Form of Reciprocal Functions	$y = \frac{a}{x - h} + k$	$Domain : Range : Asymptotes : R - {h} R - {k} x = h and y = k$

Extra

Reciprocal vs. Inverse of a Function

It is important not to confuse the reciprocal of a function with the inverse of a function. For numbers, $a^{- 1}$ refers to the reciprocal $\frac{1}{a},$ while the notation $f^{- 1} (x)$ is commonly used to refer to the inverse of a function.

Function, $f$	Reciprocal, $\frac{1}{f}$	Inverse, $f^{- 1}$
$f (x) = 2 x + 10$	$\frac{1}{f ( x )} = \frac{1}{2 x + 1 0}$	$f^{- 1} (x) = \frac{1}{2} (x - 10)$
$f (x) = x^{3} - 5$	$\frac{1}{f ( x )} = \frac{1}{x ^{3} - 5}$	$f^{- 1} (x) = 3 x + 5$

Example

Correcting the Error

Mark wants to identify the asymptotes, domain, and range of the reciprocal function shown as follows.

Mark states the characteristics of the function.

I . II . III . IV . The vertical asymptote is x = 3 . The domain is all real numbers except 3 . The horizontal asymptote is y = 4 . The range is all real numbers except 4 .

Select the correct statement(s).

Hint

Start by identifying any $x -$ values for which $f (x)$ is undefined.

Solution

Consider the given reciprocal function.

f (x) = \frac{5}{x + 3} + 4

Start by identifying any

x -

values for which

f (x)

is undefined.

x + 3 = 0 \Leftrightarrow x = - 3

The function is not defined when

x = - 3,

which means that there is a vertical asymptote at

x = - 3 .

This can also be confirmed from the given graph.

The branch of the graph to the left of $x = - 3$ shows that the $y -$ values approach $4 .$ This is true for the other branch as well. Therefore, there is a horizontal asymptote at $y = 4 .$ Draw the asymptotes on the given coordinate plane.

In the graph, it can be seen that the domain is all real numbers except $- 3,$ and the range is all real numbers except $4 .$ Therefore, Mark's correct statements are III and IV.

Pop Quiz

Identifying Characteristics of Reciprocal Functions

For the indicated reciprocal function, identify the asymptotes or the value that makes the function undefined.

Discussion

Graphing a Reciprocal Function

The identifiable asymptotes of the general form of reciprocal functions can be used to simplify the process of graphing them. As an example, consider graphing the following function.

f (x) = \frac{- 1}{x - 3} + 2

To graph this reciprocal function, these steps can be followed.

Draw the Asymptotes

Comparing the function rule to the general form of a reciprocal function

y = \frac{a}{x - h} + k

makes it possible to identify the asymptotes

x = h

and

y = k .

f (x) = \frac{a}{x - 3} + 2

For

f (x),

the vertical asymptote is

x = 3

and the horizontal asymptote is

y = 2 .

These can be drawn in a coordinate plane.

Plot Points Around the Vertical Asymptote

Plot some points both to the left and the right of the vertical asymptote. For the example, it would be appropriate to use the $x -$ values $1,$ $2,$ $2.5,$ $3.5,$ $4,$ and $5$ for this function.

$x$	$\frac{- 1}{x - 3} + 2$	$f (x) = \frac{- 1}{x - 3} + 2$
$1$	$\frac{- 1}{1 - 3} + 2$	$2.5$
$2$	$\frac{- 1}{2 - 3} + 2$	$3$
$2.5$	$\frac{- 1}{2 . 5 - 3} + 2$	$4$
$3.5$	$\frac{- 1}{3 . 5 - 3} + 2$	$0$
$4$	$\frac{- 1}{4 - 3} + 2$	$1$
$5$	$\frac{- 1}{5 - 3} + 2$	$1.5$

The points $(x, f (x))$ can be added to the coordinate plane to begin to see the behavior of $f .$

Draw the Graph

The graph can now be drawn by connecting the points with two smooth curves. They must approach but not intercept both the horizontal and vertical asymptotes.

Example

Modeling the Average Cost With a Reciprocal Function

Mark's school purchased a new math software program.

The program has a costs of $$ 360 .$ Additionally, the school has to pay $$ 15$ per student who uses the software.

a If

n

represents the number of students using the software, write a function

A (n)

that models the average cost to the school per student who uses the software.

b Graph the function written in Part A.

Answer

a Function:

A (n) = \frac{1 5 n + 3 6 0}{n}

b Graph:

Hint

a The average cost is the sum of the total cost divided by the number of students.

b Identify the asymptotes. Make a table of values, then plot and connect the points.

Solution

a Begin by writing an equation that represents the average cost. Let

A

be the average cost to each student and

n

be the number of students that use the program.

Verbal Expression	Algebraic Expression
Initial cost	$360$
Cost for $n$ number of students	$15 n$
Total cost	$15 n + 360$
Average cost $A$	$A = \frac{1 5 n + 3 6 0}{n}$

Since

A

is a function of

n,

it can be substituted with

A (n) .

A (n) = \frac{1 5 n + 3 6 0}{n}

b Recall the general form of reciprocal functions.

f (x) = \frac{a}{x - h} + k

The domain of this type of function is all real numbers except

h .

The range is all real numbers except

k .

The asymptotes are the lines

x = h

and

y = k .

To find the key characteristics of the graph of the given situation, it must first be rewritten so that it has the same form.

A (n) = \frac{1 5 n + 3 6 0}{n}

▼

Rewrite

WriteSumFrac

Write as a sum of fractions

A (n) = \frac{1 5 n}{n} + \frac{3 6 0}{n}

SimpQuot

Simplify quotient

A (n) = 15 + \frac{3 6 0}{n}

CommutativePropAdd

Commutative Property of Addition

A (n) = \frac{3 6 0}{n} + 15

Now that it has been rewritten, the values of

a,

h,

and

k

for the given function can be identified.

A (x) = \frac{3 6 0}{n} + 15 ⇓ A (x) = \frac{3 6 0}{n - 0} + 15

In this function,

a = 360,

h = 0,

and

k = 15 .

Therefore, the asymptotes are the lines

x = 0

and

y = 15 .

The domain of the function is all real numbers except $0$ and the range is all real numbers except $15 .$ With this in mind, make a table of values. Be sure to include numbers on both sides of the vertical asymptote!

$n$	$\frac{3 6 0}{n} + 15$	$A (n) = \frac{3 6 0}{n} + 15$
$- 40$	$\frac{3 6 0}{- 4 0} + 15$	$6$
$- 30$	$\frac{3 6 0}{- 3 0} + 15$	$3$
$- 20$	$\frac{3 6 0}{- 2 0} + 15$	$- 3$
$- 10$	$\frac{3 6 0}{- 1 0} + 15$	$- 21$
$10$	$\frac{3 6 0}{1 0} + 15$	$51$
$20$	$\frac{3 6 0}{2 0} + 15$	$33$
$30$	$\frac{3 6 0}{3 0} + 15$	$27$
$40$	$\frac{3 6 0}{4 0} + 15$	$24$

Next, plot and connect the points from the table. Remember that the asymptotes are $x = 0$ and $y = 15$ and that the graph will have two branches.

Since $n$ represents the number of students using the software, it would not make sense for $n$ to be negative. Therefore, only the part of the graph in the first quadrant should be considered.

Explore

Effects of the Values of $h$ and $k$

Mark's teacher showed his class an applet to illustrate the graphs of the reciprocal functions of the form

f (x) = \frac{1}{x - h} + k .

Mark and his friends observe how the graph changes as the values of

h

and

k

change.

What transformation does the change in these values represent? Does any of the asymptotes change? If yes, how?

Discussion

Translation of Reciprocal Functions

The transformations that Mark's class observed are translations. In these transformations, the whole graph is moved without changing its shape or orientation. A translation can be vertical or horizontal. A vertical translation is achieved by adding some number to every output value of a function rule. Consider the parent reciprocal function

f (x) = \frac{1}{x} .

Parent Function y = \frac{1}{x} Vertical Translation y = \frac{1}{x} + k

k

is a positive number, the translation is performed upwards. Conversely, if

k

is negative, the translation is performed downwards. If

k = 0,

then there is no translation. This transformation can be shown on a coordinate plane.

Notice that as

k

changes, the horizontal asymptote

y = k

moves but the vertical one is unaffected. A horizontal translation is instead achieved by subtracting a number from every input value.

Parent Function y = \frac{1}{x} Horizontal Translation y = \frac{1}{x - h}

In this case, if

h

is a positive number, the translation is performed to the right. Conversely, if

h

is negative, the translation is performed to the left. If

h = 0,

then there is no translation.

Notice that this transformation affects the vertical asymptote $x = h$ but not the horizontal one. The vertical and horizontal translations of the graph of a function $f$ can be summarized in a table.

Transformations of $y = \frac{1}{x}$
$\underline{Vertical Translation} y = \frac{1}{x} + k$	Translation up $k$ units, $k > 0$
$\underline{Vertical Translation} y = \frac{1}{x} + k$	Translation down $k$ units, $k < 0$
$\underline{Horizontal Translation} y = \frac{1}{x - h}$	Translation to the right $h$ units, $h > 0$
$\underline{Horizontal Translation} y = \frac{1}{x - h}$	Translation to the left $h$ units, $h < 0$

Example

Finding Domain and Range After Translation

Mark now has a rough idea about how to translate reciprocal functions. His class concluded that the asymptotes of a reciprocal function are also translated when the function is translated. Mark thought about how its domain and range were affected by this transformation. Consider the parent reciprocal function.

How will the domain and the range of the function $y = \frac{1}{x}$ change after the translation of its graph by $4$ units down and $3$ units to the left?

Answer

Domain: All real numbers except $- 3$
Range: All real numbers except $- 4$

Hint

When a function is translated, its graph keeps its original shape. Therefore, if it has any asymptotes, the asymptotes are translated in the same way.

Solution

To determine the domain and the range of the function after translating it, start by considering some possible transformations.

Transformations of $y = \frac{1}{x}$
$\underline{Vertical Translation} y = \frac{1}{x} + k$	Translation up $k$ units, $k > 0$
$\underline{Vertical Translation} y = \frac{1}{x} + k$	Translation down $k$ units, $k < 0$
$\underline{Horizontal Translation} y = \frac{1}{x - h}$	Translation to the right $h$ units, $h > 0$
$\underline{Horizontal Translation} y = \frac{1}{x - h}$	Translation to the left $h$ units, $h < 0$

Note that if the graph of the function is translated, the asymptotes are also translated the same distance and direction. A combination of transformations can be applied to the same parent function. Consider the given translations again.

Vertical translation down $4$ units
Horizontal translation $3$ units left

Therefore,

k = - 4

and

h = - 3 .

y = \frac{1}{x - ( - 3 )} + (- 4) \Leftrightarrow y = \frac{1}{x + 3} - 4

These transformations can be applied one at a time. Start by translating the parent function

f (x) = \frac{1}{x}

down

4

units.

The second transformation is a horizontal translation $3$ units left.

Finally, look at the graph of the given function and its asymptotes alone.

As shown in the graph, the vertical asymptote is the line

x = - 3

and the horizontal asymptote is the line

y = - 4 .

This information can be used to state the domain and range of the function.

Domain : Range : All real numbers except - 3 All real numbers except - 4

Explore

Effects of the Value of $a$

The graph of the reciprocal function $f (x) = \frac{a}{x}$ is shown in the coordinate plane. Observe how the graph changes as the value of $a$ changes.

How does the graph of the function change when

∣ a ∣ > 1

0 < ∣ a ∣ < 1 ?

What happens when

a

is negative?

Discussion

Stretch and Shrink of Reciprocal Functions

The graph of a function is vertically stretched or shrunk by multiplying the output of a function rule by some positive constant

a .

Consider the reciprocal parent function

y = \frac{1}{x} .

Parent Function y = \frac{1}{x} Vertical Stretch / Shrink y = a (\frac{1}{x}) = \frac{a}{x}

a

is greater than

1,

the graph is vertically stretched by a factor of

a .

Conversely, if a is less than

1

but greater than

0

(or

0 < a < 1),

the graph is vertically shrunk by a factor of

a .

a = 1,

then there is no stretch nor shrink. All vertical distances from the graph to the

x -

axis are changed by the factor

a .

Multiplying the graph of the reciprocal parent function

Similarly, a reciprocal function graph is horizontally stretched or shrunk by multiplying the input of a function rule by some positive constant

b .

Parent Function y = \frac{1}{x} Horizontal Stretch / Shrink y = \frac{1}{b x}

In this case, if

b

is greater than

1,

the graph is horizontally shrunk by a factor of

b .

Conversely, if

b

is less than

1

but greater than

0

(or

0 < b < 1),

the graph is horizontally stretched by a factor of

b .

b = 1,

then there is neither a stretch nor shrink of the graph.

Notice that these stretches and shrinks do not affect the asymptotes. The table below summarizes these transformations.

Transformations of $y = \frac{1}{x}$
$\underline{Vertical Stretch or Shrink} y = \frac{a}{x}$	Vertical stretch, $a > 1$
$\underline{Vertical Stretch or Shrink} y = \frac{a}{x}$	Vertical shrink, $0 < a < 1$
$\underline{Horizontal Stretch or Shrink} y = \frac{1}{b x}$	Horizontal stretch, $0 < b < 1$
	Horizontal shrink, $b > 1$

Discussion

Reflection of Reciprocal Functions

A function is reflected in the

x -

axis by changing the sign of all output values. Consider the following reciprocal function.

Function y = \frac{1}{x - 3} - 1 Reflection in the x - axis y = - (\frac{1}{x - 3} - 1)

Graphically, all points on the graph move to the opposite side of the

x -

axis while maintaining their distance to the

x -

axis. As such, the

x -

intercepts and vertical asymptotes are preserved.

Reflection of a Reciprocal Function in the x-axis

A reflection in the

y -

axis is instead achieved by changing the sign of every input value. The reflection of the graph of another reciprocal function is shown below.

Function y = \frac{1}{x - 2} - 3 Reflection in the y - axis y = \frac{1}{- x - 2} + 3

Notice that the

y -

intercept and horizontal asymptote are preserved.

Reflection of a Reciprocal Function in the y-axis

The table below summarizes the different types of reflections.

Transformations of $y = \frac{1}{x}$
Reflections	$In the x - axis y = - (\frac{1}{x})$
Reflections	$In the y - axis y = \frac{1}{- x}$

Notice that $- (\frac{1}{x}) = \frac{1}{- x} .$ Reflecting the parent reciprocal function in the $x -$ axis and in the $y -$ axis will produce the same graph.

Reflection of the parent reciprocal function in the x- and y-axes

Example

Comparing Graphs of Reciprocal Functions

a Mark, Paulina, and Tiffaniqua were each given a pair of reciprocal functions to compare the graphs of the given functions.

	Pair of Functions
Mark	$y = \frac{1}{x}$	$y = - 3 (\frac{1}{x})$
Paulina	$y = \frac{1}{x}$	$y = \frac{1}{x} - 6$
Tiffaniqua	$y = \frac{1}{x}$	$y = \frac{1}{x + 5}$

Help them identify the transformations, asymptotes, domain, and range of the reciprocal functions.

b Without making a table of values, sketch a graph of the function shown.

y = - 3 (\frac{1}{x + 5}) - 6

Answer

a See solution.

b Graph:

Hint

a What transformation does the number

- 3

represent? What does the number subtracted from

\frac{1}{x}

represent? What does the number added in the denominator of the last equation represent?

b Notice that the function has all the transformations applied in the previous part.

Solution

a The graphs of each pair of functions will compared one by one.

Mark's Functions

Mark was asked to compare the following pair of reciprocal functions.

y = \frac{1}{x} and y = - 3 (\frac{1}{x})

Here both graphs have a vertical asymptote at

x = 0

and a horizontal asymptote at

y = 0 .

However, the second equation is multiplied by

- 3,

which represents a vertical stretch by a factor of

∣ - 3 ∣ .

Since this factor is negative, the graph should also be reflected in the

x -

axis.

Paulina's Functions

The second pair functions was given to Paulina.

y = \frac{1}{x} and y = \frac{1}{x} - 6

Notice that the number

- 6

in the right-hand side equation represents a translation

6

units down of the graph

y = \frac{1}{x} .

Therefore, both graphs have the same vertical asymptote,

x = 0 .

The horizontal asymptote of the second function will be the line

y = - 6 .

Tiffaniqua's Functions

Finally, consider the pair of functions assigned to Tiffaniqua. Notice that the

5

added in the denominator of the second equation represents a translation

5

units to the left.

y = \frac{1}{x} and y = \frac{1}{x + 5}

The first graph has a vertical asymptote at

x = 0

and a horizontal asymptote at

y = 0 .

The second graph is translated

5

units to the left, has a vertical asymptote at

x = - 5,

and a horizontal asymptote at

y = 0 .

The characteristics of the graphs discussed above can be seen in the table.

	Domain	Range	Vertical Asymptote	Horizontal Asymptote
$y = \frac{1}{x}$	$R - {0}$	$R - {0}$	$x = 0$	$y = 0$
$y = - \frac{3}{x}$	$R - {0}$	$R - {0}$	$x = 0$	$y = 0$
$y = \frac{1}{x} - 6$	$R - {0}$	$R - {- 6}$	$x = 0$	$y = - 6$
$y = \frac{1}{x + 5}$	$R - {- 5}$	$R - {0}$	$x = - 5$	$y = 0$

b Notice that the given function is a combination of all the functions seen in the previous part.

y = - 3 (\frac{1}{x + 5}) - 6

To sketch its graph, the following transformations can be applied to the graph of

y = \frac{1}{x} .

First, notice that $- 3$ is a combination of two transformations, a vertical stretch and a reflection. Stretch the graph of $y = \frac{1}{x}$ vertically by a factor of $∣ - 3 ∣$ and reflect the graph across the $x -$ axis.
Next, translate the graph of the previous step $5$ units to the left so the graph of $y = - 3 (\frac{1}{x + 5})$ is drawn.
Finally, translate the graph $6$ units down to get the graph of $y = - 3 (\frac{1}{x + 5}) - 6 .$

The graph of the given equation is found by following these steps.

Extra

Does the Order of Transformations Matter?

When it is asked to apply two or more transformations to a function, it is often necessary to apply them in a certain order. Applying transformations in the order shown below will help to find a correct solution.

Stretch or shrink
Reflection
Horizontal translation
Vertical translation

Consider two different composition of transformation of $y = \frac{1}{x} .$

Composition I: A vertical stretch of $3$ followed by a vertical translation of $6$ up
Composition II: A vertical translation of $6$ up followed by a vertical stretch of $3 .$

The functions shown below will be obtained after applying each composition.

\underline{Composition I} y = \frac{3}{x} + 6 \underline{Composition II} y = 3 (\frac{1}{x} + 6)

Therefore, Composition I will not produce the same graph as Composition II.

Closure

Graphing a Reciprocal Function Using Transformations

Several techniques such as making a table of values and using transformations of a parent function can be used to graph a function. In this lesson, the parent reciprocal function was transformed to graph other reciprocal functions. Now the function presented at the beginning of the lesson will be graphed using transformations.

f (x) = \frac{3 x - 1}{x - 2}

a Use long division to rewrite the function in the form

y = \frac{a}{x - h} + k .

b Identify the asymptotes, domain, and range of the function. Write these values on a table.

c Plot some points around the asymptotes by making a table of values and draw the function.

d Graph the function by transforming the graph of

\frac{1}{x} .

Answer

f (x) = \frac{5}{x - 2} + 3

b Table:

	$f (x) = \frac{5}{x - 2} + 3$
Domain	All real numbers except $2$
Range	All real numbers except $3$
Vertical Asymptote	$x = 2$
Horizontal Asymptote	$y = 3$

c Graph:

d Graph:

Hint

a Check if the dividend and divisor are in standard form and if they have any missing terms.

b For what value of

x

is the denominator equal to

0 ?

What does this value mean for the domain of the function?

c Make a table of values using values around the vertical asymptote.

d Start by drawing the graph of the parent function

f (x) = \frac{1}{x} .

Solution

a The dividend and divisor are in standard form and all the terms are present.

\frac{3 x - 1}{x - 2} \to Dividend \to Divisor

To rewrite the fraction, polynomial long division can be used.

x - 2 3 x - 1

▼

Divide

$\frac{3 x}{x} = 3$

3 x - 2 3 x - 1

Multiply $term$ by $divisor$

3 x - 2 3 x - 1 3 x - 6

Subtract down

3 x - 2 5

The quotient is

3

with a remainder of

5 .

f (x) = \frac{5}{x - 2} + 3

The function is now in general form.

b Key characteristics about the graphs of a reciprocal function can be determined by looking at its function rule.

f (x) = \frac{a}{x - h} + k

The domain of the function is all real numbers except

h .

The range is all real numbers except

k .

The asymptotes are the lines

x = h

and

y = k .

Now consider the general form of the given function.

f (x) = \frac{5}{x - 2} + 3

The characteristics of this graph are shown in the table.

	$f (x) = \frac{5}{x - 2} + 3$
Domain	All real numbers except $2$
Range	All real numbers except $3$
Vertical Asymptote	$x = 2$
Horizontal Asymptote	$y = 3$

c The asymptotes of the function were identified in the previous part. To graph the function, start by drawing the asymptotes on a coordinate plane.

Recall that the domain of the function is all real numbers except $2 .$ Keeping this in mind, make a table of values to find points that fall on the graph of the function. Be sure to include numbers both above and below $2 .$

$x$	$\frac{5}{x - 2} + 3$	$f (x) = \frac{5}{x - 2} + 3$
$- 3$	$\frac{5}{- 3 - 2} + 3$	$2$
$- 1$	$\frac{5}{- 1 - 2} + 3$	$\approx 1.33$
$1$	$\frac{5}{1 - 2} + 3$	$- 2$
$3$	$\frac{5}{3 - 2} + 3$	$8$
$5$	$\frac{5}{5 - 2} + 3$	$\approx 4.67$
$7$	$\frac{5}{7 - 2} + 3$	$4$

Plot and connect the points found in the table. Remember that the asymptotes are $x = 2$ and $y = 3$ and that the graph will have two branches.

d To draw the graph of the given function, start by considering some possible transformations.

Transformations of $f (x) = \frac{1}{x}, x \neq = 0$
$\underline{Vertical Translation} y = \frac{1}{x} + k$	Translation up $k$ units, $k > 0$
$\underline{Vertical Translation} y = \frac{1}{x} + k$	Translation down $k$ units, $k < 0$
$\underline{Horizontal Translation} y = \frac{1}{x - h}$	Translation right $h$ units, $h > 0$
$\underline{Horizontal Translation} y = \frac{1}{x - h}$	Translation left $h$ units, $h < 0$
$\underline{Vertical Stretch or Shrink} y = \frac{a}{x}$	Vertical stretch, $a > 1$
$\underline{Vertical Stretch or Shrink} y = \frac{a}{x}$	Vertical shrink, $0 < a < 1$

Note that if the graph of the function is translated, the asymptotes are also translated the same distance and direction. Consider the function again.

f (x) = \frac{5}{x - 2} + 3

The given function is a combination of transformations.

Horizontal translation $2$ units right
Vertical stretch by a factor of $5$
Vertical translation $3$ units up

These transformations will be applied one at a time. Start by translating the parent function, $f (x) = \frac{1}{x},$ $2$ units right.

Now, apply a vertical stretch by a factor of $5 .$

The last transformation is a vertical translation $3$ units up.

Finally, the graph can be cleaned up to show only the given function and its asymptotes.

Level 1

Level 2

Level 3

Graphing Reciprocal Functions

Exercise 2.1

Izabella's school is planing to spend $$ 840$ on student academic achievement awards.

The value of each award is the same and at least 6 awards will be given. Additionally, each award should not be less than $$ 20 .$

Write a function

C (n)

that models the cost of one award if

n

awards are given.

What is the graph of the function?

What is the domain and range of the function?

A . B . C . D . Domain x > 0 6 < x < 20 x > 6 6 \leq x \leq 42 Range y > 0 42 < y < 840 y > 20 20 \leq y \leq 140

We know that n represents the number of awards and that each award costs C dollars. The total cost of all awards should amount to $840. Therefore, the product of n and C should equal 840. n * C= 840 Since the number of awards is the independent variable and the cost per award is the dependent variable, we should isolate C by dividing both sides of the equation by n. n * C =840 ⇔ C=840/n We can see that C is a function of n. C(n)=840/n

Recall the general form of reciprocal functions. f(x)=a/x- h + k The asymptotes are the lines x= h and y= k. To find the key characteristics of the graph of the function found in Part A, we have to rewrite it. C(n)&= 840/n &⇕ C(n)&= 840/n- 0 + 0 For this function, a= 840, h= 0, and k= 0. Therefore, the asymptotes are the lines x= 0 and y= 0.

Before we make a table of values to graph the function, we need to consider the restrictions on the number of awards and their costs. We know that the minimum number of awards is 6. By substituting n=6, we can determine the maximum cost per award.

The point (6,140) gives the minimum number of awards and the maximum cost per award. We also know that the value of an award should not be less than $20. If we substitute C(n)=20, we can determine the maximum number of awards that can be purchased at this price.

The maximum number of awards we can have is 42. Each award then costs $20. This gives us a second cut-off point at (42,20). With this in mind, we can make a table of values. We found that n is between 6 and 42.

n	840/n	C(n)=840/n
6	840/6	140
10	840/10	84
20	840/20	42
30	840/30	28
42	840/42	20

We can now plot the points and connect them.

The domain describes the number of awards and the range is the cost per award. We found that the points (6,140) and (42,20) are the endpoints of the graph.

Let's write the domain and range as compound inequalities. Domain:& 6 ≤ n ≤ 42 Range:& 20 ≤ C ≤ 140

Consider the following function.

x y + 5 = 0

Which graph is the graph of the function?

What transformations must be applied to the graph of

y = \frac{1}{x}

to get the graph of the function?

I . II . III . IV . Vertical stretch by a factor of 5 Horizontal translation 5 units right Reflection in the x - axis Vertical translation 5 units up

We will first rewrite the given function. xy + 5 = 0 ⇔ xy = - 5 If either x or y is equal to 0, the product of x and y would be 0, not - 5. Therefore, neither x nor y can be 0. With that said, we can rewrite the equation by isolating the y-variable. xy=- 5 ⇔ y=- 5/x Let's now draw its graph. To do so, we will make a table including positive and negative values. Remember that x cannot be zero!

x	- 5/x	y=- 5/x
- 5	- 5/- 5	1
- 2	- 5/- 2	2.5
- 1	- 5/- 1	5
1	- 5/1	- 5
2	- 5/2	- 2.5
5	- 5/5	- 1

Because x cannot be zero the graph will not cross the y-axis. Therefore, we need two curves to connect the points.

We will review possible transformations of the function f(x)= 1x.

Transformations of f(x)= 3x, x≠ 0
Reflection	In thex-axis [0.4em] y= - ( 1/x )
Horizontal Translation [0.4em] y=1/x- h	Translation right h units, h>0
Horizontal Translation [0.4em] y=1/x- h	Translation left h units, h<0
Vertical Translation [0.4em] y=1/x+ k	Translation up k units, k>0
Vertical Translation [0.4em] y=1/x+ k	Translation down k units, k < 0
Vertical Stretch or Shrink [0.4em] y=a/x	Vertical stretch, a>1
Vertical Stretch or Shrink [0.4em] y=a/x	Vertical shrink, 0< a< 1

Let's now consider the function. y =- 5/x ⇔ y =- 5/x- 0+ 0 We can see that a= - 5, h= 0, and k= 0. Therefore, the graph of this function can be described as a transformation of the graph of y = 1x as follows.

Transformation	Function Rule
1. Reflection in the x-axis	y=1/x → y= - 1/x
2. Vertical stretch by a factor of 5	y=- 1/x → y=- 5/x

This means the answer is I and III.

Magdalena's plans for next year include cycling $4000$ kilometers. Suppose that $y$ represents the number of kilometers she rides each day and $x$ represents the number of days.

Write a function that models the situation.

Which graph is the graph of the function?

If Magdalena decides to ride her bike every day of the year, how many kilometers should she ride? Round the answer to the nearest integer.

We know that Magdalena wants to ride her bike 4000 miles. It is also given that x represents the number of days she rides the bike. If we divide 4000 by x, we will find the distance in kilometers Magdalena should ride each day. 4000/x We are told that y also represents the number of kilometers she rides each day. Setting it equal to the expression above, we can write an equation. y=4000/x

Before we graph the function, we need to identify its domain, range, and find the asymptotes. Then, we will make a table of values and graph the function.

Domain

Recall that the domain is the set of all possible x-values. Since x, in this case, represents the number of days, it cannot have negative values. Moreover, x is the denominator of the fraction, so it cannot be equal to 0 either. x > 0

Range

The range of the function is the set of all possible values of the dependent variable. In our case, the dependent variable is y, which represents the distance Magdalena should ride each day. Hence, it cannot be a negative value. What about 0? 0 ? = 4000/d A fraction with a non-zero numerator can never be equal to 0. Therefore, y ≠ 0. We conclude that y can only have positive values. y > 0

Asymptotes

We can see that the function we wrote is a vertical stretch of the funcion y = 1x by a factor of 4000. y = 1/x [factor of4000]Stretch by a y = 4000/x Since stretching and shrinking do not change the asymptotes, the asymptotes of y = 4000x are the same as those of y = 1x. Asymptotes x = 0 and y = 0

Table of Values

Let's now make a table of values, which will help us to graph the function. We can choose some positive values of x, and substituting them into the function, calculate the values of the function y.

x	4000/x	y
25	4000/25	160
50	4000/50	80
75	4000/75	≈ 53
100	4000/100	40
125	4000/125	32

The pairs (x,y) are ordered pairs and points on the graph. Let's plot these points and draw a smooth curve through them.

This graph matches with option B.

There are 365 days in a year. Let's substitute 365 for x into the function. This way we will calculate how many kilometers Magdalena should ride every day.

Magdalena should ride about 11 kilometers per day.

Consider the following functions.

f (x) = \frac{1}{x} and g (x) = ∣ ∣ ∣ ∣ ∣ \frac{1}{x} ∣ ∣ ∣ ∣ ∣

Draw the graph of each function to determine which statement(s) is correct.

I . II . III . IV . The point (1, 1) is the only point of intersection . The graph of g (x) is a reflection of the graph of f (x) across the x - axis . For x > 0, the graphs of the functions are the same . The graphs have infinitely many points of intersection .

What differs the functions is the absolute value in the second equation. f(x) = 1/x and g(x) = | 1/x | Let's start by graphing the first reciprocal function. We know the characteristics of this graph, which are shown in the table.

	f(x) = 1/x
Domain	All real numbers except 0
Range	All real numbers except 0
Vertical Asymptote	x=0
Horizontal Asymptote	y=0

Let's draw it!

To graph the second function g(x), let's recall that the absolute value turns every negative value into a positive value. Therefore, to graph the second function we have to reflect the part of f(x)= 1x where f(x) is negative across the x-axis.

In the first quadrant, the graphs coincide and thus have an infinite number of intersections. Therefore, among the given statements, III and IV are correct.