Understanding Graphing of Reciprocal Functions: Asymptotes, Domain and Range

Asymptotes

Domain

Range

Function I

Function II

Function III

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	$y = \frac{a}{x - h} + k$	$Domain : Range : Asymptotes : R - {h} R - {k} x = h and y = k$

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

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

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

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

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

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

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}

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

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

Transformations of

y = \frac{1}{x}

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

Translation up

k

units,

k > 0

Translation down

k

units,

k < 0

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

Translation to the right

h

units,

h > 0

Translation to the left

h

units,

h < 0

Transformations of

y = \frac{1}{x}

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

Translation up

k

units,

k > 0

Translation down

k

units,

k < 0

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

Translation to the right

h

units,

h > 0

Translation to the left

h

units,

h < 0

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

Transformations of

y = \frac{1}{x}

\underline{Vertical Stretch or Shrink} y = \frac{a}{x}

Vertical stretch,

a > 1

Vertical shrink,

0 < a < 1

\underline{Horizontal Stretch or Shrink} y = \frac{1}{b x}

Horizontal stretch,

0 < b < 1

Horizontal shrink,

b > 1

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

Transformations of

y = \frac{1}{x}

Reflections

In the x - axis y = - (\frac{1}{x})

In the y - axis y = \frac{1}{- x}

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

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}

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

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

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

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.

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Catch-Up and Review

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

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

Reciprocal Functions

Extra

Correcting the Error

Hint

Solution

Identifying Characteristics of Reciprocal Functions

Graphing a Reciprocal Function

Modeling the Average Cost With a Reciprocal Function

Answer

Hint

Solution

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

Translation of Reciprocal Functions

Finding Domain and Range After Translation

Answer

Hint

Solution

Effects of the Value of $a$

Stretch and Shrink of Reciprocal Functions

Reflection of Reciprocal Functions

Comparing Graphs of Reciprocal Functions

Answer

Hint

Solution

Mark's Functions

Paulina's Functions

Tiffaniqua's Functions

Extra

Graphing a Reciprocal Function Using Transformations

Answer

Hint

Solution

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