Sign In
| 16 Theory slides |
| 10 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
Identify the asymptotes, domain, and range of each function.
Asymptotes | Domain | Range | |
---|---|---|---|
Function I | |||
Function II | |||
Function III |
f(x)=x1,x=0
The graph of the function f(x)=x1 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.
The graph of y=x1 can be used to graph other reciprocal functions. This can be done by applying different transformations.
Name | Equation | Characteristics |
---|---|---|
Parent Reciprocal Function | y=x1 | Domain:Range:Asymptotes: R−{0} R−{0} x- and y-axes
|
Inverse Variation Functions | y=xa | |
General Form of Reciprocal Functions | y=x−ha+k | Domain:Range:Asymptotes: R−{h} R−{k} x=h and y=k
|
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 a1, while the notation f -1(x) is commonly used to refer to the inverse of a function.
Function, f | Reciprocal, f1 | Inverse, f-1 |
---|---|---|
f(x)=2x+10 | f(x)1=2x+101 | f -1(x)=21(x−10) |
f(x)=x3−5 | f(x)1=x3−51 | f -1(x)=3x+5 |
Mark wants to identify the asymptotes, domain, and range of the reciprocal function shown as follows.
Start by identifying any x-values for which f(x) is undefined.
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.
For the indicated reciprocal function, identify the asymptotes or the value that makes the function undefined.
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 | x−3-1+2 | f(x)=x−3-1+2 |
---|---|---|
1 | 1−3-1+2 | 2.5 |
2 | 2−3-1+2 | 3 |
2.5 | 2.5−3-1+2 | 4 |
3.5 | 3.5−3-1+2 | 0 |
4 | 4−3-1+2 | 1 |
5 | 5−3-1+2 | 1.5 |
The points (x,f(x)) can be added to the coordinate plane to begin to see the behavior of f.
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.
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.
Verbal Expression | Algebraic Expression |
---|---|
Initial cost | 360 |
Cost for n number of students | 15n |
Total cost | 15n+360 |
Average cost A | A=n15n+360 |
Write as a sum of fractions
Simplify quotient
Commutative Property of Addition
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 | n360+15 | A(n)=n360+15 |
---|---|---|
-40 | -40360+15 | 6 |
-30 | -30360+15 | 3 |
-20 | -20360+15 | -3 |
-10 | -10360+15 | -21 |
10 | 10360+15 | 51 |
20 | 20360+15 | 33 |
30 | 30360+15 | 27 |
40 | 40360+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.
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=x1 | |
---|---|
Vertical Translationy=x1+k
|
Translation up k units, k>0 |
Translation down k units, k<0 | |
Horizontal Translationy=x−h1
|
Translation to the right h units, h>0 |
Translation to the left h units, h<0 |
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=x1 change after the translation of its graph by 4 units down and 3 units to the left?
Domain: All real numbers except -3
Range: All real numbers except -4
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.
To determine the domain and the range of the function after translating it, start by considering some possible transformations.
Transformations of y=x1 | |
---|---|
Vertical Translationy=x1+k
|
Translation up k units, k>0 |
Translation down k units, k<0 | |
Horizontal Translationy=x−h1
|
Translation to the right h units, h>0 |
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.
The second transformation is a horizontal translation 3 units left.
Finally, look at the graph of the given function and its asymptotes alone.
The graph of the reciprocal function f(x)=xa is shown in the coordinate plane. Observe how the graph changes as the value of a changes.
Transformations of y=x1 | |
---|---|
Vertical Stretch or Shrinky=xa
|
Vertical stretch, a>1 |
Vertical shrink, 0<a<1 | |
Horizontal Stretch or Shrinky=bx1
|
Horizontal stretch, 0<b<1 |
Horizontal shrink, b>1 |
The table below summarizes the different types of reflections.
Transformations of y=x1 | |
---|---|
Reflections | In the x-axisy=-(x1)
|
In the y-axisy=-x1
|
Notice that -(x1)=-x1. Reflecting the parent reciprocal function in the x-axis and in the y-axis will produce the same graph.
Pair of Functions | |||
---|---|---|---|
Mark | y=x1 | y=-3(x1) | |
Paulina | y=x1 | y=x1−6 | |
Tiffaniqua | y=x1 | y=x+51 |
Help them identify the transformations, asymptotes, domain, and range of the reciprocal functions.
Domain | Range | Vertical Asymptote | Horizontal Asymptote | |
---|---|---|---|---|
y=x1 | R−{0} | R−{0} | x=0 | y=0 |
y=-x3 | R−{0} | R−{0} | x=0 | y=0 |
y=x1−6 | R−{0} | R−{-6} | x=0 | y=-6 |
y=x+51 | R−{-5} | R−{0} | x=-5 | y=0 |
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.
Consider two different composition of transformation of y=x1.
f(x)=x−25+3 | |
---|---|
Domain | All real numbers except 2 |
Range | All real numbers except 3 |
Vertical Asymptote | x=2 |
Horizontal Asymptote | y=3 |
x3x=3
Multiply term by divisor
Subtract down
f(x)=x−25+3 | |
---|---|
Domain | All real numbers except 2 |
Range | All real numbers except 3 |
Vertical Asymptote | x=2 |
Horizontal Asymptote | y=3 |
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 | x−25+3 | f(x)=x−25+3 |
---|---|---|
-3 | -3−25+3 | 2 |
-1 | -1−25+3 | ≈1.33 |
1 | 1−25+3 | -2 |
3 | 3−25+3 | 8 |
5 | 5−25+3 | ≈4.67 |
7 | 7−25+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.
Transformations of f(x)=x1,x=0 | |
---|---|
Vertical Translationy=x1+k
|
Translation up k units, k>0 |
Translation down k units, k<0 | |
Horizontal Translationy=x−h1
|
Translation right h units, h>0 |
Translation left h units, h<0 | |
Vertical Stretch or Shrinky=xa
|
Vertical stretch, a>1 |
Vertical shrink, 0<a<1 |
These transformations will be applied one at a time. Start by translating the parent function, f(x)=x1, 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.
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.
What is the graph of the function?
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
Which graph is the graph of the function?
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 |
Translation left h units, h<0 | |
Vertical Translation [0.4em] y=1/x+ k | Translation up k units, k>0 |
Translation down k units, k < 0 | |
Vertical Stretch or Shrink [0.4em] y=a/x | Vertical stretch, a>1 |
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.
Which graph is the graph of the function?
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.
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
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
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
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.
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.