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| 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.
Which graph is the graph of the given function?
Let's start by recalling the general equation of a reciprocal function. g(x)=a/x- h+ k The asymptotes of a function written in this form are the lines x= h and y= k. We can now identify the values of a, h, and k for the given function. g(x)=3/x- 0+ 4 We can see that a= 3, h= 0, and k= 4. Therefore, the asymptotes of our function are the lines x= 0 and y= 4. With this in mind, we will make a table of values. Be sure to include different values of x, some of which are less and and some are greater than 0.
x | 3/x+4 | g(x)=3/x+4 |
---|---|---|
- 3 | 3/- 3+4 | 3 |
- 2 | 3/- 2+4 | 2.5 |
- 1 | 3/- 1+4 | 1 |
1 | 3/1+4 | 7 |
2 | 3/2+4 | 5.5 |
3 | 3/3+4 | 5 |
We can now plot and connect the obtained points. Remember that the asymptotes are x=0 and y=4, and that the graph will have two branches.
This graph corresponds to option A.
Let's recall the general equation of a rational function one more time and identify the characteristics of our function.
g(x)=a/x-h+k | g(x)=3/x-0+4 | |
---|---|---|
Domain | R - { h } | R - { 0 } |
Range | R - { k } | R - { 4 } |
Vertical Asymptote | x = h | x = 0 |
Horizontal Asymptote | y = k | y = 4 |
Therefore, among the given statements , only statement IV is correct.
Consider the graph of a reciprocal function.
We are given the graph of a reciprocal function. We know that reciprocal functions have one horizontal and one vertical asymptote. Let's identify the asymptotes!
Looking at the given graph, we see that x = - 5 is the vertical asymptote and y= - 6 is the horizontal asymptote.
Recall the general equation of a reciprocal function.
y = a/x- h + k
The asymptotes are the lines x= h and y= h. Previously, we have found the asymptotes of the given graph, x = - 5 and y= - 6. In other words, the h- and k-values for the equation of the given function are - 5 and 6, respectively. Let's use them to write the equation.
y = a/x- ( - 5) + ( - 6)
To find the value of a, we need to identify a point that lies on the graph.
Let's substitute its coordinates (- 4,- 4) into the equation and solve it for a.
Finally, we can write the equation with the known parameters. y = 2/x + 5 -6
We will start by labeling both reciprocal functions to differentiate between them. y_1 =1/x y_2 = 1/2x Note that the function y_1 = 1x is the parent reciprocal function. We can rewrite the function y_2 to see how this one relates to y_1.
As we can see y_2 is equal to y_1 scaled by a factor of 12. In other words, the values of y_2 are those of y_1 but reduced by one-half. This means that y_2 is a vertical shrink of y_1 by a factor of 12. Therefore, the answer is option B.
Write an equation for the translation of y=x5 that has the following asymptotes.
Let's start by recalling the general form of a reciprocal function. y=a/x- h+ k The vertical asymptote of the function is the line x= h, and its horizontal asymptote is y= k. Let's now write an equation for the translation of y= 5x that has asymptotes x= 4 and y= - 3. Note that in our function we have that a= 5. y=5/x- 4+( - 3) ⇕ y=5/x-4-3 Let's see what the graph of this function looks like!
Consider the general form of a reciprocal function one more time. y=a/x-h+k We know that the asymptotes occur at x = h and y=k. With this in mind, we can write an equation for the translation of y= 5x that has asymptotes x = - 5 and y= 15. Note that in our function we have that a=5. y=5/x-( - 5)+ 15 ⇕ y=5/x+5+15 Let's see what the graph of this function looks like!
We want to rewrite the given function so that it is in the form g(x)= ax-h+k. To do so, we will start by dividing the numerator of the function by the denominator.
The quotient is 7 with a remainder of - 3. Let's rewrite the function using this information. h(x)=7x + 11/x+2 ⇔ h(x)=- 3/x + 2 + 7
Let's start by analyzing possible transformations of the function f(x)= 3x.
Transformations of f(x)=3/x, x≠ 0 | |
---|---|
Reflection | In thex-axis [0.4em] y= - ( 3/x ) |
Horizontal Translation [0.4em] y=3/x- h | Translation right h units, h>0 |
Translation left h units, h<0 | |
Vertical Translation [0.4em] y=3/x+ k | Translation up k units, k>0 |
Translation down k units, k < 0 |
Let's now consider the equation found in Part A. h(x)&=- 3/x+2+7 &⇕ h(x)&=- 3/x-( -2 )+ 7 We can see that a= - 3, h= - 2, and k= 7. Therefore, the graph of this function can be described as a transformation of the graph of y = 3x as follows.
Transformation | Function Rule |
---|---|
1. Reflection in the x-axis | y=3/x → y= - 3/x |
2. Horizontal translation 2 units left | y=- 3/x → y=- 3/x-( - 2) |
3. Vertical translation 7 units up | y=- 3/x+2 → y=- 3/x+2 +7 |
We will use this information to draw the graph.