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| 14 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.
A rational function is a function that contains a rational expression. Any function that can be written as the quotient of two polynomial functions p(x) and q(x) is a rational function.
f(x)=q(x)p(x),q(x)=0
The graph of a rational function can be a smooth continuous curve, or it can have jumps, breaks, or holes. By looking at the graph of a function, functions can be categorized into two groups: continuous or discontinuous.
A point of discontinuity of a function is a point with the x-coordinate that makes the function value undefined. A point of discontinuity can also be considered as an excluded value of the function rule.
Write as a power
a2−b2=(a+b)(a−b)
Simplify quotient
holein the graph.
When a function cannot be redefined so that the point of discontinuity becomes a valid input, it is called a non-removable discontinuity. Consider, for example, y=x+11. Since x=-1 is not in the domain of the function, there is a point of discontinuity at x=-1.
Total Annual Income | Population |
---|---|
I(t)=0.02t+15000t+100000 | P(t)=0.36t+10800 |
Substitute expressions
ba/c=b⋅ca
Multiply parentheses
Add terms
C(t) is undefined | |
---|---|
0.02t+1=0 | 0.36t+10800=0 |
t=-50 | t=-30000 |
On the way to the pencil factory, LaShay noticed that road maintenance work is being carried out to fill a hole
on the road.
Factor out 3x
Factor out x2
Factor out (x+1)
Notice that it is also a removable discontinuity because x+1 is the common factor in the numerator and denominator.
This graph also has a removable discontinuity at x=-1 because x+1 is the common factor in the numerator and denominator of the rational function.
An asymptote of a graph is an imaginary line that the graph gets close to as x goes to plus or minus infinity or a particular number. For example, the graph of the rational function f(x)=x1 has two asymptotes — the x-axis and the y-axis.
Analyzing the diagram, the following can be observed.
To identify the vertical asymptotes, the function should be in its simplest form. That is, if p(x) and q(x) have common factors, the function should be simplified first. The vertical asymptotes will occur at the zeros of the denominator. Check out the following examples.
The degrees of polynomials in the numerator and denominator of a rational function will determine if the graph has a horizontal asymptote.
f(x)=q(x)p(x)=bxn+…axm+… | |
---|---|
Horizontal Asymptote | If m>n, there is no horizontal asymptote. |
If m<n, the horizontal asymptote is y=0. | |
If m=n, the horizontal asymptote is y=ba. |
Vertical Asymptotes: t=-2 and t=-1
Horizontal Asymptote: y=0
Factor the numerator. What does the Factor Theorem state?
Factor Theorem |
The expression x−a is a factor of a polynomial if and only if the value a is a zero of the related polynomial function. |
t | t3−t2−10t−8 | p(t)=t3−t2−10t−8 |
---|---|---|
1 | 13−(1)2−10(1)−8 | -18 |
-1 | -13−(-1)2−10(-1)−8 | 0 |
2 | 23−(2)2−10(2)−8 | -24 |
-2 | -23−(-2)2−10(-2)−8 | 0 |
4 | 43−6(4)2−7(4)+60 | 0 |
t3−t2−10t−8=(t+1)(t+2)(t−4)
Cross out common factors
Cancel out common factors
y=bxnaxm | Asymptote |
---|---|
m<n | y=0 |
m>n | none |
m=n | y=ba |
An asymptote can be neither vertical nor horizontal. It can be a slanted line. In such cases, it is called an oblique asymptote or slant asymptote.
4x2x3=4x
Multiply term by divisor
Subtract down
The students spent about 3 hours at the factory. At the end of this period, LaShay's teacher made a survey about the trip and drew a graph relating the students' interest in the trip and the time elapsed at the factory. The greater the S(t) value, the greater the interest of the students.
Vertical Asymptotes: t=-1
Oblique Asymptote: O(t)=-2t+2
Start by subtracting the rational expressions on the right-hand side of the function.
ba=b⋅2(t+1)a⋅2(t+1)
Multiply
Subtract fractions
Distribute 2t
Commutative Property of Addition
2t2-t3=-2t
Multiply term by divisor
Subtract down
2t24t2=2
Multiply term by divisor
Subtract down
Cross out common factors
Cancel out common factors
holeat x=3. Also, the graph has a vertical asymptote at x=4 because 4 is not included in the domain.
y=bxn+…axm+… | Asymptote | Asymptote Type |
---|---|---|
m<n | y=0 | Horizontal |
m>n | None | None |
m=n | y=ba | Horizontal |
m−n=1 | y= the quotient of the polynomials with no remainder | Oblique |
xx2=x
Multiply term by divisor
Subtract down
x9x=9
Multiply term by divisor
Subtract down
The given function has one vertical asymptote. Therefore, its graph will consist of two parts, one to the left and the other to the right of the vertical asymptote. These parts may lie above or below the oblique asymptote.
f(x)=0
LHS⋅(x−4)=RHS⋅(x−4)
Use the Zero Product Property
(I): LHS+1=RHS+1
(II): LHS−6=RHS−6
Some additional points are needed to get a good sense of the shape of the graph. Make sure to only use values included in the domain of the function.
x | x−4(x−1)(x+6) | f(x)=x−4(x−1)(x+6) |
---|---|---|
-11 | -11−4(-11−1)(-11+6) | -4 |
2 | 2−4(2−1)(2+6) | -4 |
6 | 6−4(6−1)(6+6) | 30 |
9 | 9−4(9−1)(9+6) | 24 |
19 | 19−4(19−1)(19+6) | 30 |
It is almost done. Plot the points and imagine how the shape of the graph should look!
As can be seen, one part of the graph will lie to the left of the vertical asymptote and below the oblique asymptote. The other part of the graph will be to the right of the vertical asymptote and above the oblique asymptote.
The graph can now be drawn by connecting the points with a smooth curve. It must approach the asymptotes. Do not forget to plot the hole at x=3!
The factory sells 6 pencils in a rectangular box. The dimensions of the box are shown in the diagram.
Multiply
Distribute 4
Distribute 2
Multiply parentheses
Add and subtract terms
Substitute expressions
Write as a difference
Factor out (x+3)
holesin the graph. Recall that if the real number a is not included in the domain, there is a vertical asymptote at x=a. In this case, there are two vertical asymptotes, one at x=-3 and the other at x=2.
y=bxn+…axm+… | Asymptote |
---|---|
m<n | y=0 |
m>n | none |
m=n | y=ba |
m−n=1 | y= the quotient of the polynomials with no remainder |
E(x)=0
x=0
Calculate power
Zero Property of Multiplication
Identity Property of Addition
ba=b/(-2)a/(-2)
Now, make a table of values to graph the given function.
x | x2+x−6x2+4x−2 | E(x)=x2+x−6x2+4x−2 |
---|---|---|
-6 | (-6)2+(-6)−6(-6)2+4(-6)−2 | ≈0.417 |
-3.5 | (-3.5)2+(-3.5)−6(-3.5)2+4(-3.5)−2 | ≈-1.364 |
-2 | (-2)2+(-2)−6(-2)2+4(-2)−2 | 1.5 |
-1 | (-1)2+(-1)−6(-1)2+4(-1)−2 | ≈0.833 |
1 | (1)2+(1)−6(1)2+4(1)−2 | -0.75 |
3 | (3)2+(3)−6(3)2+4(3)−2 | ≈3.167 |
4 | (4)2+(4)−6(4)2+4(4)−2 | ≈2.143 |
5 | (5)2+(5)−6(5)2+4(5)−2 | ≈1.792 |
Finally, the graph of the function can be drawn by plotting the found points and connecting them with a smooth curve. Recall that a rational function can cross the horizontal asymptote but cannot cross the vertical asymptotes. In this case, the horizontal asymptote is crossed.
Recall the Factor Theorem.
p(x) has a factor (x−a) if and only if p(a)=0. |
Substitute values
Identity Property of Multiplication
Calculate power
-a(-b)=a⋅b
Add terms
Split into factors
a⋅b=a⋅b
Calculate root
Factor out 2
Simplify quotient
Interpretation: In about 150 years, there will be no water left in the pond.
Equation: y=-15x+10
Now, push GRAPH to draw the function.
It appears that only one part of the graph is shown, but it is not possible to know that until the size of the viewing window is changed.
The graph does not seem to have any zeros. However, note that as x increases in the first quadrant, the graph gets closer to the x-axis and could cross it. To check this, zoom in on this part by changing the window settings. Push WINDOW and change the settings as shown below. Then push GRAPH once more to draw the equation with these new settings.
Now it can be seen that the graph intersects the x-axis and there is a zero. To find it, use the zero option in the calculator. It can be found by pressing 2ND and then CALC.
After selecting the zero
option, choose left and right boundaries for the zero. Finally, the calculator asks for a guess where the zero might be. After that, it will calculate the exact point.
The function intersects the x-axis at about (150,0). Since x represents the numbers of years that have passed since pumping starts, the x-intercept means that in about 150 years, there will be no water left in the pond.
y=bxn+…axm+… | Asymptote | Asymptote Type |
---|---|---|
m<n | y=0 | Horizontal |
m>n | None | None |
m=n | y=ba | Horizontal |
m−n=1 | y= the quotient of the polynomials with no remainder | Oblique |
15x2-x3=-15x
Multiply term by divisor
Subtract down
15x2150x2=10
Multiply term by divisor
Subtract down
y=bxn+…axm+… | Asymptote | Asymptote Type |
---|---|---|
m<n | y=0 | Horizontal |
m>n | None | None |
m=n | y=ba | Horizontal |
m−n=1 | y= the quotient of the polynomials with no remainder | Oblique |
x0.2x2=0.2x
Multiply term by divisor
Subtract down
x5x=5
Multiply term by divisor
Subtract down
A(x)=0
LHS⋅x=RHS⋅x
Use the Quadratic Formula: a=0.2,b=5,c=500
Calculate power and product
Subtract term
A table of values will be used to get a rough idea of the shape of the graph.
x | x0.2x2+5x+500 | A(x)=x0.2x2+5x+500 |
---|---|---|
-75 | -750.2(-75)2+5(-75)+500 | ≈-16.7 |
-25 | -250.2(-25)2+5(-25)+500 | -20 |
-10 | -100.2(-10)2+5(-10)+500 | -47 |
10 | 100.2(10)2+5(10)+500 | 57 |
25 | 250.2(25)2+5(25)+500 | 30 |
75 | 750.2(75)2+5(75)+500 | ≈26.7 |
Plot the points and imagine how the shape of the graph should look!
Finally, draw the graph by connecting the points. It must approach, but not cross, the asymptotes.
As shown, the range does not contain all real numbers. It appears that the minimum point of the part of the graph in the first quadrant is (50,25). Notice also that the graph is symmetric about the point of intersection of the asymptotes, (0,5). Using this fact, the maximum point of the other part can be identified.
A graphing calculator can be used to check the answers. The graph of the function will be drawn first. Press the Y= button and type the equation in the first row.
By pushing GRAPH, the calculator will draw the equation. For this function to be visible on the screen, re-size the standard window by pushing the WINDOW button. Change the settings to a more appropriate size and then push GRAPH.
Next, the local maximum and local minimum will be found. To do so, push 2nd, then TRACE, and choose the maximum
option.
When using the maximum
feature, choose the left and right bounds. The calculator will then provide a best guess as to where the maximum might be.
The maximum point of the graph in the third quadrant is (-50,-15). To find the minimum, repeat the same process but choose the minimum
option.
The minimum point of the part in the first quadrant is (50,25). As a result, the range is (∞,-15]∪[25,∞), and the domain is all real numbers except 0.
Let's start by reviewing what a rational function is and its main characteristics.
Rational Function | |
---|---|
Form | f(x)=P(x)/Q(x), where P(x) and Q(x) are polynomial functions |
Domain | All real numbers for which Q(x) ≠ 0 |
Point of Discontinuity | x=a if Q(a)=0 |
In a rational function, vertical asymptotes occur at the points of discontinuity. Also, if Q(x) has no real zeros, we can guarantee that f(x) will not have a vertical asymptote. Additionally, if Q(x) has a real zero that is a removable discontinuity, it will not have a vertical asymptote either. Let's review what these are and when they happen.
f(x) = P(x)/Q(x) | ||
---|---|---|
Assumptions | Let (x-a)^m and p(x) be factors of P(x) and (x-a)^n and q(x) factors of Q(x) with m ≥ n. Additionally, p(x) and q(x) do not have any common factors. | |
Factored Form | f(x) = (x-a)^m p(x)/(x-a)^n q(x), where m ≥ n | |
Removable Discontinuity | x = a | |
Example | g(x) = (x-3)^1(x+3)/(x-3)^1(x+5) ⇓ Removable Discontinuity x = 3 because 1≥ 1 |
This is because we can be simplify the quotient by canceling out common factors, which returns an equivalent expression that is not undetermined at x=a. However, the function would still have the gap in the value for which x =a. With this information, we will examine the given functions one by one.
We are given the following rational function. y = x- 9x- 9 In this case P( 9)=0 and Q( 9)=0. Since they have a common zero, this function has a removable discontinuity at x= 9. We can confirm this by looking at its graph.
We can cancel out common factors to redefine the given function. y = x-9/x-9 ⇔ y = 1 , x ≠ 9 As we can see, we could make the function continuous by defining y(9) = 1. This means that the function has no vertical asymptotes.
Now, we will analyze a function where m>n. y = (x- 9)^4/(x- 9)^2 In this case, both P( 9)=0 and Q( 9)=0. However, the factor causing this zero has an exponent with a greater value in the numerator. As we can see from its graph, there is a removable discontinuity at x=9 and the graph has no vertical asymptotes.
Again, we can write the function by removing common factors. y = (x-9)^2(x-9)^2/(x-9)^2 [0.5em] ⇕ [0.5em] y = (x-9)^2, x ≠ 9 As we can see, we can redefined y(9) = 0 to make the function continuous. Therefore, the function has no vertical asymptotes.
Now, consider the function given in option C. y = 9x/x^2+9 We are given a function where Q(x) has no real zeros. Its zeros are x = ± 3i. Since there are no x-values that make the denominator zero, it guarantees that the function will not have a vertical asymptote.
Finally, we will analyze the last function. y = (x-9)^2/(x-9)^4 In this function, n > m and both P( 9)=0 and Q( 9)=0. However, the factor causing this zero has a lower exponent in the numerator. Therefore, there will be a non-removable discontinuity at x=9.
Although the function can be simplified a bit by removing common factors, the obtained function is not continuous at x=9. y = (x-9)^2/(x-9)^2(x-9)^2 [0.5em] ⇕ [0.5em] y = 1/(x-9)^2, x ≠ 9 The function has vertical asymptote at x=9. Therefore, the answer is D.
Following the ideas discussed above, we can conclude that a rational function will have no vertical asymptotes when the denominator has no real zeros. Or, if it has them, whenever the factors causing them can be divided out with a similar factor from the numerator polynomial.