1. Functions and Modeling
FSA- Algebra 2
Functions and Modeling

2. Transformation
MAFS.912.F-BF.2.3
Identify the effect on the graph (transformations) of replacing 𝑓(𝑥) by 𝑓(𝑥) + 𝑘, 𝑓(𝑘𝑥) and 𝑓(𝑥 + 𝑘) for given values of 𝑘.

3. Given the function 𝑓 𝑥 = −2 𝑥− , describe how each of the following will shift the location of the vertex. Use complete sentences.
𝑓 𝑥+4
𝑓 𝑥 +4
𝑓 4𝑥
4∙𝑓(𝑥)
Example
Shift the vertex to the left 4 units. It would go from (5, 4) to (1, 4)
Shift the vertex up 4 units. It would go from (5, 4) to (5, 8)
Transform the x value of the vertex to 𝟓 𝟒 ,𝟒
It would stretch the parabola making it more narrow.

4. Example The function 𝑓 is defined as 𝑓 𝑥 = 𝑥 2 −4𝑥. Part A
Write an expression that defines 𝑓(𝑥+5).
𝒇 𝒙+𝟓 = 𝒙 𝟐 −𝟒𝒙
= 𝒙+𝟓 𝟐 − 𝟒 𝒙+𝟓
=𝒙 𝟐 + 𝟏𝟎𝒙+𝟐𝟓−𝟒𝒙−𝟐𝟎
=𝒙 𝟐 + 𝟏𝟎𝒙+𝟐𝟓−𝟒𝒙−𝟐𝟎
=𝒙 𝟐 +𝟔𝒙+𝟓
Part B Describe the transformation that maps the graph of 𝑓(𝑥) to 𝑓(𝑥+5). Justify your answer algebraically or by using key features of the graphs.
The transformation is a translation 5 units to the left. I can show this algebraically by looking at the vertex form of each function.
Completing the square
𝒇(𝒙)= 𝒙 𝟐 −𝟒𝒙
𝟒= 𝒙 𝟐 −𝟒𝒙+𝟒
𝒇 𝒙+𝟓 = 𝒙−𝟐 𝟐 −𝟒
= 𝒙+𝟓−𝟐 𝟐 −𝟒
𝟒= 𝒙−𝟐 𝟐
= 𝒙+𝟑 𝟐 −𝟒
𝒚= 𝒙−𝟐 𝟐 −𝟒
The two functions have the same leading coefficient. The vertex of 𝒇(𝒙) is (𝟐,−𝟒) and the vertex of 𝒇(𝒙+𝟓) is (−𝟑,−𝟒). Therefore the translation is 5 units to the left.

5. Inverse Function
MAFS.912.F-BF.2.4 
An inverse function is a function that "reverses" another function.

6. Determine the inverse function 𝑓 −1 (𝑥) of 𝑓 𝑥 = 40𝑥 10−𝑥
Determine the inverse function 𝑓 −1 (𝑥) of 𝑓 𝑥 = 40𝑥 10−𝑥 . Show your work.
Example
To find the inverse you must first replace the 𝒇(𝒙) with 𝒚.
Solve for the 𝒚.
𝟏𝟎−𝒚 𝒙= 𝟒𝟎𝒚 𝟏𝟎−𝒚 𝟏𝟎−𝒚
𝒇 𝒙 = 𝟒𝟎𝒙 𝟏𝟎−𝒙
𝒙≠𝟎
𝟏𝟎−𝒚 𝒙=𝟒𝟎𝒚
𝒚= 𝟒𝟎𝒙 𝟏𝟎−𝒙
𝟏𝟎𝒙−𝒙𝒚=𝟒𝟎𝒚
+ 𝒙𝒚
+ 𝒙𝒚
𝟏𝟎𝒙=𝟒𝟎𝒚+ 𝒙𝒚
To find the inverse function, interchange the 𝒚 and the 𝒙.
𝟏𝟎𝒙= 𝟒𝟎+𝒙 𝒚
𝟏𝟎𝒙 𝟒𝟎+𝒙 =𝒚
𝒙= 𝟒𝟎𝒚 𝟏𝟎−𝒚
𝒏𝒐𝒘 𝒚≠𝟏𝟎
𝒚= 𝟏𝟎𝒙 𝟒𝟎+𝒙 is now the inverse with constrains 𝒙≠𝟒𝟎 and 𝒚≠𝟏𝟎

7. Given 𝑓 𝑥 = 𝑥 + 3 5 , solve for 𝑓 −1 (7).
Example
A. 2
B. 10
C. 32
D. 38
Solve for the 𝒚.
𝟓∙𝒙= 𝒚+ 𝟑 𝟓 ∙𝟓
𝟓𝒙=𝒚+ 𝟑
− 𝟑
− 𝟑
𝟓𝒙−𝟑=𝒚
To find the inverse you must first replace the 𝒇(𝒙) with 𝒚.
Solve for 𝒇 −𝟏 (𝟕)
𝒇 𝒙 = 𝑥 + 3 5
𝒇 −𝟏 𝟕 =𝟓𝒙−𝟑
𝒚= 𝒙 + 𝟑 𝟓
=𝟓(𝟕)−𝟑
=𝟑𝟓−𝟑
To find the inverse function, interchange the 𝒚 and the 𝒙.
=𝟑𝟐
𝒙= 𝒚+ 𝟑 𝟓

8. Graph Features
MAFS.912.F-IF.2.4 
For a function that models a relationship between two quantities, Interpret key features of graphs and tables.

9. Example
Use the following table to find the average rate of change for the given function from 𝑥=−1 to 𝑥=11. x y -1 4 3 6 5 7 11 10
A. -½
C. 6
D. -12
Slope formula:
𝑚= 𝒚 𝟐 − 𝒚 𝟏 𝒙 𝟐 − 𝒙 𝟏
= 𝟏𝟎−𝟒 𝟏𝟏−(−𝟏)
= 𝟔 𝟏𝟐
The average rate of change implies the slope between the two given points. The two points are:
= 𝟏 𝟐
𝒙 𝟏 , 𝒚 𝟏 = −𝟏,𝟒
𝒙 𝟐 , 𝒚 𝟐 = 𝟏𝟏,𝟏𝟎

10. Example
Marisa is collecting stamps. Her collection starts off right as her grandfather has given her 3 stamps. She is determined to collect 3 stamps per month to build up her collection. Which of the following graphs represents Marisa’s stamp collection?
𝒚=𝟑𝒙+𝟑
𝒚=𝟏𝒙+𝟑

11. Example 𝒎= 𝒚 𝟐 − 𝒚 𝟏 𝒙 𝟐 − 𝒙 𝟏 = −𝟑−𝟏 𝟏−(−𝟏) = −𝟒 𝟐 =−𝟐
Use the graph below to answer the following question:  What is the average rate of change from 𝑥 = – 1 to 𝑥 = 1 ?
To solve for rate of change, you must first identify the points at which x = -1 and x = 1.
A. 2 -2
C. 0
D. -1
(−𝟏, 𝟏) and (𝟏, −𝟑)
Use these points to find the rate of change or slope by using the formula:
𝒎= 𝒚 𝟐 − 𝒚 𝟏 𝒙 𝟐 − 𝒙 𝟏
= −𝟑−𝟏 𝟏−(−𝟏)
= −𝟒 𝟐
=−𝟐

12. Example 𝟐𝒙 𝟐 + 𝟓𝒙 − 𝟑 𝒙+𝟒 𝟐𝒙 𝟑 +𝟏𝟑 𝒙 𝟐 +𝟏𝟕𝒙−𝟏𝟐 − 𝟐𝒙 𝟑 − 𝟖𝒙 𝟐 𝟐𝒙 𝟑
The polynomial 𝑝(𝑥) = 2 𝑥 𝑥 𝑥 − 12 has (𝑥 + 4) as a factor. Factor the polynomial into three linear terms. Describe the steps you would use to sketch the graph of the function defined by this polynomial. Identify all intercepts and describe the end behavior of the graph.
The graph passes through the 𝒙−𝒊𝒏𝒕𝒆𝒓𝒄𝒆𝒑𝒕𝒔
𝟐𝒙 𝟐
+ 𝟓𝒙
− 𝟑
𝒙= 𝟏 𝟐
𝒙=−𝟑
𝒙=−𝟒
𝒙+𝟒
𝟐𝒙 𝟑 +𝟏𝟑 𝒙 𝟐 +𝟏𝟕𝒙−𝟏𝟐
− 𝟐𝒙 𝟑 − 𝟖𝒙 𝟐
𝟐𝒙 𝟑
+ 𝟖𝒙 𝟐
The 𝒚−𝒊𝒏𝒕𝒆𝒓𝒄𝒆𝒑𝒕𝒔 is:
𝟓𝒙 𝟐
+ 𝟏𝟕𝒙
𝒚=𝟐(𝟎) 𝟑 + 𝟏𝟑(𝟎) 𝟐 +𝟏𝟕(𝟎)−𝟏𝟐
− 𝟓𝒙 𝟑 −𝟐𝟎𝒙
𝟓𝒙 𝟐
+ 𝟐𝟎𝒙
𝒚=−𝟏𝟐
−𝟑𝒙
− 𝟏𝟐
When polynomial has an odd degree, the ends of the graph go in opposite directions. As the coefficient of the 𝒙 𝟑 term is positive, that means as 𝒙 goes to ∞, the graph will go to −∞.
+ 𝟑𝒙+𝟏𝟐
−𝟑𝒙
− 𝟏𝟐 𝟎
𝟐 𝒙 𝟐 +𝟓𝒙−𝟑 𝟔𝒙 𝟐𝒙 −𝟏 −𝒙
Similarly , as x goes to −∞, the graph will go to −∞, the graph will go to negative infinity. 𝒙 𝟑
𝟔𝒙−𝒙=𝟓𝒙
𝟐𝒙−𝟏 𝒙+𝟑 (𝒙+𝟒)

13. Equivalent Expressions
MAFS.912.F-IF.2.4 
Equivalent expressions are expressions that are the same, even though they may look a little different. If you plug in the same variable value into equivalent expressions, they will each give you the same value when you simplify.

14. Which of the following represents the zero of the function:
𝑓 𝑥 = 𝑥 3 +4 𝑥 2 −7𝑥−10 and the end behavior of the graph.
Example
Use the Descartes’ Rule of signs to determine the number of real zeroes.
A. -1, -5, 2; left side down/right side up
B. -5, 1, 2; left side down/right side up
C. -1, -5, 2; left side up/right side down
D. -2, 1, 5; left side down/right side down
𝒇 𝒙 = 𝒙 𝟑 +𝟒 𝒙 𝟐 −𝟕𝒙−𝟏𝟎 No
change 1
change No
change
There will be 1 positive solution.
Look at the multiple choice answers and see which one has 1 positive and 2 negative solutions.
For the negative case, substitute –x into the function and simplify.
Choice A and C have the same zeros listed with a different end behavior.
𝒇 −𝒙 = −𝒙 𝟑 +𝟒 −𝒙 𝟐 −𝟕 −𝒙 −𝟏𝟎
Now to decide how to determine the end behavior, substitute in a number to the right of 2.
𝒇(𝒙)=− 𝒙 𝟑 +𝟒 𝒙 𝟐 +𝟕𝒙−𝟏𝟎
𝒇 𝟑 = 𝟑 𝟑 +𝟒 𝟑 𝟐 −𝟕 𝟑 −𝟏𝟎 1
change No
change 1
change
=𝟐𝟕+𝟑𝟔−𝟐𝟏−𝟏𝟎
There will be 2 or 0 negative solutions.
=𝟑𝟐
The point (3, 32) is on the graph and going up, so A is the answer.

15. Example
When looking at a rational function. Charles and Bobby have two different thoughts. Charles says that the function is defined at 𝑥=2, 𝑥=3, and 𝑥=5. Bobby says that the function is undefined at those 𝑥 values. Describe a situation where Bobby is correct. Is it possible for a situation to exist where they are both correct? Justify your reasoning.
Charles would be correct if the denominator of a function does not include (x+2), (x-3), or (x-5).
𝒇 𝒙 = (𝒙+𝟐)(𝒙−𝟑) (𝒙+𝟏)(𝒙+𝟑)
Example:
; 𝒖𝒏𝒅𝒆𝒇𝒊𝒏𝒆𝒅 𝒂𝒕 𝒙=−𝟏 𝒂𝒏𝒅 𝒙=−𝟑
Bobby would be correct if the denominator was a composite polynomial of those binomials.
𝒇 𝒙 = 𝒈(𝒙) (𝒙+𝟐)(𝒙−𝟑)(𝒙−𝟓)
Example:
It is not possible for them both to be correct. A function is either defined at a point or it is not. Even if the expression can be simplified so one of the factors is no longer in the denominator, you would still not say the original function was defined there.

16. What is the oblique asymptote of the function below?
𝑓 𝑥 = 𝑥 2 +5𝑥+6 𝑥−4
Example
An oblique asymptote of a function exists when the highest exponent is in the numerator. To find the oblique asymptote; divide the numerator by the denominator and ignore any remainder.
One way is to divide synthetically and the other way is to divide by long division. After the division is complete, then ignore the remainder.
𝒙−𝟒=𝟎
The vertical Asymptotes are found by setting the denominator equal to zero and solving.
𝒙=𝟒
The vertical asymptote would be x = 4. 𝟒
𝟏 𝟓 𝟔
Horizontal Asymptotes, (HA), exist when the highest exponent for the numerator is equal to the highest exponent of the denominator or the highest exponent of the function is in the denominator. 𝟒 𝟑𝟔 𝟏 𝟗 𝟒𝟐
The quotient is x + 9 with a remainder of 42. The oblique asymptote is:
𝒇 𝒙 = 𝟑 𝒙 𝟐 +𝟑𝒙−𝟕 𝟑 𝒙 𝟑
Example:
;𝒕𝒉𝒆𝒏 𝑯𝑨 𝒊𝒔 𝒚=𝟎;𝑽𝑨 𝒊𝒔 𝒙=𝟎
𝒇 𝒙 = 𝟓 𝒙 𝟐 +𝟑𝒙−𝟕 𝟑 𝒙 𝟐
;𝒕𝒉𝒆𝒏 𝑯𝑨 𝒊𝒔 𝒚= 𝟓 𝟑 ;𝑽𝑨 𝒊𝒔 𝒙=𝟎
𝒚=𝒙+𝟗
Example:

17. What is the graph of the following function: −𝑥 2 − 2𝑥 − 2 𝑥 − 2
Example
Find the asymptotes to help sketch the graph.
Hints: Vertical Asymptotes are found by setting the denominator equal to zero and solving.
𝒙−𝟐=𝟎
𝒙=𝟐
The vertical asymptote would be x = 2. 𝟐
−𝟏 −𝟐 −𝟐 −𝟐 −𝟖 −𝟏 −𝟒
−𝟏𝟎
The quotient is –𝒙−𝟒 with a remainder of −𝟏𝟎. The oblique asymptote is:
𝒚=−𝒙−𝟒

18. Logarithms
MAFS.912.F-IF.3.8
a quantity representing the power to which a fixed number (the base) must be raised to produce a given number.

19. Evaluate 𝐥𝐨𝐠 𝟔𝟒 𝟒 𝒚=𝐥𝐨𝐠 𝟔𝟒 𝟒 𝟒= 𝟔𝟒 𝒚 𝟒 𝟏 = 𝟒 𝟑𝒚 𝟏=𝟑𝒚 𝟏 𝟑 =𝒚 Example
𝒚=𝐥𝐨𝐠 𝟔𝟒 𝟒
𝟒= 𝟔𝟒 𝒚
𝟒 𝟏 = 𝟒 𝟑𝒚
𝟏=𝟑𝒚
𝟏 𝟑 =𝒚
So, log = 1 3

20. Based on a model, the solution to the equation below gives the number of years it will take for the population of country A to reach 50 million.  What is the solution to the equation expressed as a logarithm?
Example
𝟓𝟎=𝟒𝟎 𝒆 𝟎.𝟎𝟐𝟕𝒕 𝟒𝟎 𝟒𝟎
𝟏.𝟐𝟓= 𝒆 𝟎.𝟎𝟐𝟕𝒕
Rewrite the expression in log form (in this case natural log form).
𝐥𝐧𝟏.𝟐𝟓= 𝐥𝐧𝒆 𝟎.𝟎𝟐𝟕𝒕
𝐥𝐧𝟏.𝟐𝟓=𝟎.𝟎𝟐𝟕𝒕
Solve for 𝒕.
𝟎.𝟎𝟐𝟕
𝟎.𝟎𝟐𝟕
𝐥𝐧 𝟏.𝟐𝟓 𝟎.𝟎𝟐𝟕 =𝒕

21. Unit Circle The "Unit Circle" is a circle with a radius of 1.
MAFS.912.F-TF.1.2
The "Unit Circle" is a circle with a radius of 1.

22. Angle 𝜃 is in Quadrant II, and sin 𝜃= 4 5 . What is the value of cos 𝜃 ?
Example
𝒄 𝟐 = 𝒂 𝟐 + 𝒃 𝟐 𝟓 𝟒
𝟓 𝟐 = 𝟒 𝟐 + 𝒃 𝟐
𝟐𝟓=𝟏𝟔+ 𝒃 𝟐
𝟗= 𝒃 𝟐 𝜽 𝟑
𝟑=𝒃
𝐜𝐨𝐬 𝜽=− 𝟑 𝟓

23. Remember that the measure of the arc is equal to the central angle.
The measure of a central angle in a circle is given in radians. If the angle measure is 𝜋 3 and the radius is 3 cm. Fins the intercepted arc length and also fins the area of the sector with this intercepted arc. Round your answer to the nearest tenth.
Example
Remember that the measure of the arc is equal to the central angle.
Arc Length = measure of the arc divide by 360 time the circumference
Area of a sector = measure of the arc divide by 360 time the area of the circle
𝑨𝑳= 𝒎𝑨𝒓𝒄 𝟑𝟔𝟎 ∙𝟐𝝅𝒓
𝑨𝑺= 𝒎𝑨𝒓𝒄 𝟑𝟔𝟎 ∙𝝅 𝒓 𝟐
= 𝝅 𝟑 𝟐𝝅 ∙𝟐𝝅(𝟑)
= 𝝅 𝟑 𝟐𝝅 ∙𝝅 (𝟑) 𝟐
= 𝝅 𝟑 ∙ 𝟑 𝟏
= 𝝅 𝟑 ∙ 𝟏 𝟐𝝅 ∙𝟗𝝅
= 𝟑∙𝟑∙𝝅∙𝝅 𝟑∙𝟐∙𝝅
= 𝟑𝝅 𝟐 𝒐𝒓 𝟒.𝟕𝟏 𝒄𝒎 𝟐
=𝝅 𝒐𝒓 𝟑.𝟏𝟒

24. Amplitude Period
MAFS.912.F-TF.2.5 
The Amplitude is the height from the center line to the peak and the Period goes from one peak to the next (or from any point to the next matching point) 

25. Horizontal translation
Phase shift of Sine and Cosine Graphs
𝒚=𝒂 𝐬𝐢𝐧 𝒃 𝒙−𝒉 +𝒌
Vertical translation
𝒉>𝟎 
𝒉<𝟎 
Amplitude
Period
Horizontal translation
𝒉>𝟎 
𝒉<𝟎 

26. Example 𝒇(𝒙) = 𝐬𝐢𝐧⁡(𝒙 + 𝟑) − 𝟐.
Select each statement that is true about the graph of
𝒇(𝒙) = 𝐬𝐢𝐧⁡(𝒙 + 𝟑) − 𝟐.
Example

27. Find the value of the amplitude, period, and midline for the sin function below. Round to the nearest whole number.
Example
Period
|𝟏−𝟒|
Amplitude = 3 𝟑
Midline
𝒚=𝟎

28. A. 𝑔(𝑥) = cos⁡(𝑥) − 2 B. 𝑔(𝑥) = cos⁡(𝑥) + 2 C. 𝑔(𝑥) = cos⁡(2𝑥) + 0
Example
The function 𝑓(𝑥)=cos⁡(𝑥). Function 𝑔(𝑥) results from a transformation on the function 𝑓(𝑥)= cos⁡(𝑥). A portion of the graph of 𝑔(𝑥) is shown.
Part A
What is the equation of 𝑔(𝑥)?
A. 𝑔(𝑥) = cos⁡(𝑥) − 2
B. 𝑔(𝑥) = cos⁡(𝑥) + 2
C. 𝑔(𝑥) = cos⁡(2𝑥) + 0
D. 𝑔(𝑥) = 2cos⁡(𝑥) + 0

29. Relative to the graph of 𝑦=3sin⁡𝑥, what is the shift of the graph of 𝑦=3sin⁡(𝑥+ 𝜋 3 )?
Example
A. 𝜋 3 right
B. 𝜋 3 left
C. 𝜋 3 up
D. 𝜋 3 down

30. Relative to the graph of 𝑦=3sin⁡𝑥, what is the shift of the graph of 𝑦=3sin⁡(𝑥+ 𝜋 3 )?
Example
A. 𝜋 3 right
B. 𝜋 3 left
C. 𝜋 3 up
D. 𝜋 3 down

31. Parabola
MAFS.912.G-GPE.1.2
A parabola is a curve where any point is at an equal distance from: a fixed point (the focus ), and. a fixed straight line (the directrix )

32. Parabola
MAFS.912.G-GPE.1.2

33. Parabola
MAFS.912.G-GPE.1.2

34. Parabola
MAFS.912.G-GPE.1.2

35. Example
Given the focus (-2, 4) and directrix 𝑦= −2. Find the equation of the parabola.
Graph the focus (-2,4) and the directrix (y = -2).
A. 𝑦=− 𝑥
B. 𝑦=− 𝑥
C. 𝑦=− 𝑥−
D. 𝑦= 𝑥
The distance between the two is 6 and the vertex is halfway between the two. The vertex needs to be at (-2, 1).
Using the standard form of a parabola to fill in the coordinates of the vertex, the remaining missing value is a.
𝒚=𝒂 𝒙−𝒉 𝟐 +𝒌 6
𝒚=𝒂 𝒙+𝟐 𝟐 +𝟏
Use the formula
𝒂= 𝟏 𝟒𝒄
The value c is the distance from the focus to the vertex which for this problem is 3. Substitute 3 into the above formula.
𝒂= 𝟏 𝟒(𝟑)
𝒚= 𝟏 𝟏𝟐 𝒙+𝟐 𝟐 +𝟏
𝒂= 𝟏 𝟏𝟐

36. Point A (1, 5) lies on the parabola graphed below
Point A (1, 5) lies on the parabola graphed below. Find the following of the parabola:
Example
𝒚=𝒂 𝒙−𝟐 𝟐 +𝟑
Vertex:
Axis of Symmetry:
Directrix:
Parabola in vertex form:.
Now the ‘a’ value can be found by substituting in the point on the graph (1, 5) and solving for ‘a’
(𝟐,𝟑)
𝟓=𝒂 𝟏−𝟐 𝟐 +𝟑
𝒙=𝟐
𝟓=𝟏𝒂+𝟑
𝟐=𝒂
𝒚=𝟑−𝟏/𝟖
To find the focus and directrix use the formula:
𝒚=𝟐.𝟖𝟕𝟓
𝒂= 𝟏 𝟒𝒄
𝟐= 𝟏 𝟒𝒄
𝟐(𝟒𝒄)= 𝟏 𝟒𝒄 (𝟒𝒄)
𝒚=𝟐 𝒙−𝟐 𝟐 +𝟑
𝟖𝒄=𝟏
𝒄= 𝟏 𝟖
The focus is 1/8 unit from the vertex. (2, 3+1/8)

37. MAFS.912.G-GPE.1.2

38. MAFS.912.G-GPE.1.2

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40. MAFS.912.F-BF.1.2

41. MAFS.912.F-BF.1.2

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43. MAFS.912.F-BF.1.2

47. MAFS.912.F-BF.2.3

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