Cycle 2 Notes - MCR3U1 PDF

MCR3U1 Cycle 2 Cycle 2 Notes MCR3U1 Graphing Quadratics Cycle 2 Lesson 1 Learning Goals...

MCR3U1 Cycle 2 Cycle 2 Notes MCR3U1 Graphing Quadratics Cycle 2 Lesson 1 Learning Goals 2 1. I can state the key characteristics of the parent function 𝑓(𝑥) = 𝑥 2. I can state the equation, graph, and key characteristics given p. 3. I understand how the key characteristic of a quadratic function changes with its transformations. This note requires prior knowledge found in notes 1.7 and 1.12 2 The quadratic parent function is 𝑓(𝑥) = 𝑥. When we graph this, we should get the following: 1. The Basics At up until Lesson 3.8, Quadratic Functions will always be given in Vertex form, which is 2 𝑓(𝑥) = 𝑎(𝑏(𝑥 + 𝑐)) + 𝑑 1 with (𝑎) being the vertical stretch (VS), ( 𝑏 ) being the horizontal stretch (HS), (− 𝑐) being the horizontal translation (HT), and (𝑑) being the vertical translation (VT). 2. The Key Characteristics a. X-Intercept The X-Intercepts are where the Parabola intersects with the X-Axis. If both the VT and VS are positive or negative, there are no X-Intercepts, as the Vertex is not on the X-Axis and the Parabola opens away from the X-Axis. MCR3U1 Graphing Quadratics Cycle 2 Lesson 1 If the VT is negative and the VS is positive or the VT is positive and the VS is negative, there are two X-Intercepts, as the Vertex is not on the X-Axis, however the Parabola opens towards the X-Axis. If there is no VT, then there is one X-intercept, as the Vertex is on the X-Axis. We will not be discussing the methods which can be used to Algebraically determine the X-Intercepts as that is a topic which will be covered in Cycle 3. However, it is possible to determine the X-Intercept of a Quadratic function by graphing the function. b. Y-Intercept The Y-Intercept is the point at which the function Intersects of the Y-Axis. The Y-Intercept can be calculated by substituting 0 for 𝑥. c. Domain, Range In any Quadratic Function, Domain will always be 𝐷[− ∞, ∞], as a Parabola does not have |𝑉𝑆| a Vertical Asymptote. Range will always be 𝑅[𝑉𝑇, 𝑉𝑆 ∞), as the VT is the Horizontal Asymptote and the VS determines the direction of the opening of the Parabola. d. Axis of Symmetry As suggested by the name, the Axis of Symmetry (AoS) is a vertical line which divides the Parabola into two symmetrical halves. The Axis of Symmetry is always the HT. e. Vertex MCR3U1 Graphing Quadratics Cycle 2 Lesson 1 The Vertex is the point of the Parabola which would be 𝑓(0) = 0 or (0, 0) on the parent, and will be the lowest 𝑦 value of the function if the VS is positive, or highest 𝑦 value if the VS is negative. This will always be located at (𝐻𝑇, 𝑉𝑇). Examples: ex. 1 Determining the Graph and Key Characteristics of a Function given the 2 Consider 𝑓(𝑥) = 0. 25(2(𝑥 + 2)) + 2 When we graph this function (by applying a HT of -2, a VT of 2, a VS of 4, and a HS of ½ to the parent), the following graph is produced: The reference points are the same ones found on the parent (after applying the transformations) ‘ We can determine that there are no X-Intercepts as the Vertex (− 2, 2) is above the X-Axis and the Parabola is Opening Upwards. We can determine that the Y-Intercept is (0, 6) either by Graphing the Function or substituting 𝑥 for 0, which results in 2 𝑓(0) = 0. 25(2(𝑥 + 2)) + 2 = 6 As with all Quadratics, the Domain is D[-∞,∞]. By considering the VT and VS, we can determine that the Range is R[2, ∞]. By considering the HT, we can determine that the Axis of Symmetry is -2. By considering the HT and VT, we can determine that the Vertex is at (− 2, 2) MCR3U1 Graphing Quadratics Cycle 2 Lesson 1 ex. 2 Determining the Equation and Key Characteristics of a Function Given its Graph Consider this Graph: Equation: By looking at the Vertex of (− 1, 1), we can determine at VT of 1 and an HT of -1. By looking at the points (− 0. 5, 1. 5) and (0, 3) or (0. 5, 0. 5)and (1, 2) in relation to the Vertex, we can determine that both the VT and HS are 0.5, producing the function 2 𝑓(𝑥) = 0. 5(2(𝑥 + 1)) + 1 Key Characteristics: - By graphing the function, we can determine that there are no X-Intercepts and a Y-intercept of 3. By considering the VS and VT of the function, we can determine a Domain of D[-∞,∞] and a Range of R[1,∞]. By considering the HS (or graphing), we can determine an Axis of Symmetry of -1. By considering the HT and VT (or graphing), we can determine the Vertex to be (− 1, 1). MCR3U1 Quadratic Transformations Cycle 2 Lesson 1 Learning Goals 1. Understanding the quadratic function’s key terms (axis of symmetry, zeroes) 2. Understanding how to list transformations through vertex form 3. Knowing how to graph functions using mapping rule of transformations 2 A quadratic function, or “parabola,” is any function originating from 𝑦 = 𝑥 2 All parabolas can be expressed in the form of 𝑦 = 𝑎𝑥 + 𝑏𝑥 + 𝑐 2 - In the function 𝑦 = 𝑥 , 𝑦 would always be positive because of the exponent on 𝑥 - Positive 𝑎 will make the parabola open upwards and negative 𝑎 makes it open downwards - When 𝑎 ≥ 1, the parabola shrinks in width - When 1 > 𝑎 > 0, the parabola becomes wider Axis of Symmetry A parabola's axis of symmetry (AOS) is a vertical line that divides the parabola −𝑏 into two perfect halves. It can be found with the formula 𝑥 = 2𝑎. - The axis of symmetry can be used to find mirroring points by calculating the distance from the 𝑥-value of Point 1 to the 𝑥-value of the AOS, and then adding the distance to find the 𝑥-value of Point 2. For example If 𝐴𝑂𝑆 = 3 and 𝑃𝑜𝑖𝑛𝑡 1 = (− 6), find the 𝑥-value of the mirroring Point 2. - The distance from the 𝑥-value of Point 1 to the 𝑥-value of the AOS is 3 − (− 6) = 9 - Point 2 = 𝐴𝑂𝑆 + 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 =3+9 = 12 Zeroes A parabola’s zeroes are its 𝑥-intercepts (𝑦 = 0). It can be found using the quadratic formula: 2 −𝑏± 𝑏 −4𝑎𝑐 2 2𝑎 where 𝑎𝑥 + 𝑏𝑥 + 𝑐 = 0. The vertex can help one determine the axis of symmetry and horizontal transformations, as the new zeros will be shifted according to the horizontal stretch. Vertex Form MCR3U1 Quadratic Transformations Cycle 2 Lesson 1 A parabola’s position on the graph is determined by its transformations. However, the transformations 2 cannot be shown in standard form (𝑦 = 𝑎𝑥 + 𝑏𝑥 + 𝑐). Therefore, one must first convert it into vertex 2 form (𝑦 = 𝑎(𝑥 − ℎ) + 𝑘) before being able to list transformations: 2 𝑦 = 𝑎𝑥 + 𝑏𝑥 + 𝑐 Standard form 2 𝑏 𝑦 = 𝑎(𝑥 + 𝑎 𝑥) + 𝑐 Factoring out the 𝑎 2 𝑏 𝑏 2 𝑏 2 𝑦 = 𝑎[𝑥 + 𝑎 𝑥 + ( 2𝑎 ) − ( 2𝑎 ) ] + 𝑐 No new values are added 2 2 2 𝑏 𝑏 𝑏 𝑦 = 𝑎(𝑥 − 𝑎 𝑥+ 2 )+𝑐− 4𝑎 4𝑎 2 𝑏 2 𝑏 −4𝑎𝑐 𝑦 = 𝑎(𝑥 + 2𝑎 ) − 4𝑎 Combining the 𝑐 into the 𝑘 equation 2 𝑏 𝑏 −4𝑎𝑐 The vertex of the parabola will be at (− 2𝑎 , − 4𝑎 ). Quadratic Transformations 2 Now that one can turn a parabola into the form 𝑦 = 𝑎(𝑏(𝑥 + ℎ)) + 𝑘, the transformations are clearly shown. Note: the horizontal transformation (𝑏-value) is already incorporated into the function when converting into vertex form - The 𝑎-value controls the vertical transformations and reflections on the 𝑥-axis (direction of opening) - The 𝑏-value controls the horizontal transformations and reflections on the 𝑦-axis. Remember to factor the equation fully before transforming. Horizontal transformations are always reciprocals of 𝑏. - The ℎ-value controls the horizontal translations (translation is always negative ℎ) - The 𝑘-value controls the vertical translations and range Order of Transformations 1. First, apply transformations that are multiplied or divided such as stretches and reflections 2. Then, apply transformations that are added or subtracted (translations) Example Scenario 1 2 Given the function 𝑦 = 2 (𝑥 − 8) + 2, list the transformations and range by the order of their 2 application from the parent function 𝑦 = 𝑥 1 1. Vertical stretch of 2 2. Horizontal translation of 8 units right (transformations on the 𝑥-value is always opposite) 3. Vertical translation of 2 units up 4. Range: 𝑦 ≥ 2 MCR3U1 Quadratic Transformations Cycle 2 Lesson 1 Graphing The transformations can be used to graph the function. Using the “mapping rule,” we can graph a function 1 2 based on its parent function. The function 𝑦 = 2 (𝑥 − 8) + 2 is shown in the following table and graph 2 1 2 Parent Function (𝑦 = 𝑥 ) Transformed function (𝑦 = 2 (𝑥 − 8) + 2) (𝑥, 𝑦) 1 (𝑥 + 8, 2 𝑦 + 2) (− 2, 4) (6, 4) (− 1, 1) 1 (7, 2 2 ) (0, 0) vertex (8, 2) the vertex can be verified directly with (− ℎ, 𝑘) (1, 1) 1 (9, 2 2 ) (2, 4) (10, 4) MCR3U1 Quadratic Functions Cycle 2 Lesson 1 Learning Goals 1. Understanding the key characteristics: vertex, axis of symmetry, x-intercepts (a.k.a zeros/roots), step pattern 2. Constructing and applying the vertex form equation to solve unknown values 3. Graphing transformations on a quadratic function using mapping rule or vertex + step pattern 4. Formulating the equation of a quadratic function based on its graph Further exploration: Expressing quadratic functions in 3 different forms: standard, intercept, and vertex Fig. 1. The Quadratic Function The aim is to utilize the vertex form of a quadratic equation to graph out a quadratic function. The parent function for quadratic functions can be modified through transformations which also change certain key characteristics. When given a graph of a quadratic function, students should also be able to express it using a vertex-form equation. Through factoring and expanding, it is beneficial to switch between the 3 forms: vertex, standard, and intercept, depending on what a certain question asks you to solve. 2 The vertex form equation for a quadratic function can be modelled as: 𝑓(𝑥) = 𝑎(𝑥 − ℎ) + 𝑘 The “a” value represents the vertical stretch/compression, typically determining the direction of opening and step pattern. “h” value represents the horizontal shift from the original parent function. “k” value represents the vertical shift from the parent function. (h, k) therefore represents the vertex (maximum/minimum point on the function) Characteristics - AoS (Axis of Symmetry) - is determined by the horizontal shift, which is “h” - Direction of opening - is determined by “a”. If the “a” value is positive, then the graph opens upward. If the “a” value is negative, the graph opens downward - Step pattern - is used to find the subsequent points on the function MCR3U1 Quadratic Functions Cycle 2 Lesson 1 When a = 1, the step pattern goes: 1, 3, 5… When “a” is a different value, the step pattern changes and each term of the original step pattern is multiplied by the “a” value to create a new pattern. - i.e. if a = -3, the step pattern goes: -3, -9, -15… 2 Example 1: Graph the quadratic function: 𝑓(𝑥) = − 2(2𝑥 + 4) − 3 2 Given this equation, we will first put it into a similar form as 𝑓(𝑥) = 𝑎(𝑥 − ℎ) + 𝑘 To do this, we will first use our factoring skills from earlier lessons to break apart the terms inside the brackets: 2 𝑓(𝑥) = − 2 (2(𝑥 + 2)) − 3 Now, we’re trying to isolate (𝑥 + 2), so we must expand, using the exponential law which states that 2 2 2 (𝑎𝑏) = 𝑎𝑏 2 𝑓(𝑥) = − 2 (4(𝑥 + 2) ) − 3 Now, we remove the outside brackets by multiplying -2 by 4, in order to isolate (𝑥 + 2) 2 𝑓(𝑥) = − 8(𝑥 + 2) − 3 2 Our modified equation is now in the form of 𝑓(𝑥) = 𝑎(𝑥 − ℎ) + 𝑘 , where the vertex is (h, k) which is (-2, -3). The “a” value is -8 Finally, we will graph using the vertex + step pattern method: 2 Fig. 2. 𝑓(𝑥) = − 2(2𝑥 + 4) − 3 a = -8, so step pattern is: − 8 × 1, − 8 × 3, − 8 × 5 Step pattern: -8, -24, -40 Vertex: (-2, -3) The subsequent points: (-1, -11) (-3, -11) MCR3U1 Quadratic Functions Cycle 2 Lesson 1 Example 2: Given the following graph, create a quadratic equation that models the function Fig. 3. Function for Example 2 From the graph, we can see that the vertex is (3, 4) and the function is opening upwards. We can work backwards using the step pattern, because as the x-value increases/decreases by 1, the y-value increases by 2. Starting from the new point, if we go through the process again, we find that another increase/decrease in x-value by 1 will lead to an increase in y-value by 6. So now, our step pattern looks like: 2, 6... It can be concluded that the “a” value is 2 because all the terms from the original step pattern have been multiplied by 2. Since we know the value of h, k, and a, we can fit them together and create our equation for this quadratic function. 2 Vertex form equation: 𝑓(𝑥) = 𝑎(𝑥 − ℎ) + 𝑘 , where 𝑎 = 2, ℎ = 3, 𝑘 = 4 2 𝑓(𝑥) = 2(𝑥 − 3) + 4 Therefore, the quadratic function that has been graphed can be modelled with the equation 2 𝑓(𝑥) = 2(𝑥 − 3) + 4 Homework For homework, remember to complete the questions in the “2.1 Problems” file. I have also attached the solutions onto the homework assignment. MCR3U1 Exponential Transformations Cycle 2 Lesson 2 Learning Goals 1. I understand the important properties of exponential functions. 2. I can identify the parent function of any given exponential function. 3. I can differentiate between exponential growth and exponential decay. 4. I can graph exponential functions using key points, mapping rules, and asymptotes. 5. I can identify and apply transformations to an exponential function. The aim of this note is to fully understand the key characteristics of exponential functions and apply them to graph a wide variety of exponential growth and decay functions. More specifically, the focus of this note is to identify and apply the correct transformations to the appropriate parent exponential function. For further learning, the student should be able to find the complete exponential function given certain transformations and characteristics of the function. I will be referencing certain skills learned in Cycle 1, such as mapping rules for the annotated questions. Introduction The general form of an exponential function is as follows: Figure 1. The form of an exponential function with transformations Properties Take any base (no transformations) exponential function, the following properties of this function are: Horizontal Asymptote: y = 0 When looking at the properties, it is crucial to understand that the key characteristics of Domain: {𝑥 ϵ 𝑅} exponential functions can help us with identifying Range: {𝑦 > 0┃𝑦 ϵ 𝑅} transformations and graphing. x-intercept: None Asymptote -> Vertical Translation y-intercept -> VS / VT / HT y-intercept: y = 1 MCR3U1 Exponential Transformations Cycle 2 Lesson 2 Parent Function 𝑥 Parent exponential functions are written in the form: 𝑦 = 𝑏. For example: 𝑥 𝑥 𝑥 Figure 2. 𝑦 = 2 Figure 3. 𝑦 = 3 Figure 4. 𝑦 = 10 *As you can see, all base exponential functions have many similarities, such as a y-intercept of (0, 1)* In order to find the parent function of any given exponential function, we only need to look at the “b” value, or the exponent base. 𝑘(𝑥−𝑐) 𝑥 Generally, for an exponential function 𝑦 = 𝑎(𝑏) + 𝑑, the parent function is in the form 𝑦 = 𝑏. 6(𝑥−1) 𝑥 Ex. 𝑦 = 3(5) +8 -> 𝑦 = 5 7(𝑥−9) 𝑥 Ex. 𝑦 = 4(2) + 1 -> 𝑦 = 2 There exists two types of exponential functions. Exponential Growth: Increasing function Can be identified with b > 1 (given that there is no horizontal reflection: k > 0). Figure 5. Graph of exponential growth MCR3U1 Exponential Transformations Cycle 2 Lesson 2 Exponential Decay: Decreasing function Can be identified with 0 < b < 1. (given that there is no horizontal reflection: k > 0). Figure 6. Graph of exponential decay Transformations & Graphing Graphing an exponential function involves using the key points and recalling from a previous lesson, applying the respective transformations with the mapping rule. The key points used to graph an exponential function are as follows: 1 (-1, 𝑏 ) (0, 1) (1, b) 1 Ex. f(x) = 2x: (-1, 2 ); (0, 1); (-1, 2) Figure 7. Graph of y = 2x with key points labelled Examples 3𝑥−6 1) Graph and find y-intercept of the function 𝑓(𝑥) = 4(2 ) + 1. 3(𝑥−2) Factoring out the horizontal stretch (k value), we get 𝑓(𝑥) = 4(2 ) + 1. To help us better understand the function, let’s find the parent function, the transformations applied, and some key characteristics. Parent Function: y = 2x. We can find the parent function by looking at the exponent base (2). Transformations: VS by 4, HS by ⅓, HT 2 right, VT 1 up Asymptote: Horizontal at y = 1 (VT 1 up). Domain: {𝑥 ϵ 𝑅} Range: {𝑦 > 1┃𝑦 ϵ 𝑅} Type: Exponential Growth - Increasing (since b > 1). MCR3U1 Exponential Transformations Cycle 2 Lesson 2 In order to graph this function f(x), we need to find the mapping rule for both x and y coordinates. We will be using this method as it is the simplest and most efficient way for graphing exponential functions. Using the four transformations that we identified, we can derive the mapping rule. (x, y) -> (⅓x + 2, 4y + 1) 1 Recall the key points of an exponential function: (-1, 𝑏 ); (0, 1); (1, b). Now, plugging the points into the mapping rule, we get: 1 5 (-1, 2 ) -> (⅓(-1) + 2, 4(½) + 1) -> ( 3 , 3) (0, 1) -> (⅓(0) + 2, 4(1) + 1) -> (2, 5) 7 (1, 2) -> (⅓(1) + 2, 4(2) + 1) -> ( 3 , 9) Since we are given the function, finding the y-intercept is simple. Recall that the y-intercept is when x is equal to 0. 3((0)−2) 𝑓(0) = 4(2 )+1 Substitute x = 0. −6 𝑓(0) = 4(2 ) + 1 Evaluate 𝑓(0) = 0. 0625 + 1 𝑓(0) = 1. 0625 Therefore, the y-intercept is at (0, 1.0625). Figure 8. Graph of f(x) with asymptote 1 1 − 𝑥−12 2) Graph and find y-intercept of the function 𝑔(𝑥) =− 3 (3 2 ) − 7. 1 1 − (𝑥+24) Factoring out the horizontal stretch, we get 𝑔(𝑥) =− 3 (3 2 ) − 7. 1 1 1 2 (𝑥+24) Converting the horizontal reflection to the base, we get 𝑔(𝑥) =− 3 ( 3 ) − 7. This step is required for identifying whether the function is exponential growth or decay. To help us better understand the function, let’s find the parent function, the transformations applied, and some key characteristics of the function. Parent Function: y = 3x. We can find the parent function by looking at the exponent base (3). MCR3U1 Exponential Transformations Cycle 2 Lesson 2 Transformations: Rx, VS by ⅓, HS by 2, HT 24 left, VT 7 down Asymptote: Horizontal at y = -7 (VT 7 down). Domain: {𝑥 ϵ 𝑅} Range: {𝑦 (2x - 24, -⅓y - 7) 1 Recall the key points of an exponential function: (-1, 𝑏 ); (0, 1); (1, b). Now, plugging the points into the mapping rule, we get: (-1, 3) -> (2(-1) - 24, -⅓(3) - 7) -> (-26, -8) 22 (0, 1) -> (2(0) - 24, -⅓(1) - 7) -> (-24, - 3 ) 64 (1, ⅓) -> (2(1) - 24, -⅓(3) - 7) -> (-22, - 9 ) y-intercept (when x is equal to 0): 1 1 1 2 ((0)+24) 𝑔(0) =− 3 ( 3 )−7 Sub x = 0 1 1 𝑔(0) =− 3 ( 531441 ) − 7 Evaluate 1 𝑔(0) =− 1594323 −7 𝑔(0) ≅ − 7. 000000627 Therefore, the y-intercept is at (0, -7.000000627). Figure 8. Graph of g(x) with asymptote 3) Given that an exponential function h(x): a) has only 2 transformations b) has a vertical stretch of 6 c) has an asymptote at y = -3 d) passes through the point (1, 21) Find the equation and graph of h(x). MCR3U1 Exponential Transformations Cycle 2 Lesson 2 In this question, we are given key information and characteristics about the function. Using these, we can directly derive both transformations: Vertical Stretch of 6 Vertical Translation 3 down - As the asymptote is only affected by VT, it can be directly translated into a transformation. Now that we have both transformations, we can write an equation for h(x): 𝑥 ℎ(𝑥) = 6(𝑏) − 3 As you can see, we are missing one variable in the equation: b - the exponent base. In order to solve for b, we can use the fact that the function passes through the point (1, 21). This method works as when the two coordinates are plugged into the equation, we are only left with only one variable. 1 21 = 6(𝑏) − 3 Substitute 1 as 𝑥 and 21 as 𝑦 24 = 6𝑏 Solve 𝑏=4 Since we now have the values of all the variables, we can write out the full equation: 𝑥 ℎ(𝑥) = 6(4) − 3 Based on the two transformations, we can find the mapping rule. (x, y) -> (x, 6y - 3) 1 Recall the key points of an exponential function: (-1, 𝑏 ); (0, 1); (1, b). Plugging the points into the mapping rule, we get: (1, 4) -> (-1, 6(4) - 3) -> (1, 21) (0, 1) -> (0, 6(1) - 3) -> (0, 3) *y-intercept* 3 (-1, ¼) -> (1, 6(¼) - 3) -> (-1, - 2 ) Figure 9. Graph of h(x) with asymptote MCR3U1 Exponential Transformations Cycle 2 Lesson 2 Learning Goals 1. I understand the important properties of exponential functions. 2. I can identify the parent function of any given exponential function. 3. I can differentiate between exponential growth and exponential decay. 4. I can graph exponential functions using key points, mapping rules, and asymptotes. 5. I can identify and apply transformations to an exponential function. The aim of this note is to fully understand the key characteristics of exponential functions and apply them to graph a wide variety of exponential growth and decay functions. More specifically, the focus of this note is to identify and apply the correct transformations to the appropriate parent exponential function. For further learning, the student should be able to find the complete exponential function given certain transformations and characteristics of the function. I will be referencing certain skills learned in Cycle 1, such as mapping rules for the annotated questions. Introduction The general form of an exponential function is as follows: Figure 1. The form of an exponential function with transformations Properties Take any base (no transformations) exponential function, the following properties of this function are: Horizontal Asymptote: y = 0 When looking at the properties, it is crucial to understand that the key characteristics of Domain: {𝑥 ϵ 𝑅} exponential functions can help us with identifying Range: {𝑦 > 0┃𝑦 ϵ 𝑅} transformations and graphing. x-intercept: None Asymptote -> Vertical Translation y-intercept -> VS / VT / HT y-intercept: y = 1 MCR3U1 Exponential Transformations Cycle 2 Lesson 2 Parent Function 𝑥 Parent exponential functions are written in the form: 𝑦 = 𝑏. For example: 𝑥 𝑥 𝑥 Figure 2. 𝑦 = 2 Figure 3. 𝑦 = 3 Figure 4. 𝑦 = 10 *As you can see, all base exponential functions have many similarities, such as a y-intercept of (0, 1)* In order to find the parent function of any given exponential function, we only need to look at the “b” value, or the exponent base. 𝑘(𝑥−𝑐) 𝑥 Generally, for an exponential function 𝑦 = 𝑎(𝑏) + 𝑑, the parent function is in the form 𝑦 = 𝑏. 6(𝑥−1) 𝑥 Ex. 𝑦 = 3(5) +8 -> 𝑦 = 5 7(𝑥−9) 𝑥 Ex. 𝑦 = 4(2) + 1 -> 𝑦 = 2 There exists two types of exponential functions. Exponential Growth: Increasing function Can be identified with b > 1 (given that there is no horizontal reflection: k > 0). Figure 5. Graph of exponential growth MCR3U1 Exponential Transformations Cycle 2 Lesson 2 Exponential Decay: Decreasing function Can be identified with 0 < b < 1. (given that there is no horizontal reflection: k > 0). Figure 6. Graph of exponential decay Transformations & Graphing Graphing an exponential function involves using the key points and recalling from a previous lesson, applying the respective transformations with the mapping rule. The key points used to graph an exponential function are as follows: 1 (-1, 𝑏 ) (0, 1) (1, b) 1 Ex. f(x) = 2x: (-1, 2 ); (0, 1); (-1, 2) Figure 7. Graph of y = 2x with key points labelled Examples 3𝑥−6 1) Graph and find y-intercept of the function 𝑓(𝑥) = 4(2 ) + 1. 3(𝑥−2) Factoring out the horizontal stretch (k value), we get 𝑓(𝑥) = 4(2 ) + 1. To help us better understand the function, let’s find the parent function, the transformations applied, and some key characteristics. Parent Function: y = 2x. We can find the parent function by looking at the exponent base (2). Transformations: VS by 4, HS by ⅓, HT 2 right, VT 1 up Asymptote: Horizontal at y = 1 (VT 1 up). Domain: {𝑥 ϵ 𝑅} Range: {𝑦 > 1┃𝑦 ϵ 𝑅} Type: Exponential Growth - Increasing (since b > 1). MCR3U1 Exponential Transformations Cycle 2 Lesson 2 In order to graph this function f(x), we need to find the mapping rule for both x and y coordinates. We will be using this method as it is the simplest and most efficient way for graphing exponential functions. Using the four transformations that we identified, we can derive the mapping rule. (x, y) -> (⅓x + 2, 4y + 1) 1 Recall the key points of an exponential function: (-1, 𝑏 ); (0, 1); (1, b). Now, plugging the points into the mapping rule, we get: 1 5 (-1, 2 ) -> (⅓(-1) + 2, 4(½) + 1) -> ( 3 , 3) (0, 1) -> (⅓(0) + 2, 4(1) + 1) -> (2, 5) 7 (1, 2) -> (⅓(1) + 2, 4(2) + 1) -> ( 3 , 9) Since we are given the function, finding the y-intercept is simple. Recall that the y-intercept is when x is equal to 0. 3((0)−2) 𝑓(0) = 4(2 )+1 Substitute x = 0. −6 𝑓(0) = 4(2 ) + 1 Evaluate 𝑓(0) = 0. 0625 + 1 𝑓(0) = 1. 0625 Therefore, the y-intercept is at (0, 1.0625). Figure 8. Graph of f(x) with asymptote 1 1 − 𝑥−12 2) Graph and find y-intercept of the function 𝑔(𝑥) =− 3 (3 2 ) − 7. 1 1 − (𝑥+24) Factoring out the horizontal stretch, we get 𝑔(𝑥) =− 3 (3 2 ) − 7. 1 1 1 2 (𝑥+24) Converting the horizontal reflection to the base, we get 𝑔(𝑥) =− 3 ( 3 ) − 7. This step is required for identifying whether the function is exponential growth or decay. To help us better understand the function, let’s find the parent function, the transformations applied, and some key characteristics of the function. Parent Function: y = 3x. We can find the parent function by looking at the exponent base (3). MCR3U1 Exponential Transformations Cycle 2 Lesson 2 Transformations: Rx, VS by ⅓, HS by 2, HT 24 left, VT 7 down Asymptote: Horizontal at y = -7 (VT 7 down). Domain: {𝑥 ϵ 𝑅} Range: {𝑦 (2x - 24, -⅓y - 7) 1 Recall the key points of an exponential function: (-1, 𝑏 ); (0, 1); (1, b). Now, plugging the points into the mapping rule, we get: (-1, 3) -> (2(-1) - 24, -⅓(3) - 7) -> (-26, -8) 22 (0, 1) -> (2(0) - 24, -⅓(1) - 7) -> (-24, - 3 ) 64 (1, ⅓) -> (2(1) - 24, -⅓(3) - 7) -> (-22, - 9 ) y-intercept (when x is equal to 0): 1 1 1 2 ((0)+24) 𝑔(0) =− 3 ( 3 )−7 Sub x = 0 1 1 𝑔(0) =− 3 ( 531441 ) − 7 Evaluate 1 𝑔(0) =− 1594323 −7 𝑔(0) ≅ − 7. 000000627 Therefore, the y-intercept is at (0, -7.000000627). Figure 8. Graph of g(x) with asymptote 3) Given that an exponential function h(x): a) has only 2 transformations b) has a vertical stretch of 6 c) has an asymptote at y = -3 d) passes through the point (1, 21) Find the equation and graph of h(x). MCR3U1 Exponential Transformations Cycle 2 Lesson 2 In this question, we are given key information and characteristics about the function. Using these, we can directly derive both transformations: Vertical Stretch of 6 Vertical Translation 3 down - As the asymptote is only affected by VT, it can be directly translated into a transformation. Now that we have both transformations, we can write an equation for h(x): 𝑥 ℎ(𝑥) = 6(𝑏) − 3 As you can see, we are missing one variable in the equation: b - the exponent base. In order to solve for b, we can use the fact that the function passes through the point (1, 21). This method works as when the two coordinates are plugged into the equation, we are only left with only one variable. 1 21 = 6(𝑏) − 3 Substitute 1 as 𝑥 and 21 as 𝑦 24 = 6𝑏 Solve 𝑏=4 Since we now have the values of all the variables, we can write out the full equation: 𝑥 ℎ(𝑥) = 6(4) − 3 Based on the two transformations, we can find the mapping rule. (x, y) -> (x, 6y - 3) 1 Recall the key points of an exponential function: (-1, 𝑏 ); (0, 1); (1, b). Plugging the points into the mapping rule, we get: (1, 4) -> (-1, 6(4) - 3) -> (1, 21) (0, 1) -> (0, 6(1) - 3) -> (0, 3) *y-intercept* 3 (-1, ¼) -> (1, 6(¼) - 3) -> (-1, - 2 ) Figure 9. Graph of h(x) with asymptote MCR3U1 Graphing Exponential Cycle 2 Lesson 2 C1 C2 Graphing Exponential Functions Learning Goals 1. Understand the key characteristics of exponential functions. 2. Graph exponential functions accurately. 3. Analyze connections between exponential functions and previous topics. An exponential function is defined as a function where a constant is raised to the power of an independent variable (typically denoted as ‘x’), particularly when the constant is represented as ‘e’. Exponential functions exhibit rapid growth or decay, as the bigger ‘x’ is, the steeper the slope therefore they are growing exponentially fast. 𝑏(𝑥+𝑐) When graphing exponential functions, they are typically represented by the form f(x)=𝑎(𝑒) + 𝑑. We can observe a correlation between previous lessons as the basic laws of graphing functions have not changed, though there have been some adjustments. 1. Find horizontal asymptote, ‘d’ - it’s defined by the vertical translation as it represents the value that the function is approaching as ‘x’ tends towards positive or negative infinity. 2. Find key points (0,1) and (1,e) - find the y-intercept and the intercept when x is 1 and apply transformations 3. Graph - plot the key points and draw the graph. Example −0.5𝑥−4 Sketch the graph of the function f(x)=4(3) −5 First, we can find the horizontal asymptote which is the vertical translation (represented by ‘d ’in the equation) as it is already given by looking at our function, we are using this method as it is the easiest and fastest way to find the horizontal asymptote. By finding the horizontal asymptote we can see the boundaries of this function and the line the graph is approaching. Looking at the equation, we get: 𝑑 =− 5 Therefore, the horizontal asymptote is at -5. MCR3U1 Graphing Exponential Cycle 2 Lesson 2 Next, by using the mapping rule taught in previous lessons we can find the transformations and key points of the graph. To do this, we can first factor -0.5x-4 to get the two missing values from the equation, ‘b’(horizontal stretch) and ‘c’(horizontal translation) we will be using factoring as it is convenient and will get us what we need in quick steps. To get values ‘b’ and ‘c’, we will: − 0. 5𝑥 − 4 =− 0. 5(𝑥 + 8) Our HS and HT is thus, -2 and 8 respectively. Once we’ve found the missing values of the equation we can plug the numbers into the mapping rule. There is no specific method used for this step, we can simply put the numbers into the places they belong. The mapping rule: 1 ( 𝑏 𝑥 − 𝑐, 𝑎𝑦 + 𝑑) →(− 2𝑥 − 8, 4𝑦 − 5) Now, we can calculate the key points after transformation by the original points into the mapping rule: (0, 1) → (− 2(0) − 8, 4(1) − 5) → (− 8, − 1) (1, 𝑒) → (1, 3) → (− 2(1) − 8, 4(3) − 5) → (− 10, 7) Therefore, the two key points of the graph will be (-8,-1) and (-10,7). Finally, we can draw the graph by plotting the key points and connecting them, making the graph go towards the horizontal asymptote. This is the final answer to the question. MCR3U1 Exponential Function Cycle 2 Lesson 2 Learning Goals 𝑥 1. Understanding the form of exponential functions involves obtaining 𝑓(𝑥) = 𝑎 × 𝑏 where 𝑎 is the initial value and 𝑏 is the base. 2. Recognizing the starting value (y-intercept) and the rate of growth or decay as determined by the base 𝑏 are fundamental elements of identifying key characteristics. 3. Being able to identify horizontal lines that the graph approaches but never touches requires knowledge of asymptotic analysis. 4. Applying transformations involves understanding how shifts, stretches, and reflections affect the graph of exponential functions. Description Exponential functions have a constant base increased to a variable exponent, signifying rapid growth or decay. They play a crucial role in the modeling of processes such as compound interest, radioactive decay, and population expansion. The parent function of an exponential function is the basic model that all other exponential functions are constructed from. 𝑥 The parent function of exponential functions is typically represented as 𝑓(𝑥) = 𝑏 where 𝑏 is the base of the exponential function, determining the rate of growth (if 𝑏 > 1) or decay (if 0 < 𝑏 < 1) 𝑥 determines the ‘horizontal’ behaviour of the exponential function, determining how quickly the graph grows or decays as 𝑥 changes. ○ If 𝑥 is a positive integer (1, 2, 3…), then it represents the number of times the base is multiplied by itself. ○ If 𝑥 is a negative integer (-1, -2, -3…), then it represents the number of times the reciprocal of the base is multiplied by itself. ○ If 𝑥 is a fraction or a decimal, it represents fractional or repeated multiplication of the base by itself. Key Characteristics Exponential functions often have a horizontal asymptote at 𝑦 = 0 or the 𝑥-axis Domain: The domain of the function is all real numbers {− ∞, + ∞} Range: The range depends on the base: ○ if 𝑏 > 1, the range is {0, + ∞) ○ if 0 < 𝑏 < 1 the range is {0 , 1} MCR3U1 Exponential Function Cycle 2 Lesson 2 Graphing Exponential Functions with Transitions 𝑥 The basic form of a exponential function with transitions is 𝑓(𝑥) = 𝑎 × (𝑏 ) + 𝑐 𝑥 1. Start with the basic exponential function 𝑓(𝑥) = 𝑎 × 𝑏 , where a is the initial value and b is the base 2. Identify and vertical or horizontal translations by noting any constants added or subtracted inside or outside the function, respectively. ○ Vertical Translations: If there's a constant 𝑐 added or subtracted outside the function, determine whether it shifts the graph up or down. This translates the function vertically. ○ Horizontal Translations: If there's a constant 𝑑 added or subtracted inside the exponent, determine whether it shifts the graph left or right. This translates the function horizontally. ○ Stretching/Shrinking: Identify any multiplication by a constant 𝑎 greater than 1 or between 0 and 1. This stretches or shrinks the graph vertically. ○ Reflection: Identify any multiplication by − 1 or − 𝑏. These reflect the graph over the 𝑥-axis or the 𝑦-axis, respectively. 3. Plot the key points including: ○ the 𝑦-intercept → (0, 𝑎) ○ Points where 𝑥 = 1 or 𝑥 =− 1 (if applicable) ○ Any additional points based on transformations 4. Identify and draw any horizontal asymptotes. For exponential growth, the 𝑥-axis ( 𝑦 = 0) serves as an asymptote. For exponential decay, the line 𝑦 = 𝑐 (where 𝑐 is the vertical translation) serves as an asymptote. 5. Connect the plotted points with a smooth curve, ensuring it reflects the behavior of the function accurately. 6. Label the 𝑥-axis and the 𝑦-axis, and provide a label for the function. 7. Review the graph to ensure all transformations have been correctly applied and that the graph accurately represents the given exponential function with transitions. Example 𝑥+1 𝑓(𝑥) = 2 × 3 −1 𝑥 The basic function is 𝑓(𝑥) = 3 where 𝑎 = 1 and 𝑏 = 3 Translations - There's a horizontal translation of + 1 (inside the exponent), shifting the graph 1 unit to the left. - There is a vertical translation of − 1 (outside the exponent), shifting the graph 1 unit downward Vertical Translation: The graph is shifted 1 unit downward. Horizontal Translation: The graph is shifted 1 unit to the left. MCR3U1 Exponential Function Cycle 2 Lesson 2 Stretching/Shrinking: There's no stretching or shrinking since there's no multiplication by a constant 𝑎 Reflections: There are no reflections Combine Transformations Horizontal Shift: + 1 Vertical Shift: − 1 Plot Key Points 1 𝑦-intercept: (0, 2 × 3 − 1) = (0, 5) 1+1 Point where 𝑥 = 1: (1, 2 × 3 − 1) = (1, 17) Extra points obtained from the exponential function's characteristics Horizontal asymptote: 𝑦 = − 1 (Draw the horizontal asymptote) Review the graph to ensure all transformations have been correctly applied.  2.3 Logarithmic Functions MCR3U1 Cycle 2 Lesson 3 Transformations Learning Goals 1. Be able to graph the transformed logarithmic function 2. Be able to name the transformations and know how they change the function 3. Be able to identify the key characteristics of the function Recall the base logarithmic function, 𝑓(𝑥) = 𝑙𝑜𝑔𝑏(𝑥), where b is the base. This is the function without transformations. Transformations 𝑓(𝑥) = 𝑎 · 𝑙𝑜𝑔𝑏((𝑘(𝑥 − 𝑐)) + 𝑑 Where: 𝑎 = vertical stretch (by a factor of |𝑎|) c = horizontal translation (left or right) b = base d = vertical translation (up or down) 1 k = horizontal stretch (by a factor of 𝑘 ) Example 1: Graph the function 𝑔(𝑥) =− 2𝑙𝑜𝑔2(𝑥) + 1 Transformations: x y x 𝑔(𝑥) Reflection in the x-axis 1 -1 1 3 2 2 Vertical stretch by 2 1 0 → 1 1 Vertical translation of 1 up 𝑥 2 1 2 -1 (𝑥, 𝑦) → ( 𝑘 + 𝑐, 𝑎𝑦 + 𝑑) 4 2 4 -3 (𝑥, 𝑦) →(𝑥, − 2𝑦 + 1) 8 3 8 -5 Table #1: Table of Table #2: Table of values values of 𝑙𝑜𝑔2(𝑥) of 𝑔(𝑥) =− 2𝑙𝑜𝑔2(𝑥) + 1  2.3 Logarithmic Functions MCR3U1 Cycle 2 Lesson 3 Transformations By using the mapping rule, we can plug in values from the base function into 𝑔(𝑥). Key characteristics of 𝑔(𝑥): Asymptote: 𝑥 = 0, the asymptote has not changed because there is no horizontal or vertical movement of the 𝑥 value. Domain: {𝑥ϵℝ|𝑥 > 0}, 𝑥 can be anything but it has to be greater than 0 because of its asymptote. Range: {𝑦ϵℝ}, 𝑦 can be anything. Increasing / Decreasing? Decreasing, because of the domain {𝑥ϵℝ|𝑥 > 0}, x has to be smaller than 0. Y-intercept: There is no y-intercept because it will never touch the y-axis (or x=0) because of its asymptote. X-intercept: To find the x-intercept, we can set 𝑔(𝑥) equal to 0, then solve for x. We do this because the x-intercept is when y is equal to 0, and y can also be written as 𝑔(𝑥), therefore when 𝑔(𝑥) is equal to 0, solving for x will give us the x-intercept. 𝑔(𝑥) = 0 − 2𝑙𝑜𝑔2(𝑥) + 1 = 0 1 𝑙𝑜𝑔2(𝑥) = 2 1 22 = 𝑥 𝑥= 2 Therefore the x-intercept is ( 2, 0). Figure 1: Graph of 𝑔(𝑥) with labelled points.  2.3 Logarithmic Functions MCR3U1 Cycle 2 Lesson 3 Transformations Example 2: Graph the function ℎ(𝑥) = 2𝑙𝑜𝑔3(2𝑥 + 4) ← Note that we need to factor out the 2 from 2𝑥 + 4 to isolate the x variable. This becomes ℎ(𝑥) = 2𝑙𝑜𝑔3(2(𝑥 + 2)) Transformations: x y x1 ℎ(𝑥) Vertical stretch of 2 1 -1 − 11 -2 1 3 6 Horizontal stretch of 2 1 0 − 3 0 2 Horizontal translation 2 units left 𝑥 3 1 − 1 2 (𝑥, 𝑦) → ( 𝑘 + 𝑐, 𝑎𝑦 + 𝑑) 2 𝑥 (𝑥, 𝑦) → ( 2 − 2, 2𝑦) 9 2 5 4 2 By using the mapping rules, we can graph the function 27 3 23 6 2 ℎ(𝑥). Table #3: Table of Table #4: Table of values Values of base of ℎ(𝑥) function 𝑙𝑜𝑔3(𝑥) Key characteristics of ℎ(𝑥): Asymptote: 𝑥 =− 2, the asymptote has been moved 2 units to the left because of the horizontal translation. Domain: {𝑥ϵℝ|𝑥 >− 2}, 𝑥 can be anything but it has to be greater than -2 because of its asymptote. Range: {𝑦ϵℝ}, 𝑦 can be anything. Increasing / Decreasing? Increasing, because of the domain {𝑥ϵℝ|𝑥 >− 2}, x has to be greater than -2. Y-intercept:  2.3 Logarithmic Functions MCR3U1 Cycle 2 Lesson 3 Transformations To find the y-intercept, we set x equal to 0 and solve for ℎ(𝑥). ℎ(𝑥) = 2𝑙𝑜𝑔3(2(𝑥 + 2)) ℎ(0) = 2𝑙𝑜𝑔3(2(0 + 2)) ℎ(0) = 2𝑙𝑜𝑔3(4) ℎ(0) ∼ 2 · 1. 263 ℎ(0) ∼ 2. 53 The y-intercept is (0, 2. 53) X-intercept: From our table of values, we have already found the x-intercept. The x-intercept is (− 1. 5, 0). Figure #2: Graph of ℎ(𝑥) with labelled points  2.3 Logarithmic Functions MCR3U1 Cycle 2 Lesson 3 Transformations  Logarithmic Functions - Graphs & MCR3U1 Cycle 2 Lesson 3 Key Characteristics Learning Goals 1. I can state the domain & range of logarithmic functions 2. I can graph a logarithmic function using its mapping rule 3. I understand what a vertical asymptote is and why it occurs with logarithmic functions 4. I understand that the logarithmic function has many parents. Everytime the base of the log changes, a new parent must be used. 𝑓(𝑥) = 𝑙𝑜𝑔2(𝑥 − 3) 𝑣𝑠. 𝑔(𝑥) = 𝑙𝑜𝑔3(𝑥 + 2) Description The general form of a logarithmic function is 𝑓(𝑥) = 𝑎𝑙𝑜𝑔𝑏(𝑐(𝑥 − 𝑑) + 𝑘, in which certain variables represent the different transformations that can occur to the function. Domain & Range: In order to graph logarithmic functions, it is important to understand and identify their domain and range. Domain: {𝑥/𝑥 ϵ 𝑅, 𝑥 > 0} All parent log functions share the same domain, in which 𝑥 must be greater than 0. This is because logarithms of negative numbers and the logarithm of zero are undefined Range: {𝑦 ϵ 𝑅} The range for all logarithmic functions, including both parent and transformed functions, is the same Key Characteristics: A major key characteristic of logarithmic functions would be the presence of vertical asymptotes. In the parent functions of log, the vertical asymptote is represented by the line 𝑥 = 0. This is a line that the 𝑥-values of the function approach, but never reach. As a result, the 𝑦-values approach negative infinity All logarithmic parent functions pass through the points (1, 0), and (𝑏, 1), in which 𝑏 represents the base of the logarithm Transformations: 1. Vertical Stretch: In the general form of a logarithmic function, the vertical stretch is represented by the variable 𝑎. If 𝑎 > 0, the function will be stretched vertically. If 0 < 𝑎 < 1, the function will be compressed vertically.  Logarithmic Functions - Graphs & MCR3U1 Cycle 2 Lesson 3 Key Characteristics Figure 1.0. Graph Comparing Logarithmic Functions with Varying Vertical Stretch Factors Figure 1.0 compares the parent function 𝑓(𝑥) = 𝑙𝑜𝑔2(𝑥) → red function, with 𝑔(𝑥) = 2𝑙𝑜𝑔2(𝑥) → black 1 function and ℎ(𝑥) = 2 𝑙𝑜𝑔2(𝑥) → blue function. It demonstrates how the vertical stretch impacts the appearance of a logarithmic function. 2. Horizontal Stretch: A horizontal stretch in a logarithmic function is represented by the variable 𝑐. If 𝑐 > 1, the function will be horizontally compressed, which means the distance between the function and the 𝑥-axis will become smaller. If 0 < 𝑐 < 1, the function will be horizontally stretched, which means the distance between the function and the 𝑥-axis will become wider. In order to view the relationship between logarithmic functions with different horizontal stretches the functions below will be graphed. 𝑓(𝑥) = 𝑙𝑜𝑔2(𝑥) → red 𝑔(𝑥) = 𝑙𝑜𝑔2(2𝑥) → black 1 ℎ(𝑥) = 𝑙𝑜𝑔2( 2 𝑥) → blue  Logarithmic Functions - Graphs & MCR3U1 Cycle 2 Lesson 3 Key Characteristics Figure 2.0. Graph Comparing Logarithmic Functions with Varying Horizontal Stretch Factors 3. Horizontal Translation: The horizontal translation of a logarithmic function is represented by the variable 𝑑. The horizontal translation controls the vertical asymptote of the function. The horizontal translation is a transformation that can cause a change in the domain of the function. For instance, in the function 𝑓(𝑥) = 𝑙𝑜𝑔2(𝑥 + 2), in which the parent function is translated to the left by two units, the domain would be {𝑥/𝑥 ϵ 𝑅, 𝑥 >− 2}. The change is domain occurs because the vertical asymptote is now 𝑥 =− 2. 4. Vertical Translation: The vertical translation is represented by the variable 𝑘. This transformation shifts the functions position along the 𝑦-axis. Mapping Rule: Using the mapping rule is an effective way of graphing logarithmic functions. They are quicker and more efficient to use in comparison to graphing using a table of values. In order to use the mapping rule, we must be able to identify the transformations that occur to the function. Once we are able to identify the transformations, we will be able to find points on the transformed function’s graph by applying its transformations to the key points of its parent function, (1, 0) & (𝑏, 1). Exponential Functions: Logarithmic parent functions are the inverse of exponential parent functions, this is only true when both functions have the same base. As we learned in lesson 2.2, exponential transformations, the different bases of exponential functions represent different parent functions. Similarly, as the base of the logarithmic  Logarithmic Functions - Graphs & MCR3U1 Cycle 2 Lesson 3 Key Characteristics function changes, the parent changes as well. An exponential parent function that is reflected over the line 𝑦 = 𝑥 creates its inverse logarithmic function. The same can be said in vice versa. 𝑥 Figure 3.0. Graph demonstrating that 𝑓(𝑥) = 2 and 𝑔(𝑥) = 𝑙𝑜𝑔2(𝑥) are reflections about the line 𝑦 = 𝑥 As exponential functions and logarithmic functions are inverses of each other, you are able to find points on a logarithmic function using its corresponding exponential function. This can be done by taking a point from the exponential function and swapping its 𝑥 & 𝑦 coordinates. For instance, we can find a point on the logarithmic parent function of 𝑓(𝑥) = 𝑙𝑜𝑔2(𝑥), by swapping the 𝑥 𝑥 & 𝑦 coordinates of a point on its inverse function, 𝑔(𝑥) = 2. Exponential → Logarithmic (𝑥, 𝑦) → (𝑦, 𝑥) (0, 1) → (1, 0) You can use points from a transformed exponential function to find corresponding points on a transformed logarithmic function, only if they both have the same transformations. Annotated Examples Question #1: State the transformations, and graph the function 𝑔(𝑥) =− 2𝑙𝑜𝑔3(2𝑥 − 6) + 2 using the mapping rule.  Logarithmic Functions - Graphs & MCR3U1 Cycle 2 Lesson 3 Key Characteristics When graphing transformed logarithmic functions, we must first identify the function’s parent. This can be done by looking at the base of the logarithm in the transformed function. In this case, 𝑏 = 3, therefore, the parent function would be 𝑓(𝑥) = 𝑙𝑜𝑔3(𝑥). In order to use the mapping rule, we must first identify the transformations, as well as the vertical asymptote of this function. In order to do so, we must first see if we can common factor the transformations within the parent. This is done in order to better interpret the horizontal translation. 𝑔(𝑥) =− 2𝑙𝑜𝑔3(2𝑥 − 6) + 2 → In this case, we are able to common factor 𝑔(𝑥) =− 2𝑙𝑜𝑔3(2(𝑥 − 3)) + 2 → Common factored function Now we can identify and note down the type and value of the transformations that occur to the parent function. Transformations: Vertical Reflection Vertical stretch by a factor of 2 Horizontal Translation right by a factor of 3 ○ This indicates that our vertical asymptote is 3 1 Horizontal stretch by a factor of 2 → As we have learned in previous lessons, we must state the horizontal stretch as its reciprocal Vertical Translation by 2 Using the transformations, we can now create the mapping rule. The mapping rule will go from the parent function to the transformed function. 1 (𝑥, 𝑦) → ( 2 𝑥 + 3, − 2𝑦 + 2) 7 (1, 0) → ( 2 , 2) → We are taking the point (1, 0), which is a key point found on all logarithmic parent functions, and transforming into a point on the function 𝑔(𝑥) =− 2𝑙𝑜𝑔3(2𝑥 − 6) + 2 , by applying its transformations onto the point. 9 (3, 1) → ( 2 , 0) We can now plot the points of the transformed function onto a graph, and connect the points using a smooth curved line. We must ensure that as the line approaches the vertical asymptote, it will continue towards it, but will never reach it.  Logarithmic Functions - Graphs & MCR3U1 Cycle 2 Lesson 3 Key Characteristics Figure 4.0. Graph of the Function 𝑔(𝑥) =− 2𝑙𝑜𝑔3(2𝑥 − 6) + 2 Question #2: State the domain and range of the function 𝑗(𝑥) = 𝑙𝑜𝑔4(− 2𝑥 + 4) Range: {𝑦 ϵ 𝑅} → The range of all logarithmic functions will always be the same. This is because as the value of 𝑥 approaches the asymptote and reaches infinitesimal values, the 𝑦-values accordingly approach negative or positive infinity. This also occurs as the values of 𝑥 further from the asymptote. The domain of a logarithmic function is influenced by its vertical asymptote, in other words, by its horizontal translation. In order to better view the horizontal translation, we must first common factor the transformations that occur within the parent. 𝑗(𝑥) = 𝑙𝑜𝑔4(− 2𝑥 + 4) 𝑗(𝑥) = 𝑙𝑜𝑔4(− 2(𝑥 − 2)) → Common factored function Now we are able to see that there is a horizontal translation right by 2, and that the vertical asymptote is 2. With this information, we can conclude that the domain will either be: {𝑥/𝑥 ϵ 𝑅, 𝑥 < 2} or {𝑥/𝑥 ϵ 𝑅, 𝑥 > 2}. In order to figure out which of the two domains belong to the function, we must identify whether or not the function has a horizontal reflection. A horizontal reflection affects a function’s domain by changing its position on the graph in relation to the 𝑦-axis. In other words, it creates a new graph that is a reflection of the original about the 𝑦-axis. In this case, the function has a horizontal reflection, which means that the values of 𝑥 must always be less than 2, in order to prevent the function from intersecting with the vertical asymptote.  Logarithmic Functions - Graphs & MCR3U1 Cycle 2 Lesson 3 Key Characteristics Figure 5.0. Graph of the Function 𝑗(𝑥) = 𝑙𝑜𝑔4(− 2𝑥 + 4) Figure 5.0 demonstrates that the values of 𝑥 must always be less than 2, the vertical asymptote. Therefore, the domain of the function 𝑗(𝑥) = 𝑙𝑜𝑔4(− 2𝑥 + 4) is {𝑥/𝑥 ϵ 𝑅, 𝑥 < 2}.  Logarithmic Functions - Graphs & MCR3U1 Cycle 2 Lesson 3 Key Characteristics Learning Goals 1. Understand the key characteristics of a logarithmic function and how they relate to: a. an equation b. a graph 2. Be able to graph a logarithmic function Defining logarithmic functions: Logarithmic functions are the inverse of exponential functions, meaning their x and y values are swapped. The parent functions are as if they were reflected in the line: 𝑦 = 𝑥. See Figure 1. Figure 1. 𝑥 𝑦 = 𝑥 (in red) 𝑦 = 2 (in green) 𝑦 = 𝑙𝑜𝑔2𝑥 (in blue) General form: A logarithmic function has the general form 𝑙𝑜𝑔𝑚𝑥. Note: if you were to see 𝑙𝑜𝑔(𝑥) where it seems there is no base, the base is 10. It is similar to radicals ( ), 4 where it is 2 by default, unless specified otherwise ( ).  Logarithmic Functions - Graphs & MCR3U1 Cycle 2 Lesson 3 Key Characteristics Restrictions: in 𝑙𝑜𝑔𝑚𝑥, 𝑥 & 𝑚 > 0 and 𝑚 ≠ 1. 𝑏 𝑚 has those restrictions because 𝑙𝑜𝑔𝑚𝑥 = 𝑏 can be written as 𝑚 = 𝑥, so if m was 1 or 0, it would be linear; a horizontal line. If m was a negative value, there would be no x-value when b is not a positive integer. 1 i.e. 𝑚 =− 4 and 𝑏 = 2. 1 1 𝑙𝑜𝑔−4(𝑥) = 2 → 𝑥 = (− 4) 2 𝑎𝑘𝑎 − 4, which is undefined. Key Characteristics and Transformations: In 𝑎𝑙𝑜𝑔(𝑏(𝑥 − 𝑐)) + 𝑑: 𝑎, the VS factor: - vertically stretches (when |𝑎| > 1) or compresses (when |𝑎| < 1) the parent. See Figure 2 and Figure 3 below. 1 Figure 2: 𝑦 = 𝑙𝑜𝑔(𝑥) in red & 𝑦 = 2𝑙𝑜𝑔(𝑥) in green. Figure 3: 𝑦 = 𝑙𝑜𝑔(𝑥) in red & 𝑦 = 2 𝑙𝑜𝑔(𝑥) in green. - when 𝑎 is negative, the parent is reflected along the x-axis. See Figure 4 below. Figure 4: 𝑦 = 𝑙𝑜𝑔(𝑥) in red & 𝑦 =− 𝑙𝑜𝑔(𝑥) in green.  Logarithmic Functions - Graphs & MCR3U1 Cycle 2 Lesson 3 Key Characteristics - if a was 0, it would be a horizontal line (y=d), and therefore not a logarithmic function. b, the HS factor: - Horizontally stretches (when |𝑏| < 1) or compresses (when |𝑏| > 1 ) the function. See Figure 5 below and Figure 6 below. 1 Figure 5: 𝑦 = 𝑙𝑜𝑔(𝑥) in red & 𝑦 = 𝑙𝑜𝑔(5𝑥) in green. Figure 6: 𝑦 = 𝑙𝑜𝑔(𝑥) in red & 𝑦 = 𝑙𝑜𝑔 ( 5 𝑥) in green. - when 𝑏 is negative, the parent is reflected along the y-axis. See Figure 7 below. Figure 7: 𝑦 = 𝑙𝑜𝑔(𝑥) in red & 𝑦 = 𝑙𝑜𝑔 (− 𝑥) in green. - 𝑏 ≠ 0 because a number multiplied by 0 is zero so the input would be the horizontal translation factor (c) and therefore have the same output and be a horizontal line. Plus if c was 0, the input would be 0, which is not in a logarithim’s domain: 𝐷{𝑐 < 𝑥 ≤+ ∞}. c, the HT factor: - when c> 0, the function moves c units to the right. See Figure 8 below. - when c < 0, the function moves c units to the left. See Figure 9 below. - it is also the vertical asymptote (x=c) which explains the domain (𝐷{𝑐 < 𝑥 ≤+ ∞}).  Logarithmic Functions - Graphs & MCR3U1 Cycle 2 Lesson 3 Key Characteristics Figure 8: 𝑦 = 𝑙𝑜𝑔(𝑥) in red & 𝑦 = 𝑙𝑜𝑔 (𝑥 − 2) in green. Figure 9: 𝑦 = 𝑙𝑜𝑔(𝑥) in red & 𝑦 = 𝑙𝑜𝑔 (𝑥 + 5) in green. d, the VT factor: - when 𝑑 > 0, the parent moves 𝑑 units up. See Figure 10 below. - when 𝑑 < 0 the parent moves 𝑑 units down. See Figure 11 below. Figure 10: 𝑦 = 𝑙𝑜𝑔(𝑥) in red & 𝑦 = 𝑙𝑜𝑔 (𝑥) + 3 in green. Figure 10: 𝑦 = 𝑙𝑜𝑔(𝑥) in red & 𝑦 = 𝑙𝑜𝑔 (𝑥) − 3 in green. - a logarithmic function does not have a horizontal asymptote, so the range is 𝑅{− ∞ ≤ 𝑦 ≤+ ∞}. Graphing: To graph, all you need is to use the mapping rule and the key points of the parent function. The parent always has the points (1,0) and (m,1) and the VA: x=0. 1 On a transformed function (𝑎𝑙𝑜𝑔#(𝑏(𝑥 − 𝑐)) + 𝑑) the points would be ( 𝑏 + 𝑐, 𝑑) and 1 ( 𝑏 𝑚 + 𝑐, 𝑎 + 𝑑) and the VA (x=c).  Trigonometric Functions - Graphs & MCR3U1 Cycle 2 Lesson 4 Key Characteristics Learning Goals 1. Identify the key characteristics of sine and cosine functions. 2. Identify the key characteristics of the tangent function. 3. Identify the key characteristics of cosecant, secant, and cotangent functions. Sine functions and cosine functions are very similar. One could be the transformation of another. For all sine and cosine functions, they have: - Period: all the trigonometric functions are periodic, which means they repeat themselves with the change in x-value. The period of a function is the least increase in x required for the function to start repeating itself. Represented by the green line in the diagram. The period of the base function is 360. - Cycle: One complete pattern. In the diagram, one cycle can be represented by a red sine wave line over a period of 360. - Amplitude: half of the height of a cycle. It is always positive. Represented by the blue line in the diagram. - Equation of the axis: the line where the sine or cosine function dances around. It is the middle line of the peak and trough. In this diagram, the equation of the axis is 𝑦 = 1 - Phase shift: The horizontal movement of the function, compared to the original function 𝑦 = 𝑠𝑖𝑛 𝑥 or 𝑦 = 𝑐𝑜𝑠 𝑥. In this diagram, the phase shift is 0. (Figure 1, explanation of sine and cosine functions. )  Trigonometric Functions - Graphs & MCR3U1 Cycle 2 Lesson 4 Key Characteristics Tangent Function: - Asymptote: The value where the function is undefined. For the base function 𝑦 = 𝑡𝑎𝑛 𝑥, the asymptotes are 𝑥 = 90 + 180𝑛 where 𝑛 ∈ 𝑍. - Equation of axis: the corresponding y value of the x value between two consecutive asymptotes. For the base tangent function, the equation of the axis is 𝑦 = 0. - Cycle: one complete pattern. In the diagram, one cycle can be represented by - Period: same as sine and cosine, except its base has a period of 180. (Figure 2, visual explanation of tangent functions. )  Trigonometric Functions - Graphs & MCR3U1 Cycle 2 Lesson 4 Key Characteristics Cosecant & secant functions: 1 1 - 𝑐𝑠𝑐 θ = 𝑠𝑖𝑛θ and 𝑠𝑒𝑐 θ = 𝑐𝑜𝑠θ because anything divided by 0 is undefined, there will be asymptotes for these functions when the denominator is 0. Each asymptote is half a period apart. Also, because the peak for sine and cosine functions is 1 and -1, the secant and cosecant functions are always greater or equal to 1 and smaller or equal to -1. - Period: just like sine and cosine functions, they both have a period of 360. - Peak: the place where it is the closest to 𝑦 = 0 or, in other words, 𝑠𝑖𝑛 θ =± 1 or 𝑐𝑜𝑠 θ =± 1. With the peak, period, and asymptotes, it is easy to graph cosecant and secant functions using the mapping rules. (red is cosecant, blue is secant, dashed lines are asymptotes, respectively.) (Figure 3, visual explanation of cosecant and secant functions)  Trigonometric Functions - Graphs & MCR3U1 Cycle 2 Lesson 4 Key Characteristics Cotangent functions: 1 𝑐𝑜𝑠θ - 𝑐𝑜𝑡 θ = 𝑡𝑎𝑛θ = 𝑠𝑖𝑛θ Therefore, it is very similar to the tangent function. - For parent function, − 𝑐𝑜𝑡 (θ ± 90) = 𝑡𝑎𝑛 θ - They both have a period of 180. - They both have an asymptote on each side of a cycle. (red is cotangent, blue is tangent, and dashed lines are asymptotes, respectively.) (Figure 4, visual explanation of cotangent functions and its asymptotes.) MCR3U1 Trigonometric Functions - Graphs & Transformations Cycle 2 Lesson 4/5 Trigonometric Functions - Graphs & Key Characteristics Learning Goals 1. I can graph sinusoidal functions given an equation using both transformations and mapping rules. 2. I understand and can identify the key characteristics of sinusoidal functions. 3. I can state the domain and range for sine, cosine and tangent functions. Trigonometric Functions is a broad term referring to functions that relate an angle in a right angle triangle to the ratio of two of its sides. Three key types of trigonometric functions are sine, cosine and tangent. A parent function is a core representation of a function without any transformations(to learn more about the difference between a function and a relation look at lesson 1.7 & 1.12) 𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑛(θ) = 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑐𝑜𝑠(θ) = 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑡𝑎𝑛(θ) = 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 Key Characteristics of a Sinusoidal Function: Figure 1.0: Parent sine function with labelled key characteristics: Maximum & Minimum, Phase Shift, Equation of Axis, Amplitude and Period. MCR3U1 Trigonometric Functions - Graphs & Transformations Cycle 2 Lesson 4/5 1. Period of a Function: ➔ Periodic functions such as sine or cosine are functions that repeat at consistent intervals. ➔ Its period is simply the distance between two repeating points on the function’s graph. ◆ The two points can be two maximum points(shown in the figure to the right), minimum points or any other points ➔ Generally, the period of a parent trigonometric function is 360° but it can change with various transformations. Figure 2.0: Sine function, with period of 720° labelled. How to find the Period of a Function: 𝑔(𝑥) = 𝑎𝑓(𝑏(𝑥 + 𝑐)) + 𝑑 ➔ The function 𝑔(𝑥) is a transformed functions from the parent function 𝑓(𝑥). In this case, the parent functions would be a sinusoidal function like 𝑠𝑖𝑛(𝑥) and 𝑐𝑜𝑠(𝑥). For more information about parent functions and transformations, look at the notes for lesson 1.7 and 1.12. ➔ If 𝑏 < 0, it simply means a horizontal reflection over the x-axis. ➔ In this simplified formula, b is the horizontal stretch or compression of the function. ➔ If it is a horizontal stretch, when 0 < |𝑏| < 1, the distance between any two repeating points increases 𝑓(𝑥) and so does the period(Look at Figure 3.0) ➔ If it is a horizontal compression, when |𝑏| > 1, the distance between any two repeating points decreases related to 𝑓(𝑥) and so does the period. (Look at Figure 4.0) 360 ➔ You can calculate period by using the formula: 𝑃𝑒𝑟𝑖𝑜𝑑 = 𝑏 Figure 3.0 and 4.0: 3 Cosine functions with different periods indicated in the legends. MCR3U1 Trigonometric Functions - Graphs & Transformations Cycle 2 Lesson 4/5 2. Amplitude of a Function: ➔ Amplitude of a trigonometric function is the vertical distance between the average value of f(x) – known as the Equation of Axis– and the maximum or minimum value of f(x). ➔ Amplitude of any parent trigonometric function is 1, however it can change with the presence of a vertical stretch. ➔ In simple terms, it can impact the range of the functions and the value between which it oscillates(goes up or down). Figure 5.0: Graph of a parent sine function, with the amplitude of a labelled. How to find the Amplitude of a Function: 𝑔(𝑥) = 𝑎𝑓(𝑏(𝑥 + 𝑐)) + 𝑑 ➔ In this simplified equation, the variable a represents the amplitude of the function. ➔ If a< 0, there is a reflection over the x-axis. ➔ a is representation of the vertical reflection; in the case of a trigonometric function, it impacts the amplitude → the amount the graph travels above and below the Equation of Axis(discussed in the following section). 𝑀𝑎𝑥 − 𝑀𝑖𝑛 ◆ From a graph, you can find the amplitude by using the formula: 2 ◆ For example, 𝑓(𝑥) = 3𝑠𝑖𝑛(10𝑥), the amplitude is 3 → therefore the maximum will be 3 and the minimum is -3. 3−(−3) 6 Proof: 2 = 2 ∴ 𝑡ℎ𝑒 𝑎𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 𝑖𝑠 3! 3. Equation of Axis of a Function: ➔ The Equation of Axis is a horizontal line in the middle of the sinusoidal function. ➔ It is the average between the maximum and minimum of the function; thus, the point from wh