Chapter Function Transformations. Horizontal and Vertical Translations A translation can move the graph of a function up or down (vertical translation) and right or left (horizontal translation). A translation moves each point on the graph b the same fied amount so that the location of the graph changes but its shape and orientation remain the same. A vertical translation of function = f () b k units is written k = f (). Each point (, ) on the graph of the base function is mapped to (, + k) on the transformed function. Note that the sign of k is opposite to the sign in the equation of the function. If k is positive, the graph of the function moves up. Eample: In 7 = f (), k = 7. Each point (, ) on the graph of = f () is mapped to (, + 7). If f () =, as illustrated, (, ) maps to (, 8). If k is negative, the graph of the function moves down. Eample: In + = f (), k =. Each point (, ) on the graph of = f () is mapped to (, ). If f () =, (, ) maps to (, 3). - 6 8 A horizontal translation of function = f () b h units is written = f ( h). Each point (, ) on the graph of the base function is mapped to ( + h, ) on the transformed function. Note that the sign of h is opposite to the sign in the equation of the function. = f () (-3, ) 8 6 (, ) (5, ) If h is positive, the graph of the function shifts to the right. = f ( + 5) = f ( - 3) Eample: In = f ( 3), h = 3. Each point - -6 - - 6 8 (, ) on the graph of = f () is mapped to ( + 3, ). If f () = -, (, ) maps to (5, ). If h is negative, the graph of the function shifts to the left. Eample: In = f ( + 5), h = 5. Each point (, ) on the graph of = f () is mapped to ( 5, ). If f () =, (, ) maps to ( 3, ). Vertical and horizontal translations ma be combined. The graph of k = f ( h) maps each point (, ) in the base function to ( + h, + k) in the transformed function. 8 6 - -3 - (, 8) (, ) (, -3) - 7 = f () = f () + = f () -8 6 978--7-7389- Pre-Calculus Student Workbook MHR
. Reflections and Stretches A reflection creates a mirror image of the graph of a function across a line of reflection. An points where the function crosses the line of reflection do not move (invariant points). A reflection ma change the orientation of the function but its shape remains the same. Vertical reflection: Horizontal reflection: = f() = f ( ) (, ) (, ) (, ) (, ) line of reflection: -ais line of reflection: -ais also known as a reflection in the -ais also known as a reflection in the -ais R RR R A stretch changes the shape of a graph but not its orientation. A vertical stretch makes a function shorter (compression) or taller (epansion) because the stretch multiplies or divides each -coordinate b a constant factor while leaving the -coordinate unchanged. A horizontal stretch makes a function narrower (compression) or wider (epansion) because the stretch multiplies or divides each -coordinate b a constant factor while leaving the -coordinate unchanged. Vertical stretch b a factor of a : Horizontal stretch b a factor of b : = a f () or a = f () = f (b) (, ) (, a) (, ) ( b, ) shorter: < a < wider: < b < taller: a > narrower: b > A A A a B B B b 978--7-7389- Pre-Calculus Student Workbook MHR 9
.3 Combining Transformations Tpes of transformations include stretches, reflections, and translations. Multiple transformations can be applied to the same function. The same order of operations followed when ou work with numbers (sometimes called BEDMAS) applies to transformations: first multiplication and division (stretches, reflections), and then addition/ subtraction (translations). k = af (b( h)) The following three-step process will help ou to keep organized. Step : horizontal stretch b a factor of followed b reflection in the -ais if b < b Step : vertical stretch b a factor of a followed b reflection in the -ais if a < Step 3: horizontal and/or vertical translations (h and k) (, ) ( b, ) ( b, a ) ( b + h, a + k ) = f () Horizontal stretch about the -ais b a factor of b Vertical stretch about the -ais b a factor of a Horizontal translation of h units and/or vertical translation of k units Reflection in the -ais if b < Reflection in the -ais if a < - k = af(b( - h)) 8 MHR Chapter 978--7-7389-
. Inverse of a Relation The inverse of a function = f () is denoted = f () if the inverse is a function. The is not an eponent because f represents a function, not a variable. You have alread seen this notation with trigonometric functions. Eample: sin (u), where f (u) = sin(u) and the variable is u. The inverse of a function reverses the processes represented b that function. For eample, the process of squaring a number is reversed b taking the square root. The process of taking the reciprocal of a number is reversed b taking the reciprocal again. To determine the inverse of a function, interchange the - and -coordinates. (, ) (, ) or 5 f () 5 f () or refl ect in the line 5 When working with an equation of a function = f (), interchange for. Then, solve for to get an equation for the inverse. If the inverse is a function, then = f (). If the inverse of a function is not a function (recall the vertical line test), restrict the domain of the base function so that the inverse becomes a function. You will see this frequentl with quadratic functions. For eample, the inverse of f () =,, is f () =. The inverse will be a function onl if the domain of the base function is restricted. Restricting the domain is necessar for an function that changes direction (increasing to decreasing, or vice versa) at some point in the domain of the function. 6 MHR Chapter 978--7-7389-
Chapter Radical Functions. Radical Functions and Transformations Base Radical Function The base radical function = has the following graph and properties: -intercept of -intercept of 8 domain: {, R} range: {, R} 6 The intercepts and domain and range suggest an endpoint at (, ), and no right endpoint. The graph is shaped like half of a parabola. The domain and range indicate that the half parabola is in the first quadrant. Transforming Radical Functions The base radical function = is transformed b changing the values of the parameters a, b, h, and k in the equation = a b( h) + k. The parameters have the following effects on the base function: = 6 8 a b h k vertical stretch b a factor of a if a is a <, the graph of = is reflected in the -ais horizontal stretch b a factor of b if b is b <, the graph of = is reflected in the -ais horizontal translation ( h) means the graph of = moves h units right. For eample, = means that the graph of = moves unit right. ( + h) means the graph of = moves h units left. For eample, = + 5 means that the graph of = moves 5 units left. This translation has the opposite effect than man people think. It is a common error to think that the + sign moves the graph to the right and the sign moves the graph to the left. This is not the case. vertical translation + k means the graph of = moves k units up k means the graph of = moves k units down 978--7-7389- Pre-Calculus Student Workbook MHR 39
. Square Root of a Function Graphing 5 f () and 5 f () To graph = f (), ou can set up a table of values for the graph of = f (). Then, take the square root of the elements in the range, while keeping the elements in the domain the same. When graphing = f (), pa special attention to the invariant points, which are points that are the same for = f () as the are for = f (). The invariant points are (, ) and (, ) because when f () =, f () =, and when f () =, f () =. 8 6 f() = + f() = + (-.5, ) (, ) -8-6 - - 6 8 - - -6-8 Domain and Range of 5 f () You cannot take the square root of a negative number, so the domain of = f() is an value for which f (). The range is the square root of an value in = f () for which = f() is defined. The Graph of 5 f () f (), f () 5, f (), f () 5 f (). = f () is undefined because ou cannot take the square root of a negative number. The graphs of = f () and = f () intersect at =. The graph of = f () is above the graph of = f (). The graphs of = f () and = f () intersect at =. The graph of = f () is below the graph of = f (). 978--7-7389- Pre-Calculus Student Workbook MHR 7
.3 Solving Radical Equations Graphicall Strateg for Solving Algebraicall Step : List an restrictions for the variable. You cannot take the square root of a negative number, so the value of the variable must be such that an operations under the radical sign result in a positive value. Step : Isolate the radical and square both sides of the equation to eliminate the radical. Then, solve for. Step 3: Find the roots of the equation (that is, the value(s) of that make the equation have a value of zero). Step : Check the solution, ensuring that it does not contain etraneous roots (solutions that do not satisf the original equation or restrictions when substituted in the original equation). Eample: _ 7 = +, Identif restrictions. _ 3 = Isolate the radical. _ 3 = ( ) Square both sides. 9 = Solve for. 3 = Check: Solution meets the restrictions. _ 7 = 3 + 7 = 9 + 7 = 7 Strategies for Solving Graphicall Method : Graph a Single Equation Method : Graph Two Equations Graph the corresponding function and Graph each side of the equation on find the zero(s) of the function. the same grid, and find the point(s) of Eample: intersection. + + = + 6 Eample: + = + + = + 6 Graph = +. Graph = + + and = + 6. = 3 = 3 978--7-7389- Pre-Calculus Student Workbook MHR 55
Chapter 3 Polnomial Functions 3. Characteristics of Polnomial Functions What Is a Polnomial Function? A polnomial function has the form f () = a n n + a n n + a n n + + a + a + a where n is a whole number is a variable the coefficients a n to a are real numbers the degree of the polnomial function is n, the eponent of the greatest power of the leading coefficient is a n, the coefficient of the greatest power of the constant term is a Tpes of Polnomial Functions Constant Function Linear Function Quadratic Function Degree Degree Degree f() f() f() 6 f() = 3 f() = + - - - - - - - - f() = - 3 - - - Cubic Function Quartic Function Quintic Function Degree 3 Degree Degree 5 f() 6 f() 6 f() f() = 5 + 3-5 3-5 + + 6 8 - - - - - - - f() = 3 + - - -6-8 - -6-8 f() = + 5 3 + 5-5 - 6 - - - -8-6 66 MHR Chapter 3 978--7-7389-
Characteristics of Polnomial Functions Graphs of Odd-Degree Polnomial Functions etend from quadrant III to quadrant I when the leading coefficient is positive, similar to the graph of = etend from quadrant II to IV when the leading coefficient is negative, similar to the graph of = = 3 + - - = = - - - - - - - - - = - 3 + + - 3 have at least one -intercept to a maimum of n -intercepts, where n is the degree of the function have -intercept a, the constant term of the function have domain { R} and range { R} have no maimum or minimum values Graphs of Even-Degree Polnomial Functions open upward and etend from quadrant II to quadrant I when the leading coefficient is positive, similar to the graph of = f() = open downward and etend from quadrant III to IV when the leading coefficient is negative, similar to the graph of = f() = - + 6 + - 5 - - - - - - - - -6-6 -8 = + 5 3 + 5-5 - 6-8 = - have from to a maimum of n -intercepts, where n is the degree of the function have -intercept a, the constant term of the function have domain { R}; the range depends on the maimum or minimum value of the function have a maimum or minimum value 978--7-7389- Pre-Calculus Student Workbook MHR 67
3. The Remainder Theorem Long Division You can use long division to divide a polnomial b a binomial: P() a = Q() + R a The components of long division are the dividend, P(), which is the polnomial that is being divided the divisor, a, which is the binomial that the polnomial is divided b the quotient, Q(), which is the epression that results from the division the remainder, R, which is the value or epression that is left over after dividing To check the division of a polnomial, verif the statement P() = ( a)q() + R. Snthetic Division a short form of division that uses onl the coefficients of the terms it involves fewer calculations Remainder Theorem When a polnomial P() is divided b a binomial a, the remainder is P(a). If the remainder is, then the binomial a is a factor of P(). If the remainder is not, then the binomial a is not a factor of P(). Working Eample : Divide a Polnomial b a Binomial of the Form a a) Divide P() = 9 + 3 b +. Epress the result in the form P() a = Q() + R a. b) Identif an restrictions on the variable. c) Write the corresponding statement that can be used to check the division. Solution a) + 3 + + 9 Wh is the order of the terms different? Wh is it necessar to include the term? & See Eample on page of Pre-Calculus for help with long division. 3 + 9 = + 78 MHR Chapter 3 978--7-7389-
3.3 The Factor Theorem Factor Theorem The factor theorem states that a is a factor of a polnomial P() if and onl if P(a) =. If and onl if means that the result works both was. That is, if a is a factor then, P(a) = if P(a) =, then a is a factor of a polnomial P() Integral Zero Theorem The integral zero theorem describes the relationship between the factors and the constant term of a polnomial. The theorem states that if a is a factor of a polnomial P() with integral coefficients, then a is a factor of the constant term of P() and = a is an integral zero of P(). Factor b Grouping If a polnomial P() has an even number of terms, it ma be possible to group two terms at a time and remove a common factor. If the binomial that results from common factoring is the same for each pair of terms, then P() ma be factored b grouping. Steps for Factoring Polnomial Functions To factor polnomial functions using the factor theorem and the integral zero theorem, use the integral zero theorem to list possible integer values for the zeros net, appl the factor theorem to determine one factor then, use division to determine the remaining factor repeat the above steps until all factors are found Working Eample : Use the Factor Theorem to Test for Factors of a Polnomial Which binomials are factors of the polnomial P() = 3 + + 6? Justif our answers. a) b) c) + d) + 3 Solution Use the factor theorem to evaluate P(a) given a. a) For, substitute = into the polnomial epression. P( ) = Since the remainder is, a factor of P(). (is or is not) 8 MHR Chapter 3 978--7-7389-
3. Equations and Graphs of Polnomial Functions Sketching Graphs of Polnomial Functions To sketch the graph of a polnomial function, use the -intercepts, the -intercept, the degree of the function, and the sign of the leading coefficient. The -intercepts of the graph of a polnomial function are the roots of the corresponding polnomial equation. Determine the zeros of a polnomial function from the factors. Use the factor theorem to epress a polnomial function in factored form. Multiplicit of a Zero If a polnomial has a factor a that is repeated n times, then = a is a zero of multiplicit n. The multiplicit of a zero or root can also be referred to as the order of the zero or root. The shape of a graph of a polnomial function close to a zero of = a (multiplicit n) is similar to the shape of the graph of a function with degree equal to n of the form = ( a) n. Polnomial functions change sign at -intercepts that correspond to odd multiplicit. The graph crosses over the -ais at these intercepts. Polnomial functions do not change sign at -intercepts of even multiplicit. The graph touches, but does not cross, the -ais at these intercepts. zero of multiplicit zero of multiplicit zero of multiplicit 3 Transformation of Polnomial Functions To sketch the graph of a polnomial function of the form = a[b( h)] n + k or k = a[b( h)] n, where n N, appl the following transformations to the graph of = n. Note: You ma appl the transformations represented b a and b in an order before the transformations represented b h and k. Parameter k h a Transformation Vertical translation up or down (, ) (, + k) Horizontal translation left or right (, ) ( + h, ) Vertical stretch about the -ais b a factor of a For a <, the graph is also reflected in the -ais (, ) (, a) 978--7-7389- Pre-Calculus Student Workbook MHR 9
b Horizontal stretch about the -ais b a factor of b For b <, the graph is also reflected in the -ais (, ) ( b, ) Working Eample : Analse Graphs of Polnomial Functions For each graph of a polnomial function, determine the least possible degree the sign of the leading coefficient the -intercepts and the factors of the function with least possible degree the intervals where the function is positive and the intervals where it is negative a) b) 6 8 8-6 - - 6 - - -8 - -6-8 - - -6 - Solution a) There are -intercepts; the are. The -intercept of multiplicit is. The -intercept of multiplicit is. The least possible degree of the graph is. The graph etends from quadrant to quadrant. The leading coefficient is. (positive or negative) The factors are. The function is positive for values of in the interval(s). The function is negative for values of in the interval(s). 9 MHR Chapter 3 978--7-7389-
Chapter Trigonometr and the Unit Circle. Angles and Angle Measure One radian is the measure of the central angle subtended in a circle b an arc equal in length to the radius of the circle. Travelling one rotation around the circumference of a circle causes the terminal arm to turn πr. Since r = on the unit circle, πr can be epressed as π, or π radians. You can use this information to translate rotations into radian measures. For eample, B r r r A full rotation (36 ) is π radians 6 rotation (6 ) is π 3 radians rotation (8 ) is π radians 8 rotation (5 ) is π radians rotation (9 ) is π radians rotation (3 ) is π 6 radians Angles in standard position with the same terminal arms are coterminal. For an angle in standard position, an infinite number of angles coterminal with it can be determined b adding or subtracting an number of full rotations. Counterclockwise rotations are associated with positive angles. Clockwise rotations are associated with negative angles. quadrant I angle positive angle > 36 negative angle The general form of a coterminal angle (in degrees) is θ ± 36 n, where n is a natural number (,,, 3, ) and represents the number of revolutions. The general form (in radians) is θ ± πn, n N. Radians are especiall useful for describing circular motion. Arc length, a, means the distance travelled along the circumference of a circle of radius r. For a central angle θ, in radians, a = θr. 978--7-7389- Pre-Calculus Student Workbook MHR 9
. The Unit Circle In general, a circle of radius r centred at the origin has equation + = r. The unit circle has radius and is centred at the origin. The equation of the unit circle is + =. All points P(, ) on the unit circle satisf this equation. P(, ) A An arc length measured along the unit circle equals the measure of the central angle (in radians). In other words, when r =, the formula a = θr simplifies to a = θ. Recall the special right triangles ou learned about previousl. 5 3 3 5 6 These special triangles can be scaled to fit within the unit circle (r = ). θ = π θ P, θ = π 3 θ 3 P, 3 MHR Chapter 978--7-7389-
.3 Trigonometric Ratios These are the primar trigonometric ratios: sine cosine tangent sin θ = r cos θ = r tan θ = For points on the unit circle, r =. Therefore, the primar trigonometric ratios can be epressed as: sin θ = = cos θ = = tan θ = P(θ) = (, ) θ A B(, ) Since cos θ simplifies to and sin θ simplifies to, ou can write the coordinates of P(θ) as P(θ) = (cos θ, sin θ) for an point P(θ) at the intersection of the terminal arm of θ and the unit circle. These are the reciprocal trigonometric ratios: cosecant secant cotangent csc θ = sec θ = sin θ cos θ cot θ = tan θ csc θ = r r sec θ = cot θ = Recall from the CAST rule that sin θ and csc θ are positive in quadrants I and II cos θ and sec θ are positive in quadrants I and IV tan θ and cot θ are positive in quadrants I and III II SIN I ALL TAN COS III IV 978--7-7389- Pre-Calculus Student Workbook MHR 9
. Introduction to Trigonometric Equations Solving an equation means to determine the value (or values) of a variable that make an equation true (Left Side = Right Side). For eample, sin θ = is true when θ = 3 or θ = 5, and for ever angle coterminal with 3 or 5. These angles are solutions to a ver simple trigonometric equation. The variable θ is often used to represent the unknown angle, but an other variable is allowed. In general, solve for the trigonometric ratio, and then determine all solutions within a given domain, such as θ < π or all possible solutions, epressed in general form, θ + πn, n I Unless the angle is a multiple of 9 or π, there will be two angles per solution of the equation within each full rotation of 36 or π. As well, there will be two epressions in general form per solution, one for each angle. It is sometimes possible to write a combined epression representing both angles in general form. If the angle is a multiple of 9 or π (that is, the terminal arm coincides with an ais), then there will be at least one angle within each full rotation that is a correct solution to the equation. Note that sin θ = (sin θ). Also, recall that sin θ and csc θ are positive in quadrants I and II cos θ and sec θ are positive in quadrants I and IV tan θ and cot θ are positive in quadrants I and III II SIN TAN ALL COS I III IV 38 MHR Chapter 978--7-7389-
Chapter 5 Trigonometric Functions and Graphs 5. Graphing Sine and Cosine Functions Sine and cosine functions are periodic or sinusoidal functions. The values of these functions repeat in a regular pattern. These functions are based on the unit circle. Consider the graphs of = sin θ and = cos θ. amplitude.5. period.5. period.5.5 amplitude -π - π π π 3π π 5π 3π θ -.5 -π - π π π 3π π 5π 3π θ -.5 -. -.5 = sin θ -. -.5 = cos θ The maimum value is +. The minimum value is. The amplitude is. The period is π. The -intercept is. The θ-intercepts on the given domain are π,, π, π, and 3π. The domain of = sin θ is {θ θ R}. The range of = sin θ is {, R}. The maimum value is +. The minimum value is. The amplitude is. The period is π. The -intercept is. The θ-intercepts on the given domain are π, π, 3π, and 5π. The domain of = cos θ is {θ θ R}. The range of = cos θ is {, R}. For sinusoidal functions of the form = a sin b or = a cos b, a represents a vertical stretch of factor a and b represents a horizontal stretch of factor. Use the following ke b features to sketch the graph of a sinusoidal function. the maimum and minimum values the amplitude, which is one half the total height of the function maimum value minimum value Amplitude = The amplitude is given b a. the period, which is the horizontal length of one ccle on the graph of a function Period = π or 36 b b Changing the value of b changes the period of the function. the coordinates of the horizontal intercepts 978--7-7389- Pre-Calculus Student Workbook MHR 9
5. Transformations of Sinusoidal Functions You can appl the same transformation rules to sinusoidal functions of the form = a sin b(θ c) + d or = a cos b(θ c) + d. A vertical stretch b a factor of a changes the amplitude to a. = a sin θ = a cos θ If a <, the function is reflected through the horizontal mid-line of the function. A horizontal stretch b a factor of 36 changes the period to or π radians. b b b = sin (bθ) = cos (bθ) If b <, the function is reflected in the -ais. For sinusoidal functions, a horizontal translation is called the phase shift. = sin (θ c) = cos (θ c) If c >, the function shifts c units to the right. If c <, the function shifts c units to the left. The vertical displacement is a vertical translation. = sin θ + d = cos θ + d If d >, the function shifts d units up. If d <, the function shifts d units down. maimum value + minimum value d = The sinusoidal ais is defined b the line = d. It represents the mid-line of the function. Appl transformations of sinusoidal functions in the same order as for an other functions: i) horizontal stretches and reflections, b ii) vertical stretches and reflections, a iii) translations, c and d The domain of a sinusoidal function is not affected b transformations. The range of a sinusoidal function, normall {, R}, is affected b changes to the amplitude and vertical displacement. Consider the graph of = sin ( π ) +. d 3 c a π b a =, so the amplitude is b =, so the period is π, or π c = π, so the graph is shifted π units right d =, so the graph is shifted unit up - π _ π_ π_ 3π π 5π 3π 7π domain: { R} range: { 3, R} 58 MHR Chapter 5 978--7-7389-
5.3 The Tangent Function The graph of the tangent function, = tan, is periodic, but it is not sinusoidal. 8 = tan 6 - π_ π 3π π 5π - -6-8 These are the characteristics of the tangent function graph, = tan : It has period π or 8. It is discontinuous where tan is undefined, that is, when = π 3π,, 5π,, π + nπ, n I. The discontinuit is represented on the graph of = tan as vertical asmptotes. The domain is ( π + nπ, R, n I ). It has no maimum or minimum values. The range is { R}. It has -intercepts at ever multiple of π:, π, π,, nπ, n I. Each of the -intercepts is a turning point, where the slope changes from decreasing to increasing. On the unit circle, ou can epress the coordinates of the point P on the terminal arm of angle θ as (, ) or (sin θ, cos θ). The slope of the terminal arm is represented b the tangent function: slope = = = = tan θ OR slope = sin θ cos θ = tan θ Therefore, ou can use the tangent function to model the slope of a line from a fied point to a moving object as the object moves through a range of angles. 978--7-7389- Pre-Calculus Student Workbook MHR 67
5. Equations and Graphs of Trigonometric Functions You can use sinusoidal functions to model periodic phenomena that do not involve angles as the independent variable. Eamples of such phenomena include wave shapes, such as a heartbeat or ocean waves pistons in a machine or the swing of a pendulum circular motion, such as a Ferris wheel You can adjust the parameters a, b, c, and d in sinusoidal equations of the form = a sin b(θ c) + d or = a cos b(θ c) + d to fit the characteristics of the phenomenon being modelled. Graphing technolog allows ou eamine how well the model represents the data. It also allows ou to etrapolate or interpolate solutions from the model. You can find approimate solutions to trigonometric equations using the graphs of the trigonometric functions. Epress solutions over a specific interval or give a general solution. Working Eample : Solve Simple Trigonometric Equations Solve each equation over the specified interval. a) sin =.5, 7 b) sin =.5, 7 Solution a) Method : Use the Unit Circle and Special Triangles θ R = sin (.5) = The solutions are =,,,, 7. 978--7-7389- Pre-Calculus Student Workbook MHR 75
Chapter 6 Trigonometric Identities 6. Reciprocal, Quotient, and Pthagorean Identities Trigonometric Identities A trigonometric identit is a trigonometric equation that is true for all permissible values of the variable in the epressions on both sides of the equation. Reciprocal Identities csc = sin Quotient Identities tan = cos sin sec = cos cos cot = sin cot = tan Pthagorean Identities There are three forms of the Pthagorean identit: Form : Derived from the Pthagorean theorem, a + b = c, and applied to a right triangle in the unit circle where angle θ is in standard position. The hpotenuse is, the adjacent side is = cos θ, and the opposite side is θ = sin θ. Therefore, cos θ + sin θ =. Form : Divide both sides of form b sin θ and appl the quotient and reciprocal identities. cos θ sin θ + sin θ sin θ = sin θ cot θ + = csc θ Form 3: Divide both sides of form b cos θ and appl the quotient and reciprocal identities. cos θ cos θ + sin θ cos θ = cos θ + tan θ = sec θ Verification and Use of Trigonometric Identities Trigonometric identities can be verified in two was: i) numericall, b substituting specific values for the variable ii) graphicall, using technolog P(cos θ, sin θ) Verifing that two sides of an equation are equal for given values, or that the appear equal when graphed, is not sufficient to conclude that the equation is an identit. You can use trigonometric identities to simplif more complicated trigonometric epressions. 88 MHR Chapter 6 978--7-7389-
6. Sum, Difference, and Double-Angle Identities Sum and Difference Identities The sum and difference identities are used to simplif epressions and to determine eact trigonometric values of some angles. Sum Identities Eamples: sin (A + B) = sin A cos B + cos A sin B sin ( + 3 ) = sin cos 3 + cos sin 3 cos (A + B) = cos A cos B sin A sin B cos ( π + π 6 ) = cos π cos π sin π sin π 6 6 tan A + tan B tan (A + B) = tan A tan B Difference Identities sin (A B) = sin A cos B cos A sin B cos (A B) = cos A cos B + sin A sin B tan (A B) = Double-Angle Identities tan A tan B + tan A tan B tan + tan 5 tan ( + 5 ) = _ tan tan 5 sin (5 33 ) = sin 5 cos 33 cos 5 sin 33 cos ( π 3 π ) = cos π 3 cos π + sin π 3 sin tan 7 tan 35 tan (7 35 ) = + tan 7 tan 35 π Double-angle identities are special cases of the sum identities when the two angles are equal. sin A = sin A cos A The double-angle identit for cosine can be epressed in three different forms: cos A = cos A sin A cos A = cos A cos A = sin A tan A = tan A tan A Special Angles and Their Eact Trigonometric Values Eamples: sin π = sin π cos π 8 8 cos = cos 7 sin 7 cos = cos 7 cos = sin 7 tan π tan π = 6 tan π Degrees Radians sin u cos u tan u 3 π 3 6 3 5 π 6 π 3 3 3 978--7-7389- Pre-Calculus Student Workbook MHR 97
6.3 Proving Identities Guidelines for Proving Identities To prove that an identit is true for all permissible values, epress both sides of the identit in equivalent forms. One or both sides of the identit must be algebraicall manipulated into an equivalent form to match the other side. There is a major difference between solving a trigonometric equation and proving a trigonometric identit: Solving a trigonometric equation determines the value that makes a particular case true. You perform equivalent operations on both sides of the equation (that is, perform operations across the = sign) to isolate the variable and solve for the variable. Proving an identit shows that the epressions on each side of the equal sign are equivalent for all values for which the variable is defined. Therefore, ou work on each side of the identit independentl, and ou do not perform operations across the = sign. Tips for Proving Identities It is easier to simplif a complicated epression than to make a simple epression more complicated, so start with the more complicated side of the identit. Use known identities to make substitutions. If a quadratic is present, consider the Pthagorean identit first. It, or one of its alternative forms, can often be used. Rewrite the epression using sine and cosine onl. Multipl the numerator and the denominator b the conjugate of an epression. Factor to simplif epressions. Verifing Identities Identities can be verified using a specific value, but this validates that the identit is true for that value onl. Graphing each side of a possible identit ma show the identit might be true, but it does not prove the identit formall. 978--7-7389- Pre-Calculus Student Workbook MHR 5
6. Solving Trigonometric Equations Using Identities Solving Trigonometric Equations Solving a trigonometric equation means to find all the possible angle values that make the equation true within the given or restricted domain. When the domain is not restricted, ou provide a general solution. Strategies for Solving Trigonometric Equations Description Isolate the trigonometric ratio, if possible. sin = sin = Eample Factor and then set each factor equal to. Common factoring: sin tan + sin = sin (tan + ) = sin = or tan = Difference of squares: sin = (sin )(sin + ) = sin = or sin = Simplif the given equation or change the given equation to one common ratio (such as sin or cos ) b using one or more of the following: reciprocal identities quotient identities Pthagorean identities double-angle identities Then, solve. Trinomial factoring: sin sin = ( sin + )(sin ) = sin =.5 or sin = cos sin + 3 = ( sin ) sin + 3 = Replace cos with sin. sin sin + = Simplif. sin + sin = (sin + )(sin ) = Factor. sin = or sin = Solve. It is important to consider all possible solutions to ensure that the are not non-permissible values. In this eample, the root = is rejected because the minimum value for sin is. Checking Trigonometric Equations The algebraic solution can be verified graphicall. Check that solutions for an equation do not include non-permissible values from the original equation. 978--7-7389- Pre-Calculus Student Workbook MHR 5
Chapter 7 Eponential Functions 7. Characteristics of Eponential Functions An eponential function models a tpe of non-linear change. These tpes of functions have the form = c, where c is a constant (c > ). All eponential functions of this form have a -intercept of. Wh does c have to be positive? Wh is the -intercept for all eponential functions of this form equal to? When c > in an eponential function of the form = c, the eponential function is increasing. 8 6 = How can ou tell from the graph that this is an increasing function? Does this situation represent growth or deca? - (, ) When c is between and (that is, < c < ) in an eponential function of the form = c, the eponential function is decreasing. 8 6 = (, ) _ ( ) How can ou tell from the graph that this is a decreasing function? Does this situation represent growth or deca? - When c = in an eponential function of the form = c, the eponential function is neither increasing nor decreasing. Eponential functions of the form = c have domain { R}, range { >, R}, no -intercepts, and horizontal asmptote at =. How does the graph of = reflect a function that is neither increasing nor decreasing? How do the graphs above reflect the domain, range, and horizontal asmptote? 978--7-7389- Pre-Calculus Student Workbook MHR 9
7. Transformations of Eponential Functions You can use transformed eponential functions to model real-world applications of eponential growth or deca. To graph an eponential function of the form = a(c) b( h) + k, appl transformations to the base function, = c, where c >. Each of the parameters, a, b, h, and k, is associated with a particular transformation. Parameter Transformation Eample a vertical stretch about the -ais b a factor of a a < results in a reflection in the -ais (, ) (, a) For a =, the equation of the transformed base function is = (3). 8 6 = 3 = (3) - - - - = -(3) -6-8 b horizontal stretch about the -ais b a factor of b b < results in a reflection in the -ais (, ) ( b, ) For b =, the equation of the transformed base function is = (3). = (3) - 8 6 = 3 = (3) - - - 38 MHR Chapter 7 978--7-7389-
Parameter Transformation Eample h horizontal translation left or right, depending on the sign: +h shifts the graph left, and h shifts the graph right (, ) ( + h, ) For h = ±, the equation of the transformed base function is = (3) ( ± ). = 3 8 6 ( + ) = (3) ( - ) = (3) -6 - - 6 - k vertical translation up or down, depending on the sign: +k shifts the graph up, and k shifts the graph down (, ) (, + k) For k = ±, the equation of the transformed base function is = (3) ±. = 3 + = 3 6 - - = 3 - - - When appling transformations, ou must appl parameters a and b before parameters h and k. Working Eample : Translations of Eponential Functions Consider the eponential function =. For each of the following transformed functions, state the parameter and describe the transformation graph the base function and the transformed function on the same grid describe an changes to the domain, range, intercepts, and equation of the horizontal asmptote eplain the effect of the transformation on an arbitrar point, (, ), on the graph of the base function a) = + 3 b) = 5 c) = + 978--7-7389- Pre-Calculus Student Workbook MHR 39
7.3 Solving Eponential Equations Strategies for Solving Eponential Equations With a Common Base Description For equations that begin with terms on both sides of the equal sign that have the same base... For equations that begin with terms on each side of the equal sign that have different bases, but that can be rewritten as the same base... Eample ( + 6) = Bases are the same. + 6 = Equate the eponents. = Solve for. = 9 ( + 3) ( + 9) = 8 3 ( + 3) = 3 ( + 9) Rewrite terms so the have the same base. ( + 3) = ( + 9) Equate the eponents. + 6 = 8 + 36 Solve for. 6 = 3 = 5 Strategies for Solving Eponential Equations That Do Not Have a Common Base Description Eample Sstematic trial Consider the equation 7 =.. Guess : = 5:. 5 = 5.378 (less than 7) Guess : = 7:. 7 =.535 (greater than 7) Guess 3: = 6:. 6 = 7.59536 (approimatel 7) So, is approimatel 6. Graphing Method : Point of Intersection Graph = 7 and =. on the same aes, and find the point of intersection. Consider the equation 7 =.. Method : -Intercept Graph =. 7, and determine the -intercept. So, = 5.7837. 978--7-7389- Pre-Calculus Student Workbook MHR 9
Chapter 8 Logarithmic Functions 8. Understanding Logarithms A logarithm is the eponent to which a fied base must be raised to obtain a specific value. Eample: 5 3 = 5. The logarithm of 5 is the eponent that must be applied to base 5 to obtain 5. In this eample, the logarithm is 3: log 5 5 = 3. Equations in eponential form can be written in logarithmic form and vice versa. Eponential Form Logarithmic Form = c = log c The inverse of the eponential function = c, c >, c, is = c or, in logarithmic form, = log c. Conversel, the inverse of the logarithmic function = log c, c >, c, is = log c or, in eponential form, = c. The graphs of an eponential function and its inverse logarithmic function are reflections of each other in the line =. For the logarithmic function = log c, c >, c, the domain is { >, R} the range is { R} the -intercept is the vertical asmptote is =, or the -ais A common logarithm has base. It is not necessar to write the base for common logarithms: log = log Working Eample : Graph the Inverse of an Eponential Function The graph of = is shown at right. State the inverse of the function. Then, sketch the graph of the inverse function and identif the following characteristics of the graph: domain and range -intercept, if it eists -intercept, if it eists the equation of an asmptotes 8 6 (, ) = (3, 8) (, ) (, ) -8-6 - - - 6 = 8 - -6-8 6 MHR Chapter 8 978--7-7389-
8. Transformations of Logarithmic Functions To represent real-life situations, ou ma need to transform the basic logarithmic function, = log b, b appling reflections, stretches, and translations. These transformations should be performed in the same manner as those applied to an other function. The effects of the parameters a, b, h, and k in = a log c (b( h)) + k on the graph of the logarithmic function = log c are described in the table. Parameter a b h k Effect Verticall stretch b a factor of a about the -ais. Reflect in the -ais if a <. Horizontall stretch b a factor of b about the -ais. Reflect in the -ais if b <. Horizontall translate h units. Verticall translate k units. Onl parameter h changes the vertical asmptote and the domain. None of the parameters changes the range. Working Eample : Translations of a Logarithmic Function a) Sketch the graph of = log ( + ) 5. b) State the domain and range -intercept -intercept equation of the asmptote Solution a) Begin with the graph of = log. Identif ke points, such as (, ), (, ), and (6, ). Identif the transformations. The graph moves units to the left and units. 978--7-7389- Pre-Calculus Student Workbook MHR 67
8.3 Laws of Logarithms Let P be an real number, and M, N, and c be positive real numbers with c. Then, the following laws of logarithms are valid. Name Law Description Product log c MN = log c M + log c N The logarithm of a product of numbers is the sum of the logarithms of the numbers. Quotient Power log c M N = log c M log c N log c M P = P log c M The logarithm of a quotient of numbers is the difference of the logarithms of the dividend and divisor. The logarithm of a power of a number is the eponent times the logarithm of the number. Man quantities in science are measured using a logarithmic scale. Two commonl used logarithmic scales are the decibel scale and the ph scale. Working Eample : Use the Laws of Logarithms to Epand Epressions Epand each epression using the laws of logarithms. a) log 3 b) log z 5 3 Solution a) log 3 z = log log = log 3 + log (log + log z) = 3 log + log log z b) log 5 3 = log 5 ( 3 ) = log 5 ( 3 ) = (log 5 + log 5 ) = (log 5 + log 5 ) = log 5 + log 5 c) log 3 Wh does log =? 978--7-7389- Pre-Calculus Student Workbook MHR 75
8. Logarithmic and Eponential Equations When solving a logarithmic equation algebraicall, start b appling the laws of logarithms to epress one side or both sides of the equation as a single logarithm. Some useful properties are listed below, where c, L, R > and c. If log c L = log c R, then L = R. The equation log c L = R can be written with logarithms on both sides of the equation as log c L = log c c R. The equation log c L = R can be written in eponential form as L = c R. The logarithm of zero or a negative number is undefined. To identif whether a root is etraneous, substitute the root into the original equation and check whether all of the logarithms are defined. You can solve an eponential equation algebraicall b taking logarithms of both sides of the equation. If L = R, then log c L = log c R, where c, L, R > and c. Then, appl the power law for logarithms to solve for an unknown. You can solve an eponential equation or a logarithmic equation using graphical methods. Man real-world situations can be modelled with an eponential or a logarithmic equation. A general model for man problems involving eponential growth or deca is number of changes Final quantit = initial quantit (change factor) Working Eample : Solve Logarithmic Equations Solve. a) log (5 + ) = log ( + 7) b) log (5) log ( ) = c) log 6 ( 3) + log 6 ( + 6) = Solution a) Since log (5 + ) = log ( + 7), 5 + = + 7. So, = 6 and =. Check = in the original equation. Left Side Right Side log (5() +) log ( + 7) = log = log Left Side = Right Side 8 MHR Chapter 8 978--7-7389-
Chapter 9 Rational Functions 9. Eploring Rational Functions Using Transformations Rational functions are functions of the form = p(), where p() and q() are polnomial q() epressions and q(). You can graph a rational function b creating a table of values and then graphing the points in the table. To create a table of values, identif the non-permissible value(s) write the non-permissible value in the middle row of the table enter positive values above the non-permissible value and negative values below the non-permissible value choose small and large values of to give ou a spread of values You can use what ou know about the base function = and transformations to graph equations of a the form = h + k. Eample: 3 For = + 5, the values of the + parameters are a = 3, representing a vertical stretch b a factor of 3 h =, representing a horizontal translation units to the left k = 5, representing a vertical translation 5 units up vertical asmptote: = horizontal asmptotes: = 5 Some equations of rational functions can be manipulated algebraicall into the form a = + k b creating a common factor in the numerator and the denominator. h Eample: = _ 3 + 6 3 + + 6 = 3 + 8 = _ 3( ) = + 8 8 = + 3 - -8-6 - - 8 6 - - -6 = 3 + 5 + = 978--7-7389- Pre-Calculus Student Workbook MHR 97
9. Analsing Rational Functions Determining Asmptotes and Points of Discontinuit The graph of a rational function ma have an asmptote, a point of discontinuit, or both. To establish these important characteristics of a graph, begin b factoring the numerator and denominator full. Asmptotes: No Common Factors If the numerator and denominator do not have a common factor, the function has an asmptote. The vertical asmptotes are identified b the non-permissible values of the function. For a function that can be rewritten in a the form = + k, the k parameter h identifies the horizontal asmptote. Points of Discontinuit: At Least One Common Factor If the numerator and denominator have at least one common factor, there is at least one point of discontinuit in the graph. Equate the common factor(s) to zero and solve for to determine the -coordinate of the point of discontinuit. Substitute the -value in the simplified epression to find the -coordinate of the point of discontinuit. Both Asmptote(s) and Point(s) of Discontinuit If a rational epression remains after removing the common factor(s), there ma be both a point of discontinuit and asmptotes. Eample: = + 3 Since the non-permissible value is = 3, the vertical asmptote is at = 3. = + 3 = 3 + 3 + 3 = 3 3 + 7 3 7 = 3 + Since k =, the horizontal asmptote is at =. ( )( + ) Eample: = + + = : the -coordinate of the point of discontinuit is. Substitute = into the simplified equation: = = = 6 point of discontinuit: (, 6) Eample: ( )( + ) = ( + )( ) ( ) = _ ( ) common factor: +, so there is a point of discontinuit at (, ) non-permissible value: =, so the vertical asmptote is at = simplified function can be rewritten as 3 = +, so the horizontal asmptote is at = 978--7-7389- Pre-Calculus Student Workbook MHR 35
9.3 Connecting Graphs and Rational Equations Solving Rational Equations You can solve rational equations algebraicall or graphicall. Algebraicall Solving algebraicall determines the eact solution and an etraneous roots. To solve algebraicall, Equate to zero and list the restrictions. Factor the numerator and denominator full (if possible). Multipl each term b the lowest common denominator to eliminate the fractions. Solve for. Check the solution(s) against the restrictions. Check the solution(s) in the original equation. Graphicall There are two methods for solving equations graphicall. Method : Use a Sstem of Two Functions Graph each side of the equation on the same set of aes. The solution(s) will be the -coordinate(s) of an point(s) of intersection. Method : Use a Single Function Rearrange the equation so that one side is equal to zero. Graph the corresponding function. The solution(s) will be the -intercept(s). Eample: ( + 6)() + ( + 6 ) ( 6 6 + 6 = 6 + + 6 =, 6 6 ( + 6) ( + + 6 ) = ( + 6)() ( + 6)() = + 6 ) + 6 + 6 = + 8 = ( + )( ) = roots: = and = 6 Eample: + 6 = 6 Graph = and = on the same aes. + 6 The points of intersection are (, 8) and (, ), so the roots are = and =. 6 Graph = + + 6. -intercepts: = and = 3 MHR Chapter 9 978--7-7389-
Chapter Function Operations. Sums and Differences of Functions You can form new functions b performing operations with functions. Sum of Functions h() = f () + g() or h() = ( f + g)() Eample f () = and g () = + 5 h() = f () + g () h() = + ( 5) h() = + 5 Difference of Functions h() = f () g () or h() = ( f g)() Eample f () = and g () = h() = f () g () h() = ( ) h() = 3 + 8 6 f() (f + g)() g() (f - g)() - - - - g() 6 - - 6-6 f() The domain of the combined function formed b the sum or difference of two functions is the domain common to the individual functions. Eample If the domain of f() is {, R} and the domain of g() is {, R}, the domain of (f + g)() is {, R}. Domain of f() Domain of g() Domain of (f + g)() The range of a combined function can be determined using its graph. 978--7-7389- Pre-Calculus Student Workbook MHR 35
. Products and Quotients of Functions New functions can be formed b performing the operations of multiplication and division with functions. Product of Functions h() = f() g() or h() = (f g)() Eample f() = + 3 and g() = h() = f() g() h() = ( 3)( ) h() = + 5 3 (f g)() g() - - f () - - -6 Quotient of Functions h() = f (), where g() g() or h() = ( f g ) (), where g() Eample f() = 3 and g() = h() = f() g() h() = _ 3, where, g() - - f - () g - f() -6 The domain of a product or a quotient of functions is the domain common to the original functions. The domain of a quotient of functions must have the restriction that the divisor cannot equal zero. That is, for h() = f(), the values of are such that g(). g() The range of a combined function can be determined using its graph. 978--7-7389- Pre-Calculus Student Workbook MHR 335
.3 Composite Functions Composite functions are functions that are formed from two functions, f () and g (), in which the output of one of the functions is used as the input for the other function. f (g()) is read as f of g of ( f g)() is another wa of writing f (g()) and is read the same wa For eample, if f () = and g() = + 3, then f (g()) is shown in the mapping diagram. f(g()) g() f() + 3 ( + 3) - The output for g() is the input for f (). g f - - - 8 6 - -6 To determine the equation of a composite function, substitute the second function into the first. To determine f (g()), f (g()) = f( 3) Substitute + 3 for g(). f (g()) = ( 3) Substitute + 3 into f () =. f (g()) = + 6 Simplif. To determine g(f()), g ( f()) = g( ) Substitute for f (). g(f()) = ( ) + 3( ) Substitute into g () = + 3. g ( f()) = 8 + + 6 6 Simplif. g ( f()) = Note that f(g()) g( f ()). The domain of f (g()) is the set of all values of in the domain of g for which g() is the domain of f. Restrictions must be considered. 978--7-7389- Pre-Calculus Student Workbook MHR 35
Chapter Permutations, Combinations, and the Binomial Theorem. Permutations The fundamental counting principle states that if one task can be performed in a was and a second task can be performed in b was, then the two tasks can be performed in a b was. For an positive integer n, n factorial or n! represents the product of all of the positive integers up to and including n. n! = n (n ) (n )... 3.! is defined as. Linear permutation is the arrangement of objects or people in a line. The order of the objects is important. When the objects are distinguishable from one another, a new order of objects creates a new permutation. The notation n P r is used to represent the number of permutations, or arrangements in a definite order, of r items taken from a set of n distinct items. A formula for permutations is n! np r = _ (n r)!, n N For permutations with repeating objects, a set of n objects with a of one kind that are identical, b of a second kind that are identical, and c of a third kind that are identical, and n! so on, can be arranged in a!b!c!... was. To solve some problems, ou must count the different arrangements in all the cases that together cover all the possibilities. Calculate the number of arrangements for each case and then add the values for all cases to obtain the total number of arrangements. Whenever ou encounter a situation with constraints or restrictions, alwas address the choices for the restricted positions first. Working Eample : Arrangements With or Without Restrictions a) A school cafeteria offers sandwiches made with fillings of ham, salami, cheese, or egg on white, whole wheat, or re bread. How man different sandwiches can be made using onl one filling? b) In how man was can five black cars and four red cars be parked net to each other in a parking garage if a black car has to be first and a red car has to be last? 36 MHR Chapter 978--7-7389-
. Combinations A combination is a selection of objects without regard to order. The notation n C r represents the number of combinations of n objects taken r at a time, where n r and r. n! A formula for combinations is n C r = (n r)!r!, n N. The number of combinations of n items taken r at a time is equivalent to the number of combinations of n items taken n r at a time; that is, n C r = n C n r. To solve some problems, count the different combinations in cases that together cover all the possibilities. Calculate the number of combinations for each case and then add the values for all cases to obtain the total number of combinations. Working Eample : Combinations and the Fundamental Counting Principle Eight female students and nine male students are running for si offices on the student council eecutive team. a) How man selections are possible? b) How man selections are possible if the eecutive team must have three females and three males? c) One of the male students is named David. How man si-member selections consisting of David, one other male, and four females are possible? Solution a) This is a combination problem because it involves choosing students out of and the is not important. Substitute n = and r = into n C r = C = = =! _ ( )!!!!! n! (n r)!r! : There are possible was of selecting the eecutive team. 37 MHR Chapter 978--7-7389-
.3 The Binomial Theorem Pascal s triangle is a triangular arra of numbers with in the first row, and and in the second row. Each row begins and ends with. Each number in the interior of an row is the sum of the two numbers above it in the preceding row. 3 3 6 In the epansion of the binomial ( + ) n, where n N, the coefficients of the terms are identical to the numbers in the (n + )th row of Pascal s triangle. You can also determine the coefficients represented in Pascal s triangle using combinations. Use the binomial theorem to epand an power of a binomial, ( + ) n, where n N. Each term in the binomial epansion has the form n C k () n k () k, where k + is the term number. Thus, the general term of a binomial epansion is t k + = n C k () n k () k. Important properties of the binomial epansion ( + ) n include the following: Write binomial epansions in descending order of the eponent of the first term in the binomial. The epansion contains n + terms. The number of objects, k, selected in the combination n C k can be taken to match the number of factors of the second variable. That is, it is the same as the eponent on the second variable. The sum of the eponents in an term of the epansion is n. 978--7-7389- Pre-Calculus Student Workbook MHR 383