FUNCTIONS Trigonometric graphs

Transcription

1 LESSON FUNCTIONS Trigonometric graphs In grade 0, ou studied the graphs of = sin, = cos AND = tan Let us recall these. The graph of = sin 60º 70º 80º 90º 90º 80º 70º 60º The graph of = sin has a period of 60, i.e. it repeats itself after ever 60. The graph has amplitude of, i.e all values lie between and. The graph goes through the origin. The graph of = cos 60º 70º 80º 90º 90º 80º 70º 60º The graph of = cos has a period of 60, i.e. it repeats itself after ever 60. The graph has amplitude of, i.e. all values lie between and. The graph cuts the -ais at =. Page 6

2 The graph of = tan is ver different from the graphs of = sin and = cos º 70º 80º 90º 90º 80º 70º 60º 0.. The graph repeats itself ever 80, i.e. the graph of = tan has a period of 80. The graph also has asmptotes at = 70 ; = 90 ; = 90 and =70. The graph goes through the origin. In Grade we look at the effect of adding p to the, i.e we look at the graphs of = sin ( + p); = cos ( + p) and = tan ( + p). Stud the pairs of graphs below and see if ou can describe the effect. Let us first look at positive values of p. 60º 70º 80º 90º 90º 80º 70º 60º The graph of = sin ( + 0) is easil sketched b plotting points obtained from a table using the CASIO ƒ-8es PLUS. Step : Step : Press Mode. Choose option : TABLE. Enter sin ( + 0), so that our screen looks as follows The screen will prompt a starting value b showing Start? Enter 60 and then the equal sign. ƒ() = sin ( + 0) Page 7 Page Lesson Algebra

3 Step : The screen will prompt an ending value b showing End? Enter 60 and then the equal sign Step : Now the screen displas Step? This is the value that represents the interval between points. Tpe 0 and then press the equal sign. The calculator now displas the following table with points 60 0, 0 0, , , , 0 0, , , , 0 0, , , , 0 0, , , , All that is left is for ou to plot these points and join them with a sine curve. You can use the calculator to get a table of points for = cos ( + ). This time use for our step. The graph of = cos( + ) Page 8

4 60º 70º 80º 90º 90º 80º 70º 60º The graph of = tan( + 60 ) 60º 00º 0º 80º 0º 60º 60º 0º 80º 0º 00º 60º If ou are using the calculator to get a table of points to plot, use 0 as our step value and not 60. Notice above that the asmptotes for = tan are = 90 ; = 70 ; = 90 and = 70. Yet, the asmptotes for = tan ( + 60 ) are = 0 ; = 0 ; = 0 and = 0. Are ou able to describe the effect of the p in the equations = sin ( + p); = cos ( + p) and = tan ( + p). If ou look carefull ou will see that in each case the shape of the original graph has remained intact although it has shifted horizontall to the LEFT!! Observe how the positions of the asmptotes above changed. What if the values of p are negative? Again stud the pairs of graphs given and see if ou can describe the effect of adding a negative p value to. The graph = tan( 60 ) Page 9 Page Lesson Algebra

5 60º 70º 80º 90º 90º 80º 70º 60º Look at the positions of the asmptotes of = tan ( 60 ). = 0 ; = 0 ; = 0 ; = 0. Instead of the graph shifting to the left it has now shifted to the RIGHT!!! The period of = tan ( 60 ) remains 80 The graph of = sin( ) 60º 70º 80º 90º 90º 80º 70º 60º Again the graph of = sin has shifted to the right b. Note that the period remains unchanged, i.e. it is 60 Also the amplitude remains unchanged, i.e. it is. Page 0

6 The graph of = cos( 7 ) 60º 70º 80º 90º 90º 80º 70º 60º B studing the graphs above we see that the graph of = cos has shifted horizontall to the right b 7. The amplitude of the graph is still and the period is still 60. In general, if p > 0 then the graph shifts horizontall to the LEFT and if p < 0 then the graph shifts horizontall to the RIGHT Activit Activit. On the same set of aes sketch the graphs of ƒ() = sin ( 0 ) and g() = cos ( + 60 ) for Î[ 60 ; 60 ]. What do ou notice? Page Page Lesson Algebra

7 . Can ou eplain our observation in,? In each case, give the value of p. = tan ( + p) 0 0º 70º 0º 0. = sin ( + p) º 0º 0º 0º 60º 80º 00º 0º 0º º Page

8 . = cos ( + p) º º 90º º 80º º. The graphs of ƒ() = bsin and g() = cos ( + k) are sketched for Î[ 80 ; 80 ] E C. Write down the values of b and k. What is the period of g?. Write down the coordinates of C if C is a turning point on g. Write down the coordinates of E. Determine the equation of ƒ if the -ais is moved 0 to the left. Page Page Lesson Algebra

9 The graphs of = sin k; = cos k; = tan k Up to now, the trig graphs ou have sketched have all had a period of 60 in the case of sine and cosine and a period of 80 in the case of a tangent graph. If the coefficient of is changed from, then the period of the graph will change from 60 (sine and cosine) and 80 (tangent) In general, The period of = sin (k) is (_ 60 k ). The period of = cos (k) is (_ 60 k ) The period of = tan (k) is (_ 80 k ) Graphs (contrar to intuition) have reduced period if k > and the period increases if 0 < k < Activit Activit Without drawing the graph, write down the period of each of the following graphs. = sin (). = cos (). = tan ( _ ) Now let us look at the graphs of the equations above. Firstl, the graph of = sin (). 0º 0º 60º 90º 0º 0º 80º 0º 0º 70º 00º 0º 60º We see that the wave repeats itself after ever 0. There are complete waves from 0 to 60. For this reason the period is one third of the period of = sin. Secondl, the graph of = cos () Page

10 0º 0º 60º 90º 0º 0º 80º 0º 0º 70º 00º 0º 60º We see that the wave repeats itself after ever 80. There are complete waves from 0 to 60. For this reason the period is one half of the period of = cos. Finall, let us look at = tan (_ ) 60º 70º 80º 90º 90º 80º 70º 60º This graph now repeats itself after ever 60. The original = tan graph repeated itself after ever 80. The positions of the asmptotes have changed and their equations are now = 80 and = 80. Drawing the graphs of = sin k; = cos k; = tan k Again, these are easil done b using the CASIO ƒ-8 ES PLUS. Suppose ou are asked to sketch the graph of = cos () for Î[ 90 ; 90 ]. Step : Mode : TABLE and Enter the cos (). Step : Start 90 and press equals Step : End 90 and press equals Step : This is the part that changes now as the period is no longer 60. To get the period of = cos () we divide 60 b and get 90. In the previous eamples we used 0 as our step. Dividing 90 b gives us,. This is the value ou should use for the step. Step : Plot the points from the table, which would look as follows: Page Page Lesson Algebra

11 90 67, 0, 0 0, 0 67, 0 90 Plotting these points we get the following graph. 90º º º 90º Activit Activit. On the same set of aes sketch the graphs of ƒ() = sin and g() = sin for Î[0 ; 80 ]. Write down ƒ g for Î[0 ; 80 ]. Use our graphs to solve the following inequalities if Î[0 ; 80 ].. (sin )(sin ) < 0 Page 6

12 .. sin > sin. In each case, give the value of k and the period of the graph.. = sin (k) 0º 0º 80º 0º. = cos (k) 70º 80º 90º 90º 80º 70º. = tan (k) 90º 60º 0º 0º 60º 90º Page 7 Page Lesson Algebra

13 . The graphs of ƒ() = acos + b and g() = tan (k) are sketched below for Î[ ; 80 ]. 0º 0º 60º 90º 0º 0º 80º. Determine the values of a, b and k. Show on our graph where ou would read off the solution to tan (k) = if Î[ ; ]. Write down the period of g. Write down the amplitude of ƒ. Write down the range of ƒ.6 The graph of g is translated units verticall. Write down the equation of the new graph..7 The graph of f is translated horizontall to the left, write down the equation of the new graph. The graph of = a( + p) + q In Grade 0, ou studied the graph of = a + q. If a is positive, i.e. a > 0, then the shape of the graph is J. If a is negative, i.e. a < 0, then the shape of the graph is L. The turning point of the graph of = a + q is (0; q). Page 8

14 To find the -intercepts, we let = 0 and solve for. In grade, we must be able to sketch = a( + p) + q. The a value still determines whether the parabola smiles or frowns. To find the -intercepts we still let = 0. The coordinates of the turning point change from (0; q) in = a + q to ( p; q) in = a( + p) + q. As a result of adding p to the before squaring it, the position of the turning point is no longer on the -ais. To find the -intercept, we let = 0. Look at the graphs below. = ( + ) = = ( ) 7 6 In the graphs above, = ( ) is obtained b shifting = horizontall to the right b units. Also, the graph of = ( + ) is obtained b shifting = horizontall to the left b units. The domain for all three graphs is ( ; ) or R. The range for all three graphs is (0; ). Look at the graphs of = + and = ( ) + The theor tells us: Equation = + = ( ) + Turning Point (0; ) (; ) -intercepts none none -intercept (0; ) (0; ) Shape J J Domain R R Range [; ) [; ) Page 9 Page Lesson Algebra

15 = = ( ) + The graph of = ( ) + is obtained b shifting horizontall the graph of = + to the right b unit. The net eample shows us that if the shape of the parabola frowns, then The same rules still appl, i.e. adding a positive p to means a shift to the left adding a negative p to means a shift to the right. The graphs of ƒ() = +, g() = ( ) + and h() = ( + ) + are drawn. h() f() g() To draw the graph of g() = ( ) +, we simpl shift the graph of ƒ() = + horizontall to the right b units. To draw the graph of h() = ( + ) +, we simpl shift the graph of ƒ() = + horizontall to the left b units. Page 0

16 SUMMARY To sketch the graph of = a( + p) + q, ) Sketch the graph of = a + q ) Appl a horizontal shift to the right or left b p units depending on the sign of p Domain = R Range = [q; ) if J Range = (- ; q] if L Eample = ( + ) and = _ ( ) + Solution To sketch = ( + ), ou must first look at the graph of =. Then appl a horizontal shift of units to the left. Eample Solution 6 It is important to label all intercepts with the aes. Note that the graph does not cut the -ais, but the graph of = ( + ) cuts the -ais at. To find the -intercept on the -ais, we let = 0 in the equation. = (0 + ) = = 0 To sketch, = _ ( ) + ou must first look at the graph of = _ +. Then appl a horizontal shift of units to the right. Page Page Lesson Algebra

17 Note that the graph does not cut the -ais, but the graph of = _ ( ) + cuts the -ais at 6 _. To find the intercept on the ais, we let = 0 in the equation. = _ (0 ) + = _ (9) + = _ 9 + _ = _ = 6 _ IMPORTANT FACT the turning point of = a( + p) + q is ( p; q) Another eample: Sketch the graph of ƒ() = ( ) +. From the equation, we get the following: Shape L Turning Point (; ) Range ( ; ] To determine the -intercept, we let = 0 = (0 ) + = (6) + = 0 To determine the -intercept, we let = 0 ( ) + = 0 \ ( ) = \ ( ) = 6 \ = or = \ = 8 or = 0 Finall, the graph is drawn below. In the eam it is important that ou label all turning points. Page

18 0 (; ) The graph of = a + b + c Sometimes the equation of the parabola is not given in the form = a( p) + q. Instead the quadratic epression is given with the brackets epanded giving an equation of the form = a + b + c. In this case one of two approaches can be used. Approach : Complete the square to get the epression in the form = a( p) + q Approach : Use = _ b to get the -value at the turning point and then find a the -value at the turning point b substituting this -value into the equation. The line = _ b is the ais of smmetr. a Let us look at eamples: Suppose ou are asked to sketch = Approach : Complete the square (Revise in Chapter on Quadratic Epressions) = = ( ) = ( ) Approach : Use = _ b a a = b = 6 c = 8 = _ b a \ = _ 6 () \ = For the -value, = 6() + 8 = = Therefore the turning point is (; ) The finding of -intercepts and -intercept remains the same. -intercepts: Let = 0 Page Page Lesson Algebra

19 6 + 8 = 0 \ ( )( ) = 0 \ = or = -intercept: Let = 0 = 0 6(0) + 8 = 8 The graph looks as follows: Finding the equation of a parabola Often the graph is drawn and the equation is required. POSSIBILITY : Turning point given Step : Start with = a( p) + q Substitute the coordinates of the turning point for p and q. Step : Find the value of a b substituting an other point on the graph into the equation Eample Eample: (; 8) 0 Page

20 Step : = a( ) 8 Step : Substitute (; 0) or an other point into = a( ) 8 0 = a( ) 8 \ a = 8 \ a = Therefore the equation is = ( ) 8 POSSIBILITY : The -intercepts are given: We find the equation b reversing the calculation for the -intercepts. Step : Start with = a( )( ) In this equation the -intercepts are and. Step : Find a b substituting an point on the graph other than the -intercept into the equation. Eample: Eample 7 6 Step : = a( + )( ). Notice how the signs of the -intercepts change when ou put them in the brackets Step : Substitute (0; ), or an other point on the parabola: = a(0 + )(0 ) \ a = \ a = Therefore the equation of the parabola is = ( + )( ). Sometimes it is necessar to multipl out the brackets. Important fact about parabolas The parabola is smmetrical about its ais of smmetr. It is often possible to find the ais of smmetr ( at turning point) b inspection. 6 8 Look at the eample below: The graph of = + b + c is drawn below. Page Page Lesson Algebra

21 ( 6; ) (; ) (; 8) Determine the values of b and c. The points ( 6; ) and (; ) are smmetrical to one another about the ais of smmetr (dotted line). Therefore the equation of the dotted line is =. Eactl halfwa between 6 and. Therefore Step : = ( + ) + q. The a value is given in the question. Step : We now will substitute (; 8) to find the value of q. = ( + ) + q \ 8 = ( + ) + q \ 8 = 8 + q \ q = 6 The equation becomes = ( + ) 6. This time it is necessar to multipl the brackets out because of the wa the question was asked. = ( + ) 6 = ( + + ) 6 = = + Therefore b = and c = Activit Activit. The graph of ƒ() is sketched. On separate sstems of aes sketch the graph of Page 6

22 . ƒ( + ). ƒ( ). ƒ( ). ƒ(). Sketch the following graphs on the same set of aes ƒ() = + 6 and g() = ( ) 8 Show all intercepts with aes and turning points.. Find the -values at the point(s) of intersection of ƒ and g. Use our graph to solve.. ƒ() < g () g().. _ ƒ() > 0. In each case the graph of = a + b + c has been sketched. Determine the values of a, b and c (; ). ( ; ) ( 7; 7) 0 Page 7 Page Lesson Algebra

23 The graph of = a_ + p + q. This is the equation of a hperbola which has been shifted to the right or to the left b p units depending on the sign of p. a_ + q is undefined for = p. + p Thus, the asmptotes are = p and = q. To find the -intercept, we let = 0 To find the -intercept, we let = 0. Look at the graph of = 0_ + below, The asmptotes are = and. (These are read from the equation). -intercept: Let = 0: 0 0 = _ + \ 0 = ( + ) \ = \ = _ -intercept: Let = 0: = 0_ 0 + = = Let us look at a second eample, = 8_. This equation can also be written as = _ 8. Asmptotes: = and = -intercept: Let = 0: = 8_ 0 = + = -intercept: Let = 0: 0 = 8_ \ ( ) = 8 \ = The hperbola will look as follows: Page 8

24 Determining the equation of a hperbola when the graph is given Step : The position of the asmptotes give us the value(s) of p and q in = a_ + p + q. Step : To find the value of a, we substitute an point on the graph into the equation. EXAMPLE: The graph of = a_ + q is sketched below. Determine the value(s) of + p a, p and q From the graph we see that, p = and q =. The equation becomes: = a_ +. Substituting (0; 6) or an other point on the graph into the equation gives: 6 = a_ \ = a 6 \ a = The graph of = a b + p + q This is the graph of an eponential function. This graph has one horizontal asmptote, i.e. = q. The easiest wa to sketch this graph is to use our calculator: Using the Casio ƒ-8 ES Plus, Step : Mode : TABLE Step : Start? Step : End? Step : Step? Step : Plot the points that the calculator returns and sketch the eponential through the points Page 9 Page Lesson Algebra

25 The eponential will be increasing OR it will be decreasing To find the -intercept, let = 0. Eample Eample: Sketch the graph of = ( _ ) + Horizontal asmptote: = -intercept: let = 0: 0 = (_ ) + \ (_ ) + = (_ ) + = (_ ) \ + = \ = Using our calculator, we get the following table: 0, If we plot these points in the Cartesian plane as well as draw the horizontal asmptote, we will be able to see if the eponential is increasing or decreasing. 0 Now fit an eponential curve through the points so that it gets closer and closer to the asmptote. Page 0

26 0 Let us look at a second eample, = () + Horizontal asmptote: = -intercept: 0 = () + \ () = \ No -intercept Table of values from calculator:, 0, Now plot the points in the table as well as draw the asmptote: 7 6 A B C 0 Page Page Lesson Algebra

27 It is now clear that the curve is an increasing curve approaching the asmptote from the left. Now fit a smooth curve through the points. 7 6 A B C 0 Note: for b >, = b is increasing and for 0 < b <, = b is decreasing. Activit Activit. The sketch shows the graph of = ƒ() which is a translation of the graph of = _ Write down the equation of ƒ. The graph of ƒ is shifted horizontall units to the left. Write down the domain and range of the new graph.. The sketch shows the graph of g() = a + b. Determine the values of a and b. Page

28 0 ( ; ). Given: ƒ() = and g() = ƒ( + ) + Sketch the graph of ƒ().. The graphs of ƒ() = cos + p and g() = cos q are sketched for g() 90 f() 90. Write down the value of p if the graph of ƒ touches the -ais.. Write down the amplitude of ƒ.. Write down the value of q if the period of g is half the period of ƒ.. Write down the minimum value of ƒ.. Use the graphs to solve: g() ƒ() 0 if Î [ 90 o ; 90 o ]. The figure below represents the graphs of the following functions ƒ() = + 6 and g() = 8_ Page Page Lesson Algebra

29 M C N D E A B. Determine the coordinates of A, B, C, D, E. If M is the turning point of ƒ(), determine the length of MN Solutions to Activities Activit. g() 60º 70º 80º 90º 90º 80º 70º 60º f(). The graphs are reflections of each other in the -ais.. sin ( 0 ) = cos (90 ( 0 )) = cos (0 ) = cos (80 (0 )) = cos (60 ) g() = cos ( + 60 ) = sin (90 ( + 60 )) = sin (0 ) = sin ( 0 ) which is reflection in ais of ƒ() = ( 0 ) Thus ƒ() = g(). p = 0. p =. p = Page

30 . a = ; b = and k =. 60. (; ) _. = cos ( 80 + ) = cos ( ) = cos = -cos = _ _ E ( 80 ; _ ). ƒ() = sin ( + 0 ) Activit 60. _ = 0. _ 60 = 80. _ 80 = 60 _ Activit.. g() f(). ƒ g = {(0 ; 0); (80 ; 0)}.. 90 < < 80.. No solution. k = 9 Period = 0. k = _ Period = 60 =080 or 70. k = Period = 60. a = b = and k =. Shown at A and B A B Page Page Lesson Algebra

31 .. [ ; 0].6 = tan () +.7 = cos ( + ) Activit. The original graph is shifted horizontall to the left b units The original graph is shifted horizontall to the right b units Page 6

32 . is replaced with, therefore it is the graph smmetrical to the -ais is replaced with, therefore it is the graph smmetrical to the -ais g() _ 0 0 f() 0 ƒ() = + 6 g() = ( ) 8 Shape: L Shape: J Turning point: Turning point: = ( ) (; 8) \ (; ) Page 7 Page Lesson Algebra

33 -intercepts: none -intercepts: ( ) 8 = 0 \ ( ) = \ = or = \ = or = intercept: = 6 intercept: = (0 ) 8 = 0 Domain = R Domain = R Range = ( ; ] Range = [ 8; ). For point(s) of intersection, we equate the two quadratic epressions: + 6 = ( ) 8 \ + 6 = ( 6 + 9) 8 \ + 6 = \ = 0 \ ( )( ) = 0 \ = _ or =.. < _ or >.. Since ƒ() < 0 for all values of, it follows that g() < 0. Therefore, < <. = a( + )( ) Substituting (; ), we get = a( + )( ) \ a = = ( + )( ) \ = ( + 6) \ = + 6 \ a = b = c = 6. = a( + ) + substituting ( 7; 7), we get 7 = a( 7 + ) + \ 9a = 9 \ a = Therefore, the equation is = ( + ) + = ( ) + = 8 a = b = 8 c = Page 8

34 Activit. ƒ() = _ +. Domain = R {0} Range = R {}. b = Substituting ( ; ) or an other point on the graph into = a +, we get: = a + \ a =. g() = + + Asmptote: = -intercept: 0 = + + \ + = Therefore no -intercepts 0 0. p =.. q =.. Since ƒ() < 0, it follows that g() > 0. Thus,. A and B : -intercepts of ƒ: + 6 = = 0 ( + )( ) = 0 = or = \ A( ; 0) B(; 0) C is the -intercept of ƒ: let = 0 C(0; 6) D is the -intercept of g: Let = 0, we get = 8_ + 6 = 7,6 \ D (0; 7,6) 0 + E is the -intercept of g: Let = 0, we get 0 = 8_ \ 0 = 8 + 6( + ) Page 9 Page Lesson Algebra

35 \ 0 = \ 6 = 8 \ = _ 9 \ E ( _ 9 ; 0 ). ƒ() = + 6 \ ƒ() = ( + ) + 8 \ M( ; 8) Thus we know that N = \ = 8_ = 8 Thus N( ; 8) Therefore MN = 0 units.(8 8 = 0) Page 60