Maths Study Guides GR11

Copyright reserved 71 72 (Solutions pg. 263)

2.1 | Sketching the parabola To sketch the parabola, you must calculate the following The x-intercepts : Let y = 0 and solve the equation. The y-intercepts : Let x = 0 (this is always the c-value if the equation is in the form y = ax? + bx +c.) The axis of symmetry and the coordinates of the turning point This can be done in two different ways. | Equation of the axis of symmetry : x = — + This is also the x-coordinate of the turning point. To find the y-coordinate of the turning point, substitute this x-value into the equation. or | Write the equation in the form y = a(x + p)? +q by completing | Equation of the axis of symmetry : x = —p. | The coordinates of the turning point are (—p;q) | the square. | X= —pPy=q | | Example 1 Sketch the graph of y = 2x? - 6x - 8 73 Copyright reserved Solution y-intercept:y=-8 [Letx =0 > y = 2(0)? - 6(0) -8 = -8] x-intercept : Let y = 0 y 1. 2x? -6x-8 = 0 1x? -3x-4=0 “. &-4)(x4+1) =0 x =4orx=-l. [Divide by 2.] Turning point: x = — 2a nye ch 263 FS —JQy 7 47 2 Substitute x = + into equation = 1 =-124 Griz) .. Turning point = (3;-124) or Write y = 2x* — 6x - 8 in the form y = a(x + p)? + q by completing the square. y = 2x? -6x-8 y= 20x? -3x-4) [Divide by2 ] oy = 20? - 3x4 (-$)?-(-4)?-4) [Add and subtract (-4)? ] “y= A@-3)?-4-4] [Factorise first three terms and square ~(-+) ] y= 2@- 4)? - 8] [Add last two terms ] Lys lee 3) - 35 [Multiply - oa by 2 in front of bracket -B)=-F ] o 7 *. Coordinates of turning point : :-4) = (4;-124 ) 74

Example 2 Draw a neat sketch graph of y = -2(x- 1)? + 8. Equation is already in the form y = a(x +p)? +q «. Coordinates of Turning Point: (—p;q) .. (-(-1);8) = (138) Axis of symmetry : x = 1 To find the x-and y-intercepts, first simplify the equation. y=-2(x-1) +8. i —2(x? — 2x +1) +8 ps2 +4r-248 Jy = 2x? + 4x46 y-intercept :y = 6 x-intercepts : Lety = 0 1 2x? +4¥+6=0 wx? -2x-3 = 7 %-3)%4+1) =0 “x= 30rx=-1 [Divide by -2 ] 22 | Determining the equation of the parabola When you are asked te find the equation of the parabola, one of the following equations are used The roots and another point are given : y = a(x — Ist root)(x — 2nd root) Turning Point and another point are given : y = a(x + py? +q Copyright reserved 75 76 Example 3 7 The diagram shows the graph (0;4) ofy = ax? +bxt+c. Find the values of a,b and c. Solution x Ga] 0 pe [Roots : 2 and-4 ] Roots are given : y = a(x — 1st root)(x — 2nd root) oy = a(x —2)(x- (-4)) oy = a(x—2)(x +4) 4 = a(0-2)(0 + 4) [(0; 4) is a point on the graph. Now let x = 0 and y = 4 and determine a } 4 = a(-2)(4) 4 = -8a -a-4 .aart Now substitute a with — + iny = a(x —2)(x+ 4) and simplify. 2 y=—t(e-2)@4+4) vey =~? + 2x - 8) we ys px axt4

Example 4 In the diagram He) = ax? + bx+e. Determine the values of a,b and c. Solution Coordinates of turning point : (1;-6) .q=-6 [Coordinates of turning point : (-p;q)] “p=-l [Note : Coordinates of turning point : (-p; gq) “p= 7 p=) Turning point is given > y = a(x+p)> +4. Substitute p with -\ and q with -6. oo y=a(x-1)?-6 1-4 =a(2-1)?-6 1-4 = a(1)? -6 1 -44+6=a “2=a (To determine a, let x = 2 andy = -4 } Now leta = 2 iny = a(x-1)?-6 “y= 2-1)? -6 wy = 2(x? -2x + 1)-6 wy = 2x? -4x4+2-6 wy = 2x? -4x-4. a=2,b=-4,c=-4 77 Copyright reserved 3. | The hyperbola defined by y = ¥ip +q The effect of a Ifa > 0 ( positive) the graph lies in the first and third quadrant. If a < 0 (negative) the graph lies in the second and fourth quadrant. ais the vertical shrinking or stretching of the graph. The effect of g The equation of the horizontal asymptote of the hyperbola is y = q. q is the vertical shift of the graph q units upwards or downwards. The effect of p Division by 0 is undefined, therefore x + p # 0. Thus, the equation of the vertical asymptote is x = —p (x +p = 0 when x = -p). pis the horizontal shift of the graph p units to the left or to the right. 3.1| Sketching the graph of the hyperbola To sketch the hyperbola, you have to calculate the following : The horizontal asymptote : The horizontal asymptote is the line y = q. The vertical asymptote : The vertical asymptote is the line x = —p_ The x-intercept : Let y = 0 and solve the equation. The y-intercept : Let x = 0 and solve the equation. If a is positive > graph in quadrant | and 3. If a is negative > graph in quadrant 2 and 4. 78

Example 5 Sketch the graph of y = 34+ +2 Solution Horizontal asymptote : y = 2 Vertical asymptote : x = —1 (Letx+1=0, x =-1] y-intercept: Let x = 0 12 “y= oy7 t? a o y= 1242 y= lh 14 x-intercept: Let y = 0 7 . 42-0 j x+1 1 124+2(¢+1)=0 [Multiply by (x + 1)] *. 124+2x+2=0 3 x =-14 .x=-7 a> 0, graph in \st and 3rd quadrant. Example 6 Sketch the graph of y = 3 -1 Copyright reserved 79 Solution Horizontal asymptote : y = —-1 Vertical asymptote : x = 2 y-intercept . x = 0 A 3 x-intercept: y= 0 -6 a yea . sh -1-0 “. 8-(x-2) =90 “ -8-x+2=0 x = 6 “x =-6 a < 0, therefore graph in 2nd and 4th quadrant. 3.2 | The axes of symmetry of the hyperbola The function fix) = $ is symmetrical about the line y = x and the line y = —x. Therefore, the equation of the axes of symmetry are y = x and y = -x. 80

The equation of the axes of symmetry of the hyperbola defined a by f(x) = xtp tare y=(xe+p)+q andy =—-(+p)+g. Example 7 Determine the equation of the axes of symmetry of Ti y= +2 pj os T2y 0) 1 Solution 7.1 Equation of axes of symmetry v= @+p)+q and y=-(x+p)+q y=(x+1)+2 9 and y=-@t+1)4+2 “yext3 “yps-x-1+2 “ysxtl «. Equation of axes of symmetry; y=x+3 and y=-x+1 72y=(x+p)t+q and y=-(x+p)+q y=(x-2)-1 and y=-(x-2)-1 “y=x-2-1 .ye-xt2-1 “A y=x-3 “y=xtl “y=x-3 and y=-x4+1 81 Copyright reserved 3.3 | Determining the equation of the hyperbola Example 8 The sketch represents the graph ofy = op +q. Find the values of a, p and q. Solution Vertical asymptote : x = —2 —p=-2 [Vertical asymptote : x = —p ] Horizontal asymptote :y = | gsi [Horizontal asymptote : y = q ] Substitute p = 2 and q = | into equation . a x+2 Ve +1 25 0 in zt 1 [Substitute x with 0 and y with 2 and determine a | *. 2(2) =at+2 [Multiply by 2 } .4-2=a a=2 82

4.| The exponential function defined by y = a.b*? +q The effect of a : a is the vertical stretching or shrinking of the graph. The effect of the base, b : The function is ascending if b > 1 and descending if 0<b<1. The effect of q : You already know that q is the vertical shift of the graph. The equation of the horizontal asymptote is y = q. The graph of the exponential function do not have a vertical asymptote. The effect of p F Sketched are the graphs of 7 5 Sx) = 2*, g(x) = 2**! and x h(x) = 27, ‘ Therefore, p is a horizontal shift of the graph. If p > 0, the graph moves to the left and if p < 0, the graph moves to the right. 4.1] Sketching the graph of the exponential function To sketch the graph of the exponential function , calculate The horizontal asymptote : The horizontal asymptote is always y = q. The y-intercept : Let x = 0 and solve the equation. The x-intercept : Let y = 0 and solve the equation. Note : Ifg > 0 there will be no x-intercept. If the base >1 the graph will be ascending and if the base < | the graph will be descending. Copyright reserved 83 84 Example 9 Sketch the graph of y = 3! ~1 Solution Horizontal asymptote :y = —1 of t y-intercept : Letx = 0 = 301] 2 y=2 a | x-intercept : Lety = 0 34 1=0 | . Bel oy 2. 3! = 39 [Remember : 1 = 3° ] “x+1=0 (Bases are equal, equate exponents | x=] Base > 1 .. function is ascending. Example 10 Draw a sketch graph of y = 2(4)*"! +1, Solution Horizontal asymptote : y= 1 y-intercept: Letx = 0 y= ydyrted

substitute 85 86 Copyright reserved

oy = 2b* +1 7 =2(b)! +1 (Letx = 1 andy =7] 7-1=2b . 6 = 2b b= 3 Equation of graph . y = 2.3*+1 5. | Translation of graphs The translation of graphs means that an existing graph is moved to the right or the left by a specific number of units, and is translated upwards or downwards by a specific number of units. The following rules apply The graph is translated p units to the right = x changes to (x — p). The graph is translated p units to the left = x changes to (x + p). The graph is translated q units upwards = y changes to ( — q). The graph is translated g units downwards => y changes to (y+ ¢). What will the equation of the new graph be if the graph of y = x? is moved 2 units to the left and translated 1 unit upwards. y-l=(x+2)? [2 left + x becomes (x + 2); | upwards > y becomes (y — 1)] “y= (x42)? 41 Therefore, the graph of y = (x + 2)? + 1 can be sketched by moving the graph of y = x? two units to the left and one unit upwards. Copyright reserved 87 Example 13 In the diagram f(x) = x?, g(x) is the translation of f(x) 2 units to the right and A(x) is the translation of f(x) 1 unit to the left and 2 units upwards. Determine the equations of g(x) and h(x). a [= oe Equation of g(x) y= (x-2)? {Translation 2 units to the right : .. YET eY gh Equation of h(x) y-2=(x+1)? “y= (x+1)? +2 Example 14 14.1 Sketch the graph of y = x? - 2x-3. 14.2 On the same set of axes, using translation, sketch the graph of f(x) = (x+2)?-1. Solution 14.1 y-intercept : y = -3 [Translation 1 unit to the left : x > (x + 1); 2 units upwards : y > y-2] 88

x-intercept : Lety = 0. x?-2x-3=0 2 («*-3)@+1) =0 .x-3=0 or x+1=0 x= 3 or x=-l Turning Point = = 22 = 2 220 2) : 5% Substitute x = 1 into equation oy = (1)? -2C1) -3 =-4 Turning Point (1;—4) 14.2 Turning Point of f(x) = (-2;-1). Therefore, a translation 3 units to the left and 3 units upwards. First move graph 3 units to the left. :. Turning Point (-2;-4) and roots x = 0 and x = —4. Now move graph 3 units upwards. .. Turning Point (-2;-1) and roots x = -1 andx = -3. 6. | Deductions from sketch graphs You must be able to make certian deductions from sketch graphs when the equation of the graph is given, i.e. determine coordinates, lengths of lines and the points of intersection of graphs . 89 Copyright reserved Example 15 In the diagram f(x) = x? - 4x - 5 and g(x) = x41. 15.1 Calculate the lengths of OA, OB and OC. 15.2 Calculate the coordinates of D, the turning point of f(x). 15.3 Calculate the length of EF. 15.4 Calculate the maximum length of GH. 15.5 What is the range of f(x)? A 15.6 Give the coordinates of the turning point of f(x + 2) and describe the translation that took place. 15.1. To calculate the lengths of OA, OB and OC, you first have to find the coordinates of A, B and C. A and B are the x-intercepts, therefore let y = 0. wx? -4y-5=0 “ &-5)@+1) =0 x=S5orx=-l A(-1;0) B(5;0) Coordinates of C(0;-5) [C is y-intercept -. x = 0.] “OA =1unit OB =5 units OC = 5 units : F Pye od, _ 4) 15.2 Coordinates of Turning Point : x = 2a ron xed x=2 Substitute x = 2 into the equation: y = (2)?-4(2)-5 “y=4-8-5 a yad 2. D(2;-9) 90

15.3 EF is parallel to the y-axis + length of EF = y-coordinate of E. First calculate the x-coordinate of E. E is the point of intersection of f(x) and g(x). Let f(x) = g(x) and solve the equation. x2 -4x-S=xt] (fx) = x? -4x~5 and g(x) = x+1] “x? -5x-6=0 “. («-6)(x+1) =0 “x=6 or x=-l butx>0 aE “ALE: x =6 Letx = 6in g(x): y=x+1 “ y=641 “Y=T :. E(6;7) «. Length of EF = 7 units 15.4 GH is the difference between g(x) and f{x). -. GH = g(x) — fix) “ GH=x+1-(x?-4x-5) “ GH=x+1—-x? +445 +. GH = -x? +5x+6 7 =, (Now complete the square ] c = -(x? - 5x - 1 GH = ~-@? - 5x4 (-3)?- (3)? - 6) GH=-{@- 3) - B-6) GH =~ 3)? - Py “ GH = -@- 3° +P 2 x-1= 2] .. Maximum length of GH = #2 = 12% Copyright reserved 91 92 15.5 Range is the y-values. The graph has a turning point at (2;-9). Therefore, the minimum value is — 9. -. Range = {y; y > -9, y © R}. 15.5 Turning Point of fix +2) = (0;-9) [Graph moves 2 units to the left | «. Translation of 2 units to the left. Example 16 The diagram represents the graphs of Aa) = -x? — 2x + 3 and g(x) = eep +q. D is the turning point of the parabola. 16.1 Determine the average gradient of Ax) between the points x = 2 and x = 3. q 16.2 Determine the values of p and q. 16.3 Hence, calculate the coordinates of E. 16.4 Give the domain and range of g(x). 16.5 Calculate g(-5) Solution . Change in y yi-y2 16.1 Average gradient = ———"—— or ===> Change in x So first find the values of y when x = 2 andx = 3. y = (2)? —2(2) +3 y = -(3)? - 23) +3 aya y= l2 “. Average gradient between (2,-5) and (3;~-12) . Average gradient = 3501) = =H ==]

OR L2) -f3) 2-3 Average gradient = — £2)? =2@) + 3)=[-G)? - 2G) +3] Des) -5 ~(-12) =| 2 -l I Mt | x 16.2 The vertical asymptote passes through the turning point D. Therefore, calculate the x-coordinate of the turning point. =o naa Se "FC x=-l pal [Vertical asymptote: x =-p, -.-p=-l, ..p=1] qg=1 [Horizontal asymptote : y = q ] 16.3 E is the y-intercept of g(x), the hyperbola. 0 ye bei +1 [p=landq=1. Letx=0] “y=-2+l yerl “. E(;-1) 16.4 Domain = the x-values = {x : x © R, x #-1} Range = the y-values = {y : y © R, y # \} 93 Copyright reserved Ae) i ith 5 i = 72 16.5 9(5) = aT +1 [Substitute x with -5 in g(x) x41 +1] = 41 =2 = y Sketched are the graphs of Ax) = +2 3 x+1 and g(x) =x+1. 17.1 Calculate the coordinates of A ™~ 7 x and B. i 17.2 Determine the coordinates of C. 17.3 If DE = 4 units, determine the coordinates of E. 17.4 Determine the equation of the axis of symmetry of f(x). 17.5 his the graph of f shifted 2 units to the right and one unit downwards. Give the equation of h(x). Solution 17.1 A is the x-intercept of f(x). Lety = 0. =—3_ 42 = i 0 cat t2 ly = 0, determine x ] 7 0=34+2(r41) [Multiply by (x +1) ] “ 0=3+2x+2 2x = [Move 2x to left-hand side | axed 2 AC 350) 94

B is the y-intercept of f(x). Let x = 0. 3 [ye ora 2 [x = 0, determine y ] “y=3+2 pss . B(0;5) 17.2 C is the point of intersection of f(x) and g(x). Let fx) = g(x). 3 5 +2=x+1 x+1 3 42(x4+1) = (%+)D@+1) [Multiply by LCD > (x +1) J 3 42x42 =x24+x4x41 [Remove brackets ] 0 =-3-2x-24x?4+x4+x41 .0=x7-4 *, 0 = (x-2)(x +2) [Factorise RHS ] “x=2 or x=-2 AtC,x > 0 “x= 2 Letx=2iny=x+1 y=2tl ys3 - C(2;3) 17.3 DE is the difference between f(x) and g(x). “. DE = f(x) - g(x) Copyright reserved 95 96 : = _ . DE = al +2-(«+1) oo 4=—3-42-2x-1 x a [DE = 4, given ] © At 1) = 34+2(%4+1)-—x(e4+1)- (+1) [Multiply by (x +1) ] 2 4x44 =342x4+2-x?-x-x-1 o 4et4-3-2e-24+x?4+x4+x4+1=0 “x44 = 0 7. x(x+4) =0 “ x=0 or x=-4 AtE,x<0 «x=-4 E is a point on g(x). Therefore, let x = —4 in g(x). wy=3 “. E(-4;-3) 17.4. Equation of axis of symmetry y=(Q@t+p)+q 9 or y=-@t+p)+q “y= +1) +2 oy =—(e+1)4+2 “ p=xt+142 wo ps=-x-14+2 Bb Videosu) Be 7 eas Il “ y=xt+3 or y=-xt1 17.5 1 unit downwards : y => y+1. > 2 units to the right : x => x-2 Ge yap il 2 A @-2)+1° 6 ye = Lys yey t2 1 1 AG) = 3241 aE Sa

In Grade 10 you have sketched the graphs of and graph 97 98 Copyright reserved

Copyright reserved 99 7.3 | The graphs of y = sin(x +p), y = cos(x + p) and y = tan(x +p)| Sketched are the graphs of fx) = cos(x), g(x) = cos(x + 30°) and h(x) = cos(x — 60°). / | oe \-180° 7 {80° of 99° 180° / 270° S oe SS a - -L +e The effect of p : g(x) is the graph of (x) translated 30° to the left while A(x) is the translation of x) 60° to the right. Therefore : p is the horizontal shift of the graph. 74 [The graphs of y = asink(x +p), y = acosk(x + p) and y = atank(x + p) You already know that : a is a vertical stretching or shrinking of the graph, kis a horizontal stretching or shrinking and pis a horizontal shift of the graph. Example 18 Sketch the graph of y = 2sin2(x — 30°) for x © [ -180°; 180°]. Solution k = 2 > Horizontal shrinking + Period = 360° +2 = 180°. a=2 > Vertical stretching + Maximum = 2, minimum = -2. p = -30° > Horizontal shift 30° to the right. You can sketch the graph of y = sinx and then do the translations, or Set up a table, 100 x | 180° | -165° | -150° | -135° | -120° | -105° | -90° | -75° fo) -1,23) -1 | 0 | 1 lia] 2 [12] 1 | 60°] -45° | -30° |-15°] 0° | 15°] 30°] 45°) 60° o | -1 [-1.73] -2 [-1.73]-1) 0 | 1 [1.73] [75° | 90° | 105° | 120°] 135°] 150° | 165°| 180° | 2 |1.73| 1 | 0 | -1 [-1.73) -2 |-1.73 Now sketch the graph by means of point-by-point plotting. y (105° ; 2) (75°32) (65°; -2) Very small intervals are used in order to obtain the values of the turning points. In the graph of y = 2 sin2(x — 30°) 3 parameters are used , a(2), k(2) and p(30°). In the examination no more than two parameters at a time will be asked simultaneously, eg. y = sink(x + p) or y = asin(x + p) or y = asin(kx) ory = sin(x+p)+q Example 19 On the same set of axes, sketch the graphs of f(x) = cos(2x — 60°) and g(x) = 2sin(x + 30°) for x € [-180°; 180°].

Solution Use your calculator to set up tables and then plot the points. y (60°; 2) g (1502: -1) f 0 60° 120° 180° (120°; -1) (-120°; -2) Example 20 Sketched are the graphs of y = sinax for x € [-120°; 120°]. y 1 Qs) -120° i “1 20.1 Determine the value of a. 20.2 What is the range of y = sinax+1 Solution 20.1 sina(30°) = 1 [(30°; 1) is a point on the graph | *, a(30°) = 90° [sin90° = 1 shift || sin anes [90° + 30° =3] [Or : The period of sinax = 120°: +. 38 = 120° a=3] x 101 Copyright reserved 20.2 Range ofy = sinax is -l<y<1l .. Range ofy = sinax+1 is -l+1<y<1+1 “0S ys2 Example 21 21.1 On the same set of axes, draw sketch graphs of f(x) = cos(x — 30°) and g(x) = sin2x for x € [-180°; 180°]. 21.2 For which values of x in the interval x € [0°;90°] is (x) = g(x). Solution : 21.1 Set up a table or sketch using translations. y -180 ad -150 -120 - Ome 30) 60 a 1 150 0 21.2 cos(x — 30°) = sin2x *. cos(x — 30°) = cos(90° — 2x) Be boewelhe Ceo rh or “. X+2x = 90° + 30° + k.360° . 3x = 120° + k.360° “x = 40° +#,120° x — 30° = 360° — (90° — 2x) <. X= 30° = 270° + 2x + k.360° “X= 2x = 270° + 30° + k.360° *. —x = 300° + k.360° “x = —300° + k.360° Let k= 1 : x = -300° + (1)360° = 60°. -. fix) = g(x) for x € [0°;90°] when x = 40° or x = 60°. Example 22 Sketch the graph of y = tan2x {-90° < x < 90°} 102

Solution Asymptotes at x = 45° + k.90°. Use intervals of 22, 5° on x-axis. y ee eee a | & ' ' i t 80 ae 225 1 ' ‘ 1 1 Example 23 Sketched are the graphs of f(x) = tanax and g(x) = cos(x + p®). Determine the values of a and p. Solution Asymptote of tanx is x = 90°, <. a(45°) = 90° [The asymptote of tanax is x = 45° ] “.@=2 [90+45 =2] cos(45° +p) = 1 [(45°; 1) point on g(x) J =) Eh “. 45° +p = 0° L| shift “p= 45° Copyright reserved 103 Example 24 The diagram shows the graphs of f(x) = atan(x + p) and g(x) = for x € [-360°; 360°]. Determine the equations of f(x) and g(x). uy (0; 3) ' ' ' p ' ' 1 1 ' \ 1 1 fl 1 ' 1 1 ' i ' ‘ 104 acosx+q 1 i ' 1 1 \ Solution ' 1 1 1 1 1 4 ' 1 1 Ee 360° 5 -28R_ -180 5420-60 0 12H 180 3 1 4 1 i i | (180; -1) t f ‘ f i 1 ‘ 4 x 00360 \ \ ' Asymptote of atan(x +p) : x = 120° > Asymptote of tanx : x = 90° -. fx) is horizontal shift of y = tanx 30° to the right and areflection of y = tanx in the x-axis. «. fix) = —tan(x — 30°) Amplitude of g(x) = 2 + Amplitude of cosx = 1. .a=2 Maximum value of 2cosx = 2 Maximum value of acosx + q = 3 - Vertical shift | unit upwards . gel “. g(x) = 2cosx+1

SUMMARY The 105 106 Copyright reserved