Yr.12 Transition Workshop 2012-2013

Yr.12 Transition Workshop 2012-2013

• 1.
  Yr.12 Maths Methods Transition Workshop
• 2.
  4 3 2 Q1: If P(x) = 3x + ax + 2x – 5x + 12 and P(-2) = 14 , find the value for a . 14 = 48 – 8a + 8 + 10 + 12 8a = 64 a=8 CAS: 4 3 2 F4 1: Define P(x) = 3x + a ´ x + 2x – 5x + 12 F2 1: solve (p(-2) = 14,{ a} ) ENTER a=8
• 3.
  4 3 2 4 Q2: If P(x) = x – 3x + 2x – 22 , Q(x) = 35 – 3x – x and 3 2 R(x) = 4x – 3x + 3x – 2 , find 3 P(x) – Q(x) + 4R(x) . By hand: 4 3 2 4 3 2 3(x – 3x + 2x – 22) – (35 – 3x – x ) + 4(4x – 3x + 3x – 2) 4 3 2 4 3 2 = 3x – 9x + 6x – 66 – 35 + 3x + x + 16x – 12x + 12x – 8 4 3 2 = 4x + 7x – 6x + 15x – 109 CAS: 4 3 2 F4 1: Define P(x) = x – 3x + 2x – 22 4 F4 1: Define Q(x) = 35 – 3x – x 3 2 F4 1: Define R(x) = 4x – 3x + 3x – 2 3 P(x) – Q(x) + 4R(x) ENTER
• 4.
  4 3 2 Q3: If Q(x) = x + ax + bx + 8x – 9 , Q(2) = -21 and Q(-3) = 174 , find the values for a and b. By hand: 4 3 2 Q(x) = x + ax + bx + 8x – 9 Q(2) = 16 + 8a + 4b + 16 – 9 = -21 Q(-3) = 81 – 27a + 9b – 24 – 9 = 174 8a + 4b = -44  -27a + 9b = 126  2a + b = -11  , (   4) -3a + b = 14  , (   9)
• 5.
  a = -5 Sub. in ƒ b = -1 CAS: 4 3 2 F4 1: Define Q(x) = x + ax + bx + 8x – 9 F2 1: solve (Q(2) = -21 and Q(-3) = 174,{ a,b} ) a = -5, b = -1
• 6.
  4 3 Q4: Find the quotient and remainder when 2x + 7x + 9x – 3 is divided by 2x – 5 . CAS: F2 1: solve ( 2x – 5 = 0,x ) (gives remainder) æ 5ö p ç ÷ ENTER è 2ø This tells us that 207 is the remainder. What is the quotient? The quotient is the expression created by dividing.
• 7.
  Q6: Find thevalues of a and b, if (x + 4) and (x – 5) are 3 2 factors of x + ax – 17x + b CAS: 3 2 F4 1: Define p(x) = x + ax – 17x + b F2 1: solve (p(-4) = 0 and p(5) = 0,{ a,b} ) a = 4, b = 60
• 8.
  Q7 (i) : 2y – 6x – 12 = 0 CAS: y1 = 3x + 6 X intercept: F2 1: solve (3x + 6 = 0, x) x = -2 (-2,0) Y intercept: F5 1: value x = 0 (0,6) Sketch graph on CAS y 5 -4 -2 2 4 x -5
• 9.
  Q8 (i) : y – y1 = m(x – x1) m= –3 (x1 , y1) = (-3 , 5) 4 y – 5 = – 3(x – -3) 4 y–5= –3x–9 4 4 y = – 3 x + 11 4 4 4y = -3x + 11 3x + 4y – 11 = 0
• 10.
  Q8 (ii) : y – y1 = m(x – x1) (x1 , y1) = (-2 , 3) (x2 ,y2) = (5, – 8) m = y2 – y1 = -8 – 3 = – 11 x2 – x1 5 – -2 7 y – 3 = – 11(x – -2) 7 y – 3 = – 11 x – 22 7 7 y = – 11 x – 1 7 7 7y = -11x + 1 11x + 7y + 1 = 0
• 11.
  Q9: f(x) =2x – 1 , x Î (-2, 6] Graph : when x = -2 , f(-2) = -5 when x = 6, f(6) = 11 y 10 5 -4 -2 2 4 6 8 x -5 CAS: y1 = 2x – 1 | x 2nd -2 and x 0 6
• 12.
  Q10: Find theequation of the straight line that passes through the point (-3,7) and is perpendicular to the line with equation x – 4y + 5 = 0 Straight line y – y1 = m(x – x1) (x1 , y1) = (-3, 7) m=? Required line is perpendicular to x – 4y + 5 = 0 -4y = – x – 5 y=1x+5 4 4
• 13.
  Recall perpendicular gradients:m1 = – 1 m2 Here m2 = 1 4  m1 = -4 So y – 7 = -4(x – -3) y – 7 = -4x – 12 y = -4x – 5 4x + y + 5 = 0
• 14.
  Q11: Use thediscriminant to determine the number of 2 intercepts with the x-axis for y = 4x – 20x + 25 2  = b – 4ac 2  = (-20) – 4  4  25 = 400 – 400 =0 One intercept with the x-axis
• 15.
  Q12: Determine theequation of the form 2 y = ±x + bx + c for the following graph. Points which belong to the graph: (-3, 0) (0, – 6) (2, 0) CAS: 2 Define f(x) = x + bx + c (min. turning point – positive graph) b = 1, c = -6 2 y=x +x–6
• 16.
  13. An objectis thrown from the top of a high-rise building with an equation of trajectory 2 H = 8x – 4x + 5, x Î [0,4] , where H is the vertical distance from the top of the building and x is the horizontal distance from the building (both distances measured in metres). y 10 1 2 3 4 x - 10 - 20 - 30
• 17.
  2 10 y H = 8x – 4x + 5, x  [0,4] , 2 x H = -4x + 8x + 5 1 2 3 4  x2 – 2x – 5  H = -4  - 10   4 - 20  (x2 – 2x + 1) – 1 – 5  H = -4   - 30  4  (x – 1) 2 – 9  H = -4    4 Turning point: (1, 9) Max. height = 9
• 18.
  3 2 14. Factorise 2x + x – 13x + 6 and sketch the graph 3 2 of y = 2x + x – 13x + 6 , showing all intercepts. CAS: 3 2 F2 2: factor (2x + x – 13x + 6,x) y = (x – 2)(x + 3)(2x – 1) X intercepts: put y = 0 (x – 2)(x + 3)(2x – 1) = 0 (2,0) (-3,0)  1,0     2  Y-intercept: put x = 0 3 2 F4 1: define f(x) = 2x + x – 13x + 6 f(0) = 6 (0,6)
• 19.
  Check shape ofgraph by sketching it on CAS (note: doesn’t ask for turning points) Sketch: y 25 20 15 10 5 -5 -4 -3 -2 -1 1 2 3 4 5 x -5 - 10
• 20.
  15. The graphdepicted below is a quartic. Determine the equation of the curve from the information given. Quartic graph: · Turning point at (4,0) · X-intercepts at (-3,0), æ 2,0 ö , (4,0) ç ÷ è5 ø
• 21.
  2 Rule: y = a(x – 4) (x + 3)(5x – 2) · y-intercept (0,96) 2 96 = a  (-4) (3)(-2) a = -1 2 y = – (x – 4) (x + 3)(2x – 5) CAS 4 3 2 F4 1: Define f(x) = ax + bx + cx + dx + e F4 1: Define g(x) = F3 1: d(f(x),x) a = 5, b = 27, c = 30, d = 256, e = 96 4 3 2 y = -5x + 27x + 30x – 256x + 96
• 22.
  4 3 2 16. If the functions f(x) = x + ax + bx + 36x + 144 4 3 2 and g(x) = x + (a + 3)x – 23x + (b + 10)x + 40 both cross the x-axis at -2 , determine the values for a and b. 4 3 2 0 = (-2) + a  (-2) + b  (-2) + 36(-2) + 144 -8a + 4b + 88 = 0 -8a + 4b = -88  4 3 2 0 = (-2) + (a + 3)(-2) – 23(-2) + (b + 10)(-2) + 40 0 = -8a – 2b – 80 -8a – 2b = 80  Solve  and ƒ simultaneously a = -3, b = -28
• 23.
  CAS: 4 3 2 F4 1: Define f(x) = x + ax + bx + 36x + 144 F2 1: solve (f(-2) = 0 and g(-2) = 0,{ a,b} ) a = -3, b = -28