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 the values 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 the equation 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 the discriminant 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 the equation 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 object is 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 of graph 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 graph depicted 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