Download - Yr.12 Transition Workshop 2012-2013
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14 = 48 – 8a + 8 + 10 + 12
8a = 64 a = 8
a = 8
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3(x4
– 3x3
+ 2x2
– 22) – (35 – 3x – x4) + 4(4x
3– 3x
2+ 3x – 2)
= 3x4
– 9x3
+ 6x2
– 66 – 35 + 3x + x4
+ 16x3
– 12x2
+ 12x – 8
= 4x4
+ 7x3
– 6x2
+ 15x – 109
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Q(x) = x4
+ ax3
+ bx2
+ 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)
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a = -5
b = -1
a = -5, b = -1
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a = 4, b = 60
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y1 = 3x + 6
x = -2 (-2,0)
x-4 -2 2 4
y
-5
5
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Q8 (i) : y – y1 = m(x – x1)
m = – 34
(x1 , y1) = (-3 , 5)
y – 5 = – 34
(x – -3)
y – 5 = – 34
x – 94
y = – 34
x + 114
4y = -3x + 11 3x + 4y – 11 = 0
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Q8 (ii) : y – y1 = m(x – x1) (x1 , y1) = (-2 , 3) (x2 ,y2) = (5, – 8)
m = y2 – y1
x2 – x1
= -8 – 35 – -2
= – 117
y – 3 = – 117
(x – -2)
y – 3 = – 117
x – 227
y = – 117
x – 17
7y = -11x + 1 11x + 7y + 1 = 0
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x-4 -2 2 4 6 8
y
-5
5
10
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Straight line y – y1 = m(x – x1)
(x1 , y1) = (-3, 7) m = ?
-4y = – x – 5
y = 1
4 x + 5
4
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m1 = -4
So y – 7 = -4(x – -3)
y – 7 = -4x – 12
y = -4x – 5
4x + y + 5 = 0
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= b2
– 4ac
= (-20)2
– 4 4 25
= 400 – 400
= 0
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(-3, 0) (0, – 6) (2, 0)
b = 1, c = -6 y = x2
+ x – 6
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x1 2 3 4
y
- 30
- 20
- 10
10
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x1 2 3 4
y
- 30
- 20
- 10
10H = 8x – 4x
2+ 5, x [0,4] ,
H = -4x2
+ 8x + 5
H = -4 x2
– 2x – 54
H = -4 (x
2– 2x + 1) – 1 – 5
4
H = -4
(x – 1)
2– 9
4
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y = (x – 2)(x + 3)(2x – 1)
(x – 2)(x + 3)(2x – 1) = 0
(2,0) (-3,0)
12
,0
x = 0
f(x) = 2x3
+ x2
– 13x + 6 f(0) = 6
(0,6)
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x- 5 - 4 - 3 - 2 - 1 1 2 3 4 5
y
- 10
- 5
5
10
15
20
25
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96 = a (-4)2(3)(-2)
a = -1
y = – (x – 4)2(x + 3)(2x – 5)
a = 5, b = 27, c = 30, d = 256, e = 96
y = -5x4
+ 27x3
+ 30x2
– 256x + 96
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0 = (-2)4
+ a (-2)3
+ b (-2)2
+ 36(-2) + 144
-8a + 4b + 88 = 0
-8a + 4b = -88
0 = (-2)4
+ (a + 3)(-2)3
– 23(-2)2
+ (b + 10)(-2) + 40
0 = -8a – 2b – 80
-8a – 2b = 80
a = -3, b = -28
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a = -3, b = -28