ADDITIONAL MATHEMATICS FORM 4: TOPIC & SUBTOPIC 3.0 QUADRATIC FUNCTIONS, Q f f(x) = ax2 + bx + c 3.1 Q f – characteristics → a≠0 , x2 , smooth symmetrical curve (parabola:- a > 0 : U & a < 0 : ∩) – relation between Graph & Types of Roots – Different → TWO – Same → ONE – No root → NOT TOUCH 3.2 Max & Min value of Q f → use Completing the Square f(x) = a(x + P)2 + Q
↔
Max @ Min = ( -P , Q )
1) f (x) = ax2 + bx + c 2) Factorise to make a = 1….[ a is friend of x2 ] 3) Double check 2
b 4) Take out ? ( b is friend of x ) 2 5) Do CtS with + &
–
→ operation = a[……….
6) Solve to form in: f(x) = a(x + P)2 + Q
↔
2
2
]
Max @ Min = ( -P , Q )
3.3 Sketch Q f – shape of graph [parabola:- a > 0 : U & a < 0 : ∩] – coordinate of MAX @ MIN point [by CtS] : (x , y) – label the y-intercept / x-intercept : (0 , y) / (x , 0) – smooth curve
3.4 Quadratic Inequalities → f(x) > @ ≥ 0 : shaded ABOVE x-axis & f(x) < @ ≤ 0 : shaded BELOW xaxis → REMEMBER: ( a > 0 : U & a < 0 : ∩) → if a > 0 follow the condition as shown: f(x) < @ ≤ 0: shaded BELOW x-axis f(x) > @ ≥ 0 : shaded ABOVE x-axis
(x – h)(x – k) < 0 h
(x – h)(x – k) ≤ 0 h≤x≤k
(x – h)(x – k) > 0 x < h or x > k
(x – h)(x – k) ≥ 0 x ≤ h or x ≥ k
BRILLIANT CORNER :
1. The relation of Q f and QE (types of roots) → Parabola, when a > 0 : graph smile U & when a < 0 : graph unsmile ∩
TYPES OF ROOTS
DIFFERENT
SAME
NO ROOT
CONDITION/EQN
b2 – 4ac > 0
b2 – 4ac = 0
b2 – 4ac < 0
TOUCH @ INTERSECT
TWO POINT
ONE POINT
NOT TOUCH
GRAPH POSITION when a > 0
GRAPH POSITION when a < 0
2. Solve QE → method: Completing the Square in the form f(x) = a(x + P)2 + Q 2
= 4 [ x2 + 2x –
7 ] 4
= 4 [ x2 + 2x + 12 12 – = 4 [ x 12 –
2
b 2 …… 12 2 2
a) f (x) = 4x2 + 8x – 7
7 ] 4
11 ] 4
= 4 x 12 – 11 ] ↔ in form f(x) = a(x + P)2 + Q ↔ Max @ Min = (- P, Q) ∴ P = 1 , Q = - 11 ↔ MIN. point = (- 1 , -11 ) b) f (x) = 5 – 6x – x2 = - x2 – 6x + 5 = - [ x2 + 6x – 5 ] = - [ x2 + 6x + 32 32 – 5 ] = - [ x 32 – 14 ] = - x 32 + 14
↔ in form f(x) = -a(x + P)2 + Q ↔ Max @ Min = (- P, Q)
∴ P = 3 , Q = 14 ↔ MAX. point = (- 3, 14)
c) f (x) = -3x2 + 2x + 5 2 5 = -3 [ x2 – x – ] 3 3 2 2 2 5 = -3 [ x2 – x + 1 1 – ] 3 3 3 3 2 16 1 = -3 [ x – ] 9 3 2 16 1 = -3 [ x + ] ↔ in form f(x) = -a(x + P)2 + Q ↔ Max @ Min = ( - P , Q) 3 3 1 16 1 16 ∴ P=- ,Q= ↔ MAX. point = ( , ) 3 3 3 3 E1 - Given the Qf : y 2( x 3) 2 9 . State → compare in form f(x) = a(x + P)2 + Q ↔ Max = ( - P , Q) (a)
the coordinates of the maximum point,
(b) the equation of the axis of symmetry → symmetry:
→ MAX. point = ( - 3 , 9 )
x=-3
E2 - Diagram shows the graph of a Qf : f ( x) 3( x p) 2 4 , where p is a constant. The curve y f (x) has the minimum point (2, q), where q is a constant Rajah menunjukkan graf fungsi kuadratik f ( x) 3( x p) 2 4 , dengan keadaan p ialah pemalar. Lengkung y f (x) mempunyai titik minimum (2, q), dengan keadaan q adalah pemalar. f(x) = a(x + P)2 + Q ↔ Min = (- P , Q)
y y = f (x)
given: f ( x) 3( x p) 2 4 ↔ MIN. point = ( -2 , q ) {a = 3, p = 2, 4 = q} (2, q) O
State:
(a)
x
the value of p and q ∴p=2 ∴q=4
→ MIN. point = ( -2 , q ) = ( -2, 4) → symmetry: (b)
x = -2
equation of the axis of symmetry. → symmetry: x = -2
E3 - The Qf : f ( x) a( x p) 2 q , where a, p and q are constants, has a maximum value of 5. The equation of the axis of symmetry is x = 3. Fungsi kuadratik f ( x) a( x p) 2 q , dengan keadaan a, p dan q adalah pemalar, mempunyai nilai minimum 5. Persamaan paksi simetri ialah x = 3. (a)
range of values of a →a>0
(b) value of p and q → MIN. point = ( 3 , 5) given: f ( x) a( x p) 2 q ∴ p = -3 ∴ q = 5
E4 - Diagram shows the graph of the function y ( x 1) 2 9 , where m is a constant. The curve touches the line y m at point A and cut the y-axis at point B. The curve also cut the x–axis at point P. Rajah 2 menunjukkan graf fungsi y ( x 1) 2 9 , dengan keadaan m ialah pemalar. Lengkung itu menyentuh garis y m di titik A dan menyilang paksi-y di titik B. Lengkung itu juga menyilang paksi-x di titik P. y
f(x) = -a(x + P)2 + Q ↔ Max = (- P , Q)
A
given: y ( x 1) 2 9 ↔ MAX. point = ( -1 , 9 ) → MAX. point = ( - 1 , 9 ) → symmetry: x = - 1
y=m
B (0, k)
P
(a)
Find the value of m and of k.
f ( x) ( x 1) 9 2
O
(b) State the coordinates of point P.
f ( x) ( x 1) 2 9
→ at B (0, k): f (0) k
→ at P (x, 0): f ( x) 0
(0 1) 9 = 8
( x 1) 2 9 = 0
∴k=8
( x 2 2 x 1) 9 = 0 x 2 2x 1 9 = 0
2
→ at y = m, MAX. point = ( - 1 , 9 ) ∴m=9
x
x 2 2x 8 = 0 (x – 2) (x + 4) = 0 x = 2 or x = -4 → since coordinate of P located at –ve x, so x = -4 ∴ P =(-4,0)
E5 - By expressing the function f ( x) 3x 2 6 x 5 in form of f ( x) a( x p) 2 q , find the minimum value of f (x) . Dengan mengungkapkan fungsi f ( x) 3x 2 6 x 5 dalam bentuk f ( x) a( x p) 2 q , cari nilai minimum bagi f (x) . → use method: Completing the Square f (x) = 3x2 – 6x + 5 5 = 3 [ x2 – 2x + ] 3 5 = 3 [ x2 – 2x + 12 12 + ] 3 2 = 3 [ x 12 + ] 3 = 3 x 12 + 2 ↔ in form f(x) = a(x + P)2 + Q ↔ Min = ( - P , Q) MIN. point = (1 , 2) ∴ Minimum value of f (x) = 2
E6 - Diagram shows the graph of a Qf : y f (x) . The straight line y 9 is a tangent to the curve y f (x) . Rajah menunjukkan graf fungsi kuadratik y f (x) . Garis lurus y 9 ialah tangen pada lengkung y f (x) . y
x=4
f(x) = a(x + P)2 + Q ↔ Min = ( - P , Q) → found: MIN. point = ( 4 , - 9 ) → f(x) = a(x – 4)2 – 9 → symmetry: x = 4
0
1
x
7
y = 9
( 4 , -9 )
(a)
Write the equation of the axis of symmetry of the curve. → symmetry: x = 4
(b)
Express f (x) in the form of ( x p) 2 q , where p and q are constants. → f(x) = a(x – 4)2 – 9
E7 - Find the maximum or minimum value for the function y 1 3x 2 x 2 . Thus, find the equation of the axis of symmetry for the function. Cari nilai maksimum atau minimum bagi fungsi y 1 3x 2 x 2 . Seterusnya, cari persamaan paksi simetri bagi graf fungsi itu. → use method: Completing the Square f (x) = 1 + 3x – 2x2 = – 2x2 + 3x + 1 3 1 = -2 [ x2 – x – ] 2 2 2 2 3 1 = -2 [ x2 – x + - 3 3 – ] 2 2 4 4 2 17 3 = -2 [ x – ] 16 4 2 17 3 = -2 x + ↔ in form f(x) = -a(x + P)2 + Q ↔ Max = ( - P , Q) 8 4 MAX. point = (
3 17 , ) 4 8
∴ Maximum value of f (x) = ∴ symmetry:
x=
3 4
17 8
E8 - Diagram shows the graph of a Qf : f ( x) px 2 4 x q , where p and q are constants. The curve y f (x) has a maximum point (1, 5). State the values of p and of q. Rajah menunjukkan graf fungsi kuadratik f ( x) px 2 4 x q , dengan keadaan p dan q adalah pemalar. Lengkung y f (x) mempunyai titik maksimum (1, 5). Nyatakan nilai p dan nilai q. y
f(x) = a(x + P)2 + Q ↔ Max = ( - P , Q)
(1, 5)
→ given MAX. point = ( -1 , 5 ) → f(x) = - a(x + 1)2 + 5 → symmetry: x = -1 O
x
→ use method: Completing the Square f (x) = px2 – 4x – q 4 q = p [ x2 – x – ] p p 2 2 4 q = p [ x2 – x + 4 4 – ] p p 2p 2p 2 q 16 4 = p [ x – ] – p 2p 4 p2 2 16 4 = p x –q ↔ – 4p 2 p
So that, P =
in form f(x) = a(x + P)2 + Q ↔ Min = ( - P , Q)
4 16 , Q=– –q 2p 4p
MIN. point = (
4 16 ,– – q ) ↔ Found Min = (-1 , 5) 2p 4p
→compare pattern:
4 = -1 2p 4 = -2p -2 = p ∴
16 –q=5 4p 16 – –q=5 4(2) 2–q=5 2–5=q -3 = q ∴ –
E9 - The function f ( x) a bx 3x 2 has a minimum value of 6 when x = 2. Find the value of a and of b. Fungsi f ( x) a bx 3x 2 mempunyai nilai maksimum 6 apabila x = 2. Carikan nilai a dan nilai b.
→ use method: Completing the Square f (x) = a – bx – 3x2 = – 3x2 – bx + a b a = -3 [ x2 – x – ] 3 3 2 2 a = -3 [ x2 – bx + - b - b – ] 3 6 6 2 b2 a b = -3 [ x – – ] 36 3 6 2 b2 b = -3 x + +a ] ↔ 12 6
in form f(x) = -a(x + P)2 + Q ↔ Max = ( - P , Q)
given: maximum value = 6 when x = 2, so MAX. point = (2 ,6)
b b2 So that, MAX. point = ( , – a ) ↔ MAX. point = ( 2 , 6 ) 6 12 →compare pattern:
b =2 6 ∴ b = 12
b2 +a=6 12 (12) 2 +a=6 12 12 + a = 6 a = 6 – 12 ∴ a=-6
E10 - Find the range of values of x for which (3x 2)( x 5) x 5.
{a = 3, b = -14, c = -5} {use EQN, then x = 5, x = -
3x2 – 15x + 2x – 10 > x – 5 3x2 – 13x – x – 10 + 5 > 0 3x2 – 14x – 5 > 0 (x – 5)(3x + 1) > 0 1 } 3
→ please help yourself to sketch and write down the answer
{a > 0 OK!}
E11 -
Find the range of values of x for which x 2 16
x2 – 16 ≤ 0 (x – 4)(x + 4) ≤ 0
{a = 1, b = 0, c = -16} {use EQN, then x = 4, x = -4 } → please help yourself to sketch and write down the answer
E12 - Find the range of values of x for which (3x 1)(2 x 1) 3 x 6x2 – 3x + 2x – 1 > 3 + x
6x2 – x – x – 1 – 3 > 0 6x2 – 2x – 4 > 0 3x2 – x – 2 > 0 (x – 5)(3x + 1) > 0
{a = 3, b = -1, c = - 2} {use EQN, then x = 1, x = -
{a > 0 OK!}
2 } 3
→ please help yourself to sketch and write down the answer
E13 - Given that the graph of quadratic function f ( x) 2 x 2 bx 8 always lies above the x-axis. Find the range of values of b. Diberi graf fungsi kuadratik f ( x) 2 x 2 bx 8 sentiasa berada di atas paksi-x. Cari julat nilai b. Always lies above the x-axis
{a = 2, b = b, c = 8}
f ( x) 2 x 2 bx 8
→ no root:
b2 – 4ac < 0 b2 – 4(2)(8) < 0 b2 < 64 b < ± √64 b<±8 ∴ -8 < b < 8
