Page 43 - Hybrid PBD 2022 Form 4 Additional Mathematics

Matematik Tambahan Tingkatan 4 Bab 2 Fungsi Kuadratik
16. Lakarkan setiap fungsi kuadratik yang berikut. SP 2.3.5 TP4
Sketch each of the following quadratic functions.

(a) f(x) = 2x + 8x + 11
2
f(x) = 4x + 10x – 6 Melakarkan graf
2
a = 4  0, titik minimum fungsi kuadratik a = 2  0, titik minimum / minimum point
minimum point Sketching the graph 2 2
NOTA of quadratic function b – 4ac = 8 – 4(2)(11)
b – 4ac = 10 – 4(4)(–6) = –24  0
2
2
= 196  0 Tiada pintasan-x / No x-intercept
Pintasan-x pada dua titik berbeza 11

2
x-intercepts at two different points f(x) = 2 x + 4x + 2 
5


f(x) = 4 x + x – 3  = 2 x + 4x + 2 – 2 + 11 
2
2
2
2
2 2 2
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2
2
5
2
5

5
3
= 4 x + x +      = 2(x + 2) + 3
–
–
2
2 4 4 2 Titik minimum: (–2, 3)

= 4 x + 5  2 – 49 Minimum point
4 4 f(0) = 2(0) + 8(0) + 11
2
5

Titik minimum: – , – 49  = 11
Minimum point 4 4
f(x)
f(x) = 0 f(x) f(x) = 2x + 8x + 11
2
2
4x + 10x – 6 = 0 f(x) = 4x + 10x – 6
2
(2x – 1)(x + 3) = 0 – 3 0 1 x 11
1
x = , x = –3 2
2 – 6
f(0) = 4(0) + 10(0) – 6
2
= –6 – (–2, 3) x
5
49
–
,
4 4 0
(b) f(x) = –(x – 2)(x + 4) (c) f(x) = 3 – (x + 1) 2
a = –1  0, titik maksimum / maximum point a = –1  0,
2
f(x) = –x – 2x + 8 titik maksimum / maximum point (–1, 3)
2
b – 4ac = (–2) – 4(–1)(8) f(x) = –x – 2x + 2
2
2
= 36  0 b – 4ac = (–2) – 4(–1)(2)
2
2
Pintasan-x pada dua titik berbeza = 12  0
x-intercepts at two different points Pintasan-x pada dua titik berbeza
f(x) = –(x + 2x – 8) x-intercepts at two different points
2
= –(x + 2x + 1 – 1 – 8) –(–2) ± 12
2
2
2
= –(x + 1) + 9 x = 2(–1)
2
Titik maksimum / Maximum point : (–1, 9) = −2.73 atau / or 0.73
f(x) = 0 f(0) = 3 –(0 + 1) 2
–(x – 2)(x + 4) = 0 = 2
x = 2, x = –4 f(x)
f(0) = –(0 – 2)(0 + 4) (–1, 3)
= 8
2
f(x)
(–1, 9)
x
8 –2.73 0 0.73
f(x) = 3 – (x + 1) 2
x
–4 0 2
f(x) = –(x – 2)(x + 4)
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