Cambridge International Mathematics

a Thex-intercepts are¡ 1 , 2 , and 4

) f(x)=a(x+ 1)(x¡2)(x¡4)

But when x=0, y=¡ 8

) a(1)(¡2)(¡4) =¡ 8

) 8 a=¡ 8

) a=¡ 1

So, f(x)=¡(x+ 1)(x¡2)(x¡4)

b The graph touches thex-axis at^23 ,

indicating a squared factor (3x¡2)^2.

Its otherx-intercept is¡ 3 ,so

f(x)=a(3x¡2)^2 (x+3)

But when x=0, y=6

) a(¡2)^2 (3) = 6

) 12 a=6

) a=^12

So, f(x)=^12 (3x¡2)^2 (x+3)

Example 3 Self Tutor

Find the equation of the cubic

with graph:

We only know two of thex-intercepts,¡ 3 and 4 ,so(x+3)and (x¡4) are linear factors.

We suppose the third linear factor is (ax+b),sof(x)=(x+ 3)(x¡4)(ax+b)

But f(0) = 6

) (3)(¡4)(b)=6

) ¡ 12 b=6

) b=¡^12 ...... (1)

and f(2) = 25

) (5)(¡2)(2a+b)=25

) 2 a¡^12 =¡^52 fusing (1)g

) 2 a=¡ 2

) a=¡ 1

Thus f(x)=(x+ 3)(x¡4)(¡x¡^12 ) or f(x)=¡(x+ 3)(x¡4)(x+^12 )

EXERCISE 23A.2

1 Find the form of the cubic function with graph:

abc

O

y

-3 (^4) x
6
()2 ¡25,
y
-1 x
12
O 23
y
x
-4
-12
3
O
O
-3 -2 x
-12
-\Qw_
y
472 Further functions (Chapter 23)
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Y:\HAESE\IGCSE01\IG01_23\472IGCSE01_23.CDR Monday, 27 October 2008 2:18:34 PM PETER