Orange Coast College MATH A115: Exam 5

VL. Dhann




oZ) Fnd #e solution in ferms of ("’fi-‘

w® (1.02) - oo, o0
[ )

L5 (4 Oé)“- 500

ot(03 (408)= log (50)
£ - log(5oo>
o? Log('f OX)

‘lJc 4o, 5%50]

4)' Tfl: (Jofi 2 (bx+ Z): £
q
02 &= 6><‘f‘°2


A6 = 6x +

14 = 6x

M = x
6

7
l5~x

,| B) Madth tach equatien “Thu Pham
x=-1 |x:0 | x:1 Exom >
2\(1.(06 1650 | 4 24q |p 4 Bl He (03

Gos|aay | 4 |am6 |[E log 10,00) x
dlo.eef | 066t | 4 |Ley |p 10" _ 40,0
403 (33603 | 4 436 |a L

4(4y) |2.857L | 4 5,77: :>E&

é) F\vu(. ‘W-L equh'm- .




I(X): a+bccx
sWe bowe He qroph passes Tuowh (0,-2)4(1,2)
¥ o= -4 because x approaches -0 > Y=-4
£ (o) :’4+beix: -2

=D-& + be}@’; -2 2

"’,2 _f()(>: _H+02gn5)k
-.Ll +b =

oy b; ol :>Lf_{ikq +°2(5


SEG): -4+ 2e’
£(4)= -4 + J48 . 2
-4 de =2
ief - b

L—ef"—fig czlnb