WRITTEN ASSIGNMENT 5 Name of the Student Name

Running head: WRITTEN ASSIGNMENT 5
WRITTEN ASSIGNMENT 5
Name of the Student
Name of the University
Author Note

6.1:
1.
y=12.25 ( 2 x
3 )
3
When x increases 2x/3 decreases and its cube also decreases. Thus the function is
exponentially decaying.
2.
y= ( 7
8 )
0.75 x
−2
When x is positive then 0.75x is less than x and hence as x is increased in positive x axis then
the value asymptotically goes to zero. However, when x is increased in a negative direction
the value of y exponentially goes to infinity. Hence, the function exponentially increasing in
negative x and asymptotically goes to -2 in positive x direction.
3.
Let, the equation of the exponential function is
y = a*(b^x)
Now, function passes through (0,21) and (2,38)
Hence, 21 = a*b^0 => a = 21
And, 38 = a*b^2 => b^2 = 38/a => b = sqrt(38/21) = 1.345.
Hence, y=21∗(1.345x)
4.
(−1, 5) and (4, 2) points are in y = a*(b^x)
Hence, 5 = a*(b^-1) => 5 = a/b => a = 5b

And, 2 = a*(b^4) => 2 = 5b*b^4 => b^5 = 2/5 => b = 0.833.
a = 5*0.833 = 4.165
Hence, the exponential model = 4.165∗(0.833x )
5.
a. Given value of investment account = 15500 ( 1+ 0.06
6 ) 90
= 37953.806
b. The initial deposit made to the account is 15500.
c. The number of years taken for the account is 90/12 = 7.5 years.
6.
A(t)=P (1+ r
n )nt
P = value in 1980 = $95000
A(t) = $133300
n = 1, t = 20
133300 = 95000( 1+r ) 20
 ( 1+r ) 20=133300 / 95000
 20ln( 1+r ¿ = ln(133300/ 95000 ¿ = 0.3387
 ln(1+r) = 0.0169
 1+r = 1.017
 r = 0.017
 r = 1.7%
Now, if the same percentage continues then the median price in the year 2020 will be
A(40) = 95000 ( 1+0.017 ) 40 = $ 186449.71 ~ $ 186450

6.2:
7.
f(x) = (-2/3)(1.25)^(3x)
Reflection about y axis is f(-x) = (-2/3)(1.25)^(-3x)
Y intercept is (0,-0.667).
8.
Given f(x) = 3x+1 – 2