College Algebra Assignment: Analysis and Solutions to Questions 2-10

This document presents a detailed solution to a college algebra assignment, addressing questions 2 through 10. The solutions cover various topics, including compound interest calculations, the multiplication of complex numbers, and determining the slope and y-intercept of a linear equation. The assignment also explores quadratic equations, providing methods to find the quadratic and linear coefficients, and the constant. Furthermore, the solution covers exponential functions, finding the roots of functions, and factorizing equations. Finally, the document includes calculations for population growth and the half-life of a substance, providing a comprehensive overview of algebra concepts.

College Algebra
Student Name
Institution Name
1

Question 2
Investment
Capital $8,000
Time 12 years
a. 3% compounded monthly
A=P ( 1+ r
n )
nt
¿ 8000∗( 1+ 0.03
12 )
12∗12
=$ 11,461.49
b. 3% compounded daily
Let 1 year equals 365 days
Then n will be 365
A=P ( 1+ r
n )
nt
¿ 8000∗( 1+ 0.03
365 )
365∗12
=$ 11,466.47
c. 3% compounded continuously
To obtain the effective interest rate compounded annually we use the formula
i= ( 1+ 0.03
12 )
12
−1=0.03
δ=ln ( 1+i ) =ln (1+0.03)
A=8000∗eδ t=$ 11,406.09
Question 3
Dividing complex numbers
5+3 i
4−3 i we multiply the denominator and the numerator both by the conjugate of the denominator
5+3 i
4−3i ∗4+3 i
4 +3 i = 20+27 i+ 9
16+9
¿ 29+27 i
25 = 29
25 + 27
25 i
Question 4
a. T
b. F
c. T
d. F
2

e. T
f. T
g. T
h. T
i. T
j. T
Question 5
a. f ( x ) =3 ( x−1 ) +(4 x+ k)
The equation is linear and when expanded gives
f ( x )=3 x−3+ 4 x +k
f ( x )=7 x +(k−3)
The slope of the equation is 7
While the y intercept is ( k −3 )
b. The points (-1,12), (1,4), and (3,13) are on the graph of the function g.
The function g is quadratic given by
y=a x2 +bx +c
Now we use the three points to form three equations
12=a ( −1 ) 2+ b ( −1 ) + c
1 4=a ( 1 ) 2 +b ( 1 ) +c
1 3=a ( 3 ) 2+ b ( 3 ) + c
Simplify
12=a−b +c
14=a+b+c
13=9 a+3 b+ c
From equation 1
c=12−a+b
Replace this value in equation 2 and 3
We get
b=1
1=8 a+4 b
a=−3
8
c=12−3
8 +1=101
8
Now theequation if the lines becomes
y=−3
8 x2 +x + 101
8
3

Now the quadratic coefficient is −3
8
Linear coefficient is 1 and the constant is 101
8
c. h ( w ) =6 ( 2 ) w−1
The equation is exponential
Initial amount is given at
h ( 0 ) =6 ( 2 ) 0−1=3
the base of the function is 2
Question 6
From the graph the rots of the function are at points
x=−1 , x=1∧x=3
This means that the function can be factorized as
(x +1)( x −1)( x−3)
Justification; the roots of the function are the points at which the function crosses the x axis. The points
can thus be applied to obtain the function in a factorized form.
Question 7
Function f ( x )=x2 + 4 x−1 and g ( y )=2 y
a. f ( 0 )=02+ 4 ( 0 ) −1=−1
b. (gof )(0)
( gof )=g ( f ( x ) ) =2(x2+4 x−1)
Now ( gof ) ( 0 ) =2∗−1=−2
c. g
f ( 0 ) = g ( o )
f ( 0 ) = 0
−1 =o
d. f
g ( o )= f ( 0 )
g ( 0 ) =−1
e. (fof )(0)
( fof )=f ( f ( x ) )= ( x2 + 4 x−1 )2
+ 4 ( x2+ 4 x−1 ) −1
This can be simplified to obtain
x4 +8 x3 +8 x2−4
Question 8
The roots of the equation are -2 and -4
Hence the equation can be derived by
( x +2 ) ( x + 4 ) =x2 +6 x +8
4

The y intercept is the point at which the value of x is zero
Hence y=0+0+8
The y intercept is (0,8)
Question 9
As at 1990 34,690
Growth rate 1.03% pa
a. City’s population at t will be given by
¿ P∗ (1+ r )t
whereP is the population at the end of 1990, r is the growth rate and t is the time.
b. At the end of 2019, t will be 29
The population is thus
P∗ ( 1+ r )t=34,690∗(1+1.03
100 )29
=46694.37
This is approximately 46,694
c. Doubling time is the time taken for the city’s population to double.
( 34690∗2 )=34,690∗(1+ 1.03
100 )t
2= ( 1+ 1.03
100 )
t
2=1.0103t
ln ( 2 ) =tln(1.0103)
t= ln ( 2 )
ln ( 1.0103 ) =67.64 years
Question 10
Safe level x
Current level 12x
A ( t ) = A ( 0 ) ∗( 1
2 ) t
h
Where h is the half life
x=12 x∗( 0.5 )
t
8
1
12=0.5
t
8
5

ln ( 1
12 ) = t
8∗ln (0.5)
t=
ln ( 1
12 )
ln ( 0.5 ) ∗8=28.68 days
That is approximately 29 days
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