Algebra Assignment: Comprehensive Solutions to Algebra Problems

This document presents a detailed solution to an algebra assignment, encompassing a variety of problems and concepts. The assignment covers topics such as the relationship between variables, average rate of change, linear functions, and the interpretation of slopes and intercepts. It includes step-by-step solutions to questions involving calculations, graphical representations, and real-world applications. The assignment also explores the application of algebraic principles to scenarios involving population trends, meeting schedules, and geometric concepts, providing a comprehensive understanding of algebraic principles and their practical uses. Furthermore, the document analyzes the relationship between different algebraic concepts and their interconnections.

Running head: ALGEBRA 1
Algebra
Name
Institution

ALGEBRA 2
Algebra
Question 1
If f ( x ) decreases with the increase in x, then it must increase as x gets smaller. Therefore, F (−2)
would be larger as compared to F (2) .
Question 2
Average rate of change¿ Change∈Y
Change∈ X = 4.9−2.9
6.1−2.2 = 2
3.9 =0.513
Question 3
∆ f ( x )
∆ x =0 at the intervals:
2.2 ≤ x ≤5.2 and 5.2 ≤ x ≤6.9
Question 4
Average rate of change¿ Change∈Y
Change∈ X = 4.9−2.9
6.1−2.2 = 2
3.9 =0.513
Question 5
The average rates of change of f and g between x=2.2∧x=6.1 are equal.
Question 6
Part a
The rate of change of f is negative

ALGEBRA 3
Part b
The rate of change of f is positive
Question 7
Line Function
A g( x )
B f (x)
C h(x )
D u( x )
E v ( x)
Question 8
Part a
g ( 4 ) −g(0)
4−0 = 2−0
4 = 1
2 =0. 5
Part b
Line AB shows the line whose gradient was calculated in part a. That is, m=0.5

ALGEBRA 4
Part c
g ( b ) −g(a)
b−a = g ( −1 ) −g (−9)
−1−−9 =−1−(−3)
−1+9 = 2
8 = 1
4 =0.25
Part d
Line CD shows the line whose gradient was calculated in part c. That is, m=0. 2 5
Question 9
Part i
f ( x )=x2 +1
f (−1 )=(−1)2 +1=2
f ( 3 ) =( 3)2 +1=1 0
The two points become,(−1,2)∧(3,10)

ALGEBRA 5
Average rate of change ¿ Change∈Y
Change∈ X = 10−2
3−−1 = 8
4 =2
Part ii
f ( x )=x2 +1
f ( a )=(a)2 +1=a2+1
f ( b )=(b)2+ 1=b2+1
The two points become,( a , a2 +1)∧(b ,b2+1)
Average rate of change:¿ Change∈Y
Change∈ X
¿( b¿¿ 2+1)−(a¿ ¿ 2+1)
b−a =(b ¿¿ 2−a2)
b−a =( b+a)(b−a)
b−a =b+a ¿ ¿ ¿
Part iii
f ( x )=x2 +1
f ( x +h )=( x +h)2 +1
The two points become,(x , x2 +1)∧(x+ h ,( x+ h)2 +1)
Average rate of change¿ Change∈Y
Change∈ X
¿(x +h)2 +1−( x ¿¿ 2+1)
x +h−x =(x+ h)2 −x2
h = x2+2 xh+h2−x2
h ¿

ALGEBRA 6
¿ 2 xh+h2
h = h (2 x +h)
h =2 x +h
Part b
In each of the three pairs of points, the average rate of change equals the sum of x-coordinate
values of the two points.
Question 10
The constant 78.9 refers to the number of SARS cases as at March 17th when t=0. On the other
hand, the constant 30.1 refers to the rate of change of SARS cases over time.
Question 11
Part a
Country A is Afghanistan while B is Sri Lanka. Notably, the trend in 1980-1990 show that the
population in Afghanistan decreased due to war.
Part b
We calculate the rate of change of population in Sri Lanka because it is linear unlike that of
Afghanistan.
Rate of change= ∆∈ population
∆∈time = 19.2−7.5
2000−1950 = 11.7
50 =0.234
The rate of change shows that the population of Sri Lanka increases at the rate of 0.234 yearly.
Part c
The population in 1988¿ Population at 1950+Rate of change ×time(¿ 1950¿1988)
¿ 7.5+0.234 × ( 1988−1950 )
¿ 7.5+0.234 × 38

ALGEBRA 7
¿ 7.5+8.892=16.392
Question 12
The graph of the function over the specified interval appears to be a straight line. The selected
area scales it to look like a straight line. However, it is a parabola due to the x2 term in the
function.
Question 13
First, we determine the coordinates of point P and point Q as follows.
Coordinates of point P:
y=x2 +1
8=x2+1
x2=7 , x=− √ 7
Thus, the coordinates of point P are P(− √ 7 , 8)
Coordinates of point Q:
y=x2 +1
y=22+1=5
Thus, the coordinates of point Q are Q(2,5)
The slope of line PQ= 5−8
2−− √ 7 =−0.646
Let the point pass through an arbitrary point with coordinates ( x , y ) ∧the point Q
The slope of line PQ= y−5
x−2 =−0.646
y−5=−0.646 ( x−2 )=−0.646 x+1.292
y=−0.646 x+1.292+5
Thus , the equation of the line , y=−0.646 x +6.292
Question 14
Part a

ALGEBRA 8
Clients= 3 hours
Co-workers=2 hours
y=¿number of client meetings
x=¿ number of co-worker meetings
Total co-worker meeting time in a week¿ 2 x
Total client meeting time in a week ¿ 3 y
The relation between x and y becomes, 2 x+3 y =60 hours
The graph of the relation is shown below:
Part b
2 x+3 y =60 hours
3 y=−2 x +60
y=−2
3 x +20
Part c

ALGEBRA 9
The slope¿−2
3 shows the negative ratio of the maximum client meetings to the maximum co-
worker meetings possible in a week.
The x−intercept , 0=−2
3 x+20 , 2 x
3 =20 , x=30 meetings
The x-intercept shows the maximum number of co-worker meetings that the consultant can hold
in a week given that no client turns up during the week.
The y−intercept , y =−2
3 ( 0 ) +20 , y=20 meetings
The y-intercept shows the maximum number of client meetings that the consultant can hold in a
week given that no co-worker meeting is held during the week.
Part d
If co-worker meetings take 1.5 hours then, the total co-worker meeting time in a week ¿ 1.5 x
The relation between x and y becomes, 1.5 x+3 y =60 hours
1.5 x+3 y =60
3 y=−1.5 x +60
y=−1.5
3 x +20=−0.5 x+ 20
The relation between x and y can be graphed a shown below.

ALGEBRA 10
It is evident that the y-intercept remains the same but the slope and the x-intercept change.
Question 15
y= x
β−3 + 1
6−β
The equation is in the form, y=mx+c where,
m= 1
β−3 and c= 1
6−β
For the slope to be positive, 1
β−3 ≥ 1. That is, β >3
For the intercept to be positive, 1
6−β ≥ 1 .That is , β<6
Therefore, 3< β <6 for the slope and intercept to be positive. Which implies that β is either equal
to 4 or 5.
Question 16
y= ( β −7 ) x−3
m1=β−7
We know, m2= 1
β−3
Since the two lines are perpendicular, m1=−1
m2
β−7=−( β−3)
Solving for β we get,
β−7=−β +3
β + β=7+3
2 β=10

ALGEBRA 11
β=5
Therefore, β=5