MAT1514
Assignment 1
Semester 2,2023
No calculator may be used: Leave answers as fractions or roots where appropiate.
Show steps in answers.
QUESTION 1
Given the graphs A, B, C and D of relations below, determine for each:
1.1 If the graph represents a function.
(4)
A – function
B – function
C – not a function
D- Function
1.2 For the graphs that represents a function determine if the function is even, odd or neither
A – Odd
B – Even
D – Neither
(3)
A.
Function - Odd
Symmetric around the origin.
B.
Function – even –
symmetric around y-axis
-1-
Open Rubric
MAT1514
Assignment 1
C.
Semester 2,2023
Not a function
D.
Function neither odd or even
[7]
QUESTION 2
2.1
Given f ( x ) 
3
x  5 and g ( x )  x 3  5 . Find and simplify f  g  x  
Answer:
f  g  x    f ( x 3  5)

3
x

3
x3
3
 5  5
x
-2-
(3)
MAT1514
Assignment 1
Semester 2,2023
x
2.2
1
Sketch the function h( x )    for 1  x  1 and determine if the function is
4
increasing or decreasing on this interval.
(5)
Answer:
Note the point at x = -1 is included while the point at x = 1 is not included.
Decreasing – function values from left to right are getting smaller.
2.3
Solve for x if 2 x  3  17
(4)
Answer:
2 x  3  17
 2 x  3  17
or
 2 x  3  17
2 x  20
2 x  14
x  10
x  7
[12]
-3-
MAT1514
Assignment 1
Semester 2,2023
QUESTION 3
3.1
Find the point of intersection of the lines f  x   3 x  4 and g ( x )  5  x .
(5)
Answer:
3.2
Write an equation for the line perpendicular to the line f  x   4 x  3 and passing
through the point  8,10  .
(5)
Answer:
-4-
MAT1514
3.3
Assignment 1
Semester 2,2023
Given the polynomial function f  x   x  4 x  45, determine the x and y intercepts
4
of f  x  .
2
(6)
Answer:
[16]
QUESTION 4
Perform the indicated operations and express the result as a simplified complex number:
4.1
 3  2i    5  2i   8  4i
(2)
4.2
2  3i
2  4i
(5)
Answer:
4.1
-5-
MAT1514
4.2
i 
Assignment 1
Semester 2,2023
7
(2)
Answer:
[9]
QUESTION 5
The kinetic energy of a moving object varies jointly with its mass and the square of its velocity.
An object with mass 40 kilograms and a velocity of 15 meters per second has a kinetic energy
of 1000 joules. Find the kinetic energy if the velocity is increased to 20 meters per second.
[6]
Answer: In these type of questions be careful not to assume what is not given in the
question.
Your physics or chemistry are not tested. We want to see if you can translate the words
correctly into mathematics.
The kinetic energy of a moving object varies jointly with its mass and the square of its velocity.
Thus K .E  mv 2 where KE = kinetic energy, m = mass, v = velocity
-6-
MAT1514
Assignment 1
Semester 2,2023
To make this an equation we must introduce a constant of proportionality, say k.
K .E  kmv 2
With the first set of conditions we can find the value of k:
K .E  kmv 2
K .E
k
mv 2

=
1000 J
 40kg  15m.s 1 
2
(the units should cancel because the constant of proportionality has no unit)
25
15
2
25
225
1

9

Then K.E after the velocity is increased will be:
K .E  kmv 2
1
2
  40  20  J
9
 40  400  J

9
16 000

J
9
You can leave the answer as a fraction.
{If it is an exact answer for example 18 000/9 = 2000 then you will give the final answer and not leave
as a fraction.}
Total: 50
©
UNISA 2023
-7-