Department of Mathematics

Dr. Mustafa El-Agamy

Dr. Mohammad Yasin
Mathematics
for Engineering and Computer Science Students (Math 103)
WorkSheet No. (1) – Solution
Problem 1:

Decide whether the relation is a function or not. If it is a function, find its domain and range.

It can be tested using Vertical Line Test (VLT) as shown below.

✓ ✓
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 1) 𝑦 is 𝐧𝐨𝐭 a function of 𝑥:
because there is at least one value of 𝑥 for which 𝑦 has more than one value.
𝐨𝐫
By using 𝐕𝐋𝐓, the line intersects the graph in more than one point.
∴ The given relation is 𝐧𝐨𝐭 a function.

2) 𝑦 is a function of 𝑥:
because for all 𝑥 ∈ [−4, 3], there is one and only one corresponding value of 𝑦.

3) 𝑦 is a function of 𝑥:
because for all 𝑥 ∈ [−4, 4], there is one and only one corresponding value of 𝑦.

4) 𝑦 is a function of 𝑥:
because for all 𝑥 ∈ [−3, 3], there is one and only one corresponding value of 𝑦.

5) 𝑦 is 𝐧𝐨𝐭 a function of 𝑥:
because there is at least one value of 𝑥 for which 𝑦 has more than one value.
𝐨𝐫
By using 𝐕𝐋𝐓, the line intersects the graph in more than one point.
∴ The given relation is 𝐧𝐨𝐭 a function.

6) 𝑦 is a function of 𝑥:
because for all 𝑥 ∈ ] − 4, 3[ , there is one and only one corresponding value of 𝑦.

Note That: We will ONLY find the domain and range of the relations which are functions.

Graph Domain Range

2) [−4, 3] [−4, 3]
3) [−4, 4] [−2, 4]
4) [−3, 3] [−4, 2]
6) ] − 4, 3[ [1, 5[

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Problem 2:
Does the relation 𝒙𝟐 + 𝒚𝟐 = 𝟏 define 𝒚 as a function of 𝒙? Why?
Solution
𝑥2 + 𝑦2 = 1 ⟹ 𝑦2 = 1 − 𝑥2 ⟹ 𝑦 = ± √1 − 𝑥 2
∵ For each one input value, we have two output values. ∴ 𝑥 2 + 𝑦 2 = 1 is not a function.

Problem 3:
Find the equation of the line such that:
(i) passes (𝟏, 𝟐) and its slope is −𝟑.
(ii) passes (−𝟏, 𝟐) and (𝟑, 𝟒).
(iii) inclines by 𝟒𝟓 ° with the positive 𝒙 −axis and its 𝒙 −intercept is −𝟑.
Solution

(i) 𝑚 = −3
∴ 𝑦 = −3𝑥 + 𝑘
for the point (1, 2):
𝑦 = −3𝑥 + 𝑘 → 2 = −3(1) + 𝑘 → 𝑘=5
𝒚 = −𝟑𝒙 + 𝟓

𝑦2 − 𝑦1 4−2 1
(ii) 𝑚 = = =
𝑥2 − 𝑥1 3 − (−1) 2
1
∴𝑦= 𝑥+𝑘
2
for the point (−1, 2):
1 1 5
𝑦= 𝑥+𝑘 → 2= (−1) + 𝑘 → 𝑘=
2 2 2
𝟏 𝟓
𝒚= 𝒙+
𝟐 𝟐

(iii) 𝑚 = tan 𝛼 = tan 45 = 1

∴𝑦 =𝑥+𝑘
for the point (−3, 0): 𝟒𝟓°

𝑦 =𝑥+𝑘 → 0 = −3 + 𝑘 → 𝑘 = 3
𝒚=𝒙+𝟑

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Problem 4:
For the line 𝟐𝒙 − 𝟑𝒚 + 𝟔 = 𝟎, find its slope, angle of inclination and the intercepts. Also, sketch the
graph of the line.
Solution
The equation 𝟐𝒙 − 𝟑𝒚 + 𝟔 = 𝟎 can be written in the standard form:

𝟐
𝒚= 𝒙+𝟐
𝟑

𝟐
Slope is
𝟑
2
Angle of inclination is 𝜶 = tan−1 ( ) = 𝟑𝟑. 𝟕𝐨
3

𝑦-intercept = 𝟐
𝜶
𝑥-intercept = −𝟑

Problem 5:

Sketch the following functions, then from the graph find the domain and the range. Find also the
𝒙 −intercept and the 𝒚 −intercept.

(a) 𝒇(𝒙) = (𝒙 − 𝟑)𝟐 + 𝟐 (b) 𝒇(𝒙) = −𝟐 + √𝒙 − 𝟏 (c) 𝒇(𝒙) = |𝒙 + 𝟒| − 𝟐

𝟏 −𝟏
(𝐝) 𝒇(𝒙) = −𝟏 (e) 𝒇(𝒙) = 𝟔 − 𝟑𝒙 (𝐟) 𝒇(𝒙) =
𝒙−𝟐 𝒙𝟐 + 𝟔𝒙 + 𝟗

𝟏
(𝐠) 𝒇(𝒙) = 𝟏 − (𝟑 − 𝒙)𝟐 (h) 𝒇(𝒙) = 𝟓 − √𝟒 − 𝒙 (𝐢) 𝒇(𝒙) = 𝟑 −
(𝟐 − 𝒙)𝟐

𝟏
(j) 𝒇(𝒙) = −𝟏 − |−𝟏 − 𝒙| (𝒌) 𝒇(𝒙) = | | (𝒍) 𝒇(𝒙) = √|𝒙|
𝒙

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Solution

(a) 𝑓 (𝑥) = (𝑥 − 3)2 + 2

2 𝑦 = (𝑥 − 3)2
𝑦=𝑥
Shift right by 3 units

𝑓 (𝑥 ) = (𝑥 − 3)2 + 2
Shift up by 2 units

𝐷𝑓 ∶ 𝑥 ∈ ℝ

𝑅𝑓 ∶ 𝑦 ∈ [2, ∞[

𝑦 −intercept : 𝑦 = 11

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(b) 𝑓(𝑥) = −2 + √𝑥 − 1

𝑦 = √𝑥 − 1
𝑦 = √𝑥
Shift right by 1 unit

𝑓(𝑥) = −2 + √𝑥 − 1 𝐷𝑓 : [1, ∞[
Shift down by 2 units

𝑅𝑓 : 𝑦 ∈ [−2, ∞[

𝑥 −intercept: 𝑥 = 5

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(c) 𝑓(𝑥) = |𝑥 + 4| − 2
𝑦 = |𝑥 + 4|
𝑦 = |𝑥|
Shift left by 4 units

𝑓 (𝑥) = |𝑥 + 4| − 2
Shift down by 2 units
𝐷𝑓 : 𝑥 ∈ ℝ

𝑅𝑓 : 𝑦 ∈ [−2, ∞[

𝑥 −intercept: 𝑥 = −6, −2

𝑦 −intercept: 𝑦 = 2

1
(d) 𝑓(𝑥) = −1
𝑥−2
1
1 𝑦=
𝑦= 𝑥−2
𝑥 Shift right by 2 units

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 1
𝑓 (𝑥 ) = −1
𝑥−2
Shift down by 1 unit

𝐷𝑓 : 𝑥 ∈ ℝ − {2}

𝑅𝑓 : 𝑦 ∈ ℝ − {−1}

𝑥 −intercept: 𝑥 = 3

𝑦 −intercept: 𝑦 = −1.5

(e) 𝑓 (𝑥) = 6 − 3𝑥

𝐷𝑓 : 𝑥 ∈ ℝ

𝑅𝑓 : 𝑦 ∈ ℝ

𝑥 −intercept: 𝑥 = 2

𝑦 −intercept: 𝑦 = 6

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 −1
(f) 𝑓(𝑥) =
𝑥2 + 6𝑥 + 9

By completing the square:

−1 −1
𝑓 (𝑥 ) = =
𝑥 2 + 6𝑥 + 9 (𝑥 + 3)2

1
1 𝑦=
𝑦= 2 (𝑥 + 3)2
𝑥 Shift Left by 3 units

−1
𝑓 (𝑥 ) =
(𝑥 + 3)2
Reflection about 𝑥 −axis

𝐷𝑓 : 𝑥 ∈ ℝ − (−3)

𝑅𝑓 : 𝑦 ∈ ] − ∞, 0[

−1
𝑦 −intercept: 𝑦 =
9

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(g) 𝑓 (𝑥 ) = 1 − (3 − 𝑥 ) 2 = 1 − (𝑥 − 3)2

𝑦 = (𝑥 − 3)2
𝑦 = 𝑥2
Shift right by 3 units

𝑦 = − (𝑥 − 3)2 𝑓 (𝑥 ) = 1 − (𝑥 − 3)2
Reflection about 𝑥−axis Shift up by 1 unit

𝐷𝑓 : 𝑥 ∈ ℝ 𝑅𝑓 : 𝑦 ∈ ]−∞, 1]

𝑥 −intercept: 𝑥 = 2, 4 𝑦 −intercept: 𝑦 = −8

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(h) 𝑓 ( 𝑥 ) = 5 − √4 − 𝑥

𝑦 = √−𝑥
𝑦 = √𝑥
Reflection about 𝑦 −axis

𝑦 = √4 − 𝑥 = √−(𝑥 − 4) 𝑦 = −√4 − 𝑥
Shift right by 4 units Reflection about 𝑥 −axis

𝑓 (𝑥) = 5 − √4 − 𝑥
Shift up by 5 units
𝐷𝑓 : 𝑥 ∈ ]−∞, 4]

𝑅𝑓 : 𝑦 ∈ ]−∞, 5]

𝑥 −intercept: 𝑥 = −21

𝑦 −intercept: 𝑦 = 3

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 1 1
(i) 𝑓 (𝑥 ) = 3 − = 3 −
(2 − 𝑥 ) 2 (𝑥 − 2)2
1
1 𝑦=
𝑦= 2 (𝑥 − 2)2
𝑥
Shift right by 2 units

1 1
𝑦=− 𝑓(𝑥) = 3 −
(𝑥 − 2) 2 (𝑥 − 2)2
Reflection about 𝑥 −axis Shift up by 3 units

𝐷𝑓 : 𝑥 ∈ ℝ − {2} 𝑅𝑓 : 𝑦 ∈ ] − ∞, 3[

1 11
𝑥 −intercept: 𝑥 = 2 ± 𝑦 −intercept: 𝑦 =
√3 4

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(j) 𝑓(𝑥) = −1 − |−1 − 𝑥| = −1 − |1 + 𝑥|

𝑦 = |1 + 𝑥|
𝑦 = |𝑥|
Shift left by 1 unit

𝑦 = −|1 + 𝑥| 𝑓 (𝑥) = −1 − |1 + 𝑥|
Reflection about 𝑥 −axis Shift down by 1 unit

𝐷𝑓 : 𝑥 ∈ ℝ 𝑅𝑓 : 𝑦 ∈ ]−∞, −1]

𝑦 −intercept: 𝑦 = −2

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 1
(𝑘 ) 𝑓 (𝑥 ) = | |
𝑥
1 1
𝑦= 𝑓 (𝑥 ) = | |
𝑥 𝑥

𝐷𝑓 : 𝑥 ∈ ℝ − {0} 𝑅𝑓 : 𝑦 ∈ ]0, ∞[

√𝑥 if 𝑥 ≥ 0
(𝑙) 𝑓(𝑥) = √|𝑥| = {
√−𝑥 if 𝑥 < 0

𝑓(𝑥) = √|𝑥|

𝐷𝑓 : 𝑥 ∈ ℝ

𝑅𝑓 : 𝑦 ∈ [0, ∞[

𝑥 −intercept: 𝑥 = 0

𝑦 −intercept: 𝑦 = 0

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Problem 6:
Given the function 𝒇(𝒙) = 𝒂 − √𝒙 + 𝒃. If the function range is ]−∞, 𝟐] and its zero is 𝒙 = 𝟓,
find the value of 𝒂 and 𝒃.
Solution

The range of the function 𝑓 (𝑥) = 𝑎 − √𝑥 + 𝑏 is ]−∞, 𝑎 ], therefore 𝒂 = 𝟐.

Since the zero of the function 𝑓(𝑥) is 𝑥 = 5, thus 𝑓 (5) = 0.

𝑓 (5) = 2 − √5 + 𝑏 = 0,

∴ 𝒃 = −𝟏

Problem 7:
Given the function 𝒇(𝒙) = 𝒙𝟐 − 𝟔 𝒙 + 𝟓.

(i) Sketch its graph.

(ii) Find its domain and range.

(iii) Find the values of 𝑥 that satisfy 𝑓(𝑥) ≥ 0.

Solution

(i) 𝑓(𝑥) = 𝑥 2 − 6 𝑥 + 5 = (𝑥 − 3)2 − 4

(ii) 𝐷𝑓 : 𝑥 ∈ ℝ

𝑅𝑓 : 𝑦 ∈ [−4, ∞[

(iii) 𝑓(𝑥) ≥ 0 when 𝑥 ∈ ] − ∞, 1] ∪ [5, ∞[

Page 15 of 16
Problem 8:
Find the equations of the following graphs.

(i) (ii)

1
𝑓 (𝑥) = |𝑥 + 1| 𝑓 (𝑥 ) = +1
(𝑥 − 2) 2

(iii) (iv)

−1
𝑓 (𝑥 ) = +1 𝑓 (𝑥) = −2𝑥 + 3
𝑥+2

(v) (vi)

𝑓(𝑥) = − (𝑥 − 1)2 + 3 𝑓 (𝑥) = − √−𝑥 + 2

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