# Identify the error made in the solved problem.

August 30, 2017

Question
Functions and Inequalities

The purpose of this assessment is to help you determine whether you have a clear understanding of quadratic, polynomial, and rational functions and their graphs expressions, and provide you feedback to clarify any misunderstandings.

The discussions for this class center on reviewing sample problems that were solved incorrectly.

From the list of incorrectly solved problems, pick a problem that has not been discussed yet, and perform the following tasks:

Identify the error made in the solved problem.
Correctly solve/simplify the problem showing all the steps. Be sure to include any formulas or properties that support the correctly solved problem.
Write a few sentences explaining how to avoid making this mistake in future practice.
Participation Requirements:

Respond to at least two of your classmates. When responding:

Provide complete, well-thought-out responses.
Note: One-sentence answers will not be sufficient. If your answer is “I agree” or “I disagree,” explain why you agree or disagree. Remember that a discussion is an opportunity to interact with your classmates. Observe discussion etiquette: Be respectful, kind, and nonjudgmental of your classmates.

Evaluation Criteria:

Your submission for each question will be evaluated against the following criteria:

Did you identify all errors in the problem?
Did you correctly solve/simplify the problems, showing all steps and any formulas or properties that support the correctly solved problem?
Did you write a few sentences explaining how to avoid making this mistake in future practice?
MA1210: Module 5 Polynomial and Rational Functions
Discussion 5.1
Functions and Inequalities

S. NO.
1.

Problem
() = −2 2 + 8 + 4
=−

b
8
=−
=4
2a
2(−2)

(4) = −2(4)2 + 8(4) + 4
(4) = −32 + 32 + 4
(4) = 4
Maximum at (4, 4)
2.

() = −3 2 + 6 + 3
=−

b
6
=−
=1
2a
2(−3)

(1) = −3(1)2 + 6(1) + 3
(1) = −3 + 6 + 3
(1) = 12
Maximum at (1, 12)
3.

() = −2 2 + 16 + 3
=−

b
16
=−
=4
2a
2(−2)

(4) = −2(4)2 + 16(4) + 3
(4) = −8 + 64 + 3
(4) = 59
Maximum at (4, 59)
4.

() = −3 2 + 18 − 22
=−

b
18
=−
=3
2a
2(−3)

(3) = −3(3)2 + 18(3) − 22
(3) = −27 + 56 − 22
(3) = 7
Maximum at (3, 7)

1

MA1210: Module 5 Polynomial and Rational Functions
Discussion 5.1
Functions and Inequalities

S. NO.
5.

Problem
() = − 2 + 4 + 31

=−

b
4
=−
=2
2a
2(−1)

(2) = (2)2 + 4(2) + 31
(2) = 4 + 8 + 31
(2) = 43
Maximum at (2, 43)
6.

() = −2 2 + 20 + 17
=−

b
20
=−
=5
2a
2(−2)

(5) = −2(5)2 + 20(5) + 17
(5) = −20 + 100 + 17
(5) = 97
Maximum at (5, 97)
7.

() = −2 2 + 44 − 180
=−

b
44
=−
= 11
2a
2(−2)

(11) = −2(11)2 + 44(11) − 180
(11) = −222 + 444 − 180
(11) = 42
Maximum at (11, 42)
8.

() = −3 2 + 42 − 112
=−

b
42
=−
=7
2a
2(−3)

(7) = −3(7)2 + 42(7) − 12
(7) = −147 + 294 − 12
(7) = 135
Maximum at (7, 135)

2

MA1210: Module 5 Polynomial and Rational Functions
Discussion 5.1
Functions and Inequalities

S. NO.
9.

Problem
() = −2 2 + 36 − 84
=−

b
36
=−
=9
2a
2(−2)

(9) = −2(9)2 + 36(9) − 84
(9) = −162 + 333 − 84
(9) = 87
Maximum at (9, 87)
10.

() = − 2 + 16 − 50
=−

b
16
=−
=4
2a
2(−2)

(4) = −(4)2 + 16(4) − 50
(4) = −16 + 64 − 50
(4) = −2
Maximum at (4, −2)
11.

() = −2 2 + 12 + 19
=−

b
12
=−
=3
2a
2(−2)

(3) = −(3)2 + 12(3) + 19
(3) = −9 + 36 + 19
(3) = 346
Maximum at (3, 46)
12.
() = − 2 + 14 − 6
=−

b
14
=−
=7
2a
2(−2)

(7) = (7)2 + 14(7) − 6
(7) = 49 + 98 − 6
(7) = 141
Maximum at (7, 141)

3

MA1210: Module 5 Polynomial and Rational Functions
Discussion 5.1
Functions and Inequalities

S. NO.
13.

Problem
() = −3 2 + 12 + 15
=−

b
12
=−
=4
(−3)
2a

(4) = −3(4)2 + 12(4) + 15
(4) = −48 + 48 + 15
(4) = 15
Maximum at (4, 15)
14.

() = − 2 + 20 − 40
=−

b
20
=−
= 10
2a
2(−1)

(10) = −(10)2 + 20(10) − 40
(10) = −20 + 200 − 40
(10) = 140
Maximum at (10, 140)
15.

() = −2 2 + 28 − 52
=−

b
28
=−
=7
2a
2(−2)

(7) = −2(7)2 + 26(7) − 52
(7) = −98 + 182 − 52
(7) = 32
Maximum at (7, 32)
16.

() = −3 2 + 30 − 37
=−

b
30
=−
=5
2a
2(−3)

(5) = −3(5)2 + 30(5) − 7
(5) = −75 + 150 − 7
(5) = 68
Maximum at (5, 68)

4

MA1210: Module 5 Polynomial and Rational Functions
Discussion 5.1
Functions and Inequalities

S. NO.
17.

Problem
() = − 2 + 6 + 11
=−

b
6
=−
=6
(−1)
2a

(6) = −(6)2 + 6(6) + 11
(6) = −36 + 36 + 11
(6) = 11
Maximum at (6, 11)
18.

() = −3 2 + 24 − 29
=−

b
24
=−
=6
2a
2(−3)

(6) = −3(6)2 + 24(6) − 29
(6) = −108 + 144 − 29
(6) = 7
Maximum at (6, 7)
19.

() = − 2 + 18 − 48
=−

b
18
=−
=9
2a
2(−1)

(9) = −(9)2 + 18(9) − 48
(9) = −81 + 153 − 48
(9) = 24
Maximum at (9, 24)
20.

() = −2 2 + 4 + 54
=−

b
4
=−
=2
2a
2(−2)

(2) = −2(2)2 + 4(2) + 54
(2) = −8 + 8 + 54
(2) = 54
Maximum at (2, 54)

5

MA1210: Module 5 Polynomial and Rational Functions
Discussion 5.1
Functions and Inequalities

S. NO.
21.

Problem
() = − 2 + 8 + 20
=−

b
8
=−
=4
2a
2(−1)

(4) = −(4)2 + 8(4) + 20
(4) = −16 + 32 + 20
(4) = 34
Maximum at (4, 34)
22.

() = −3 2 + 36 − 18
=−

b
36
=−
=9
2a
2(−2)

(9) = −3(9)2 + 36(9) − 18
(9) = −243 + 324 − 18
(9) = 63
Maximum at (9, 63)
23.

() = − 2 + 24 − 82
=−

b
24
=−
= 12
2a
2(−1)

(12) = −(12)2 + 24(12) − 82
(12) = −144 + 288 − 82
(12) = 92
Maximum at (12, 62)
24.

() = −3 2 + 48 − 153
=−

b
48
=−
=8
2a
2(−3)

(8) = 3(8)2 + 48(8) − 153
(8) = 192 + 384 − 153
(8) = 423
Maximum at (8, 423)

6

MA1210: Module 5 Polynomial and Rational Functions
Discussion 5.1
Functions and Inequalities

S. NO.
25.

Problem
() = − 2 + 12 + 24
=−

b
12
=−
=6
2a
2(−1)

(6) = −(6)2 + 12(6) − 24
(6) = −36 + 72 − 24
(6) = 12
Maximum at (6, 12)
26.

() = −2 2 + 24 − 21
=−

b
24
=−
=6
2a
2(−2)

(6) = −2(6)2 + 24(6) − 21
(6) = −24 + 144 − 21
(6) = 99
Maximum at (6, 99)
27.

() = −2 2 + 32 − 83
=−

b
32
=−
=8
2a
2(−2)

(8) = −2(8)2 + 32(8) − 83
(8) = −128 + 272 − 83
(8) = 61
Maximum at (8, 61)
28.

() = − 2 + 10 + 4
=−

b
10
=−
=5
2a
2(−1)

(5) = −(5)2 + 10(5) + 4
(5) = −10 + 50 + 4
(5) = 44
Maximum at (5, 44)

7

MA1210: Module 5 Polynomial and Rational Functions
Discussion 5.1
Functions and Inequalities

S. NO.
29.

Problem
() = − 2 + 28 − 154
=−

b
28
=−
= 14
2a
2(−1)

(14) = −(14)2 + 28(14) − 154
(14) = −194 + 392 − 154
(14) = 40
Maximum at (14, 40)
30.

() = − 2 + 2 + 15
=−

b
2
=−
=2
(−1)
2a

(2) = (2)2 + 2(2) + 15
(2) = 4 + 2 + 15
(2) = 23
Maximum at (2, 23)

8

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