Determining the equation of the quadratic functions

1. Lesson 1
Determines the equation of a quadratic
function given:
a. table of values,
b. graphs,
c. zeros

2. Objectives
At the end of the lesson, students should be
able to:
1. Derive the equation of a quadratic function
given the table of values, graphs and zeros.
2. Participate actively in virtual class
discussion

3. Quadratic Function
𝒚 = 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 𝒐𝒓 𝒇(𝒙) = 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄
The table of values and the graph below described
quadratic functions.

4. The functions may obtain into different procedure:
Example 1:
Find the quadratic functions whose zeros are 2 and
-7?
Solution:
Since zeros are 2 and -7, the we can say
therefore that 𝑥 = 2, 𝑎𝑛𝑑 𝑥 = −7.
Equating to zero, then . . . 𝑥 − 2 = 0 𝑎𝑛𝑑 𝑥 + 7 = 0.
Multiply the two quantity… 𝑓(𝑥) = (𝑥 − 2)(𝑥 + 7)
The quadratic function . . . . 𝑓(𝑥) = 𝑥2 + 5𝑥 − 14

5. Example 2.
Find the quadratic function represented by table of values
below.
X -3 -2 -1 0 1 2 3
Y 24 16 10 6 4 4 6
Standard form of a quadratic function was
𝒚 = 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄.
By substitution from 𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐, then 6 = 𝑎(0)2 + 𝑏(0) + 𝑐.
Using , 𝑐 = 6 means 𝑥 = 1 𝑎𝑛𝑑 𝑦 = 4
By substitution from 𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐, then 4 = 𝑎(1)2 + 𝑏(1) + 6.
eq. 1 𝑎 + 𝑏 = −2

6. By substitution from 𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐, then
6 = 𝑎(3)2 + 𝑏(3) + 6.
eq. 2 9𝑎 + 3𝑏 = 0
Use eq. 1 and eq. 2: 𝑎 + 𝑏 = −2, 𝑎𝑛𝑑 9𝑎 + 3𝑏 = 0, 𝑏𝑦 𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛
eq. 1 𝑎 + 𝑏 = −2 . . multiply by -3. . −3𝑎 − 3𝑏 = 6
eq. 2 9𝑎 + 3𝑏 =0 . . .. . . . . . . . …… 9𝑎 + 3𝑏 = 0
6𝑎 = 6 . 𝑎 = 1
Substitute a from 𝑎 + 𝑏 = −2 . . 1 + 𝑏 = −2 . . . 𝑏 = −3
Thus, 𝒂 = 𝟏, 𝒃 = −𝟑 𝒂𝒏𝒅 𝒄 = 𝟔, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡ℎ𝑒 𝑞𝑢𝑎𝑑𝑟𝑎𝑡𝑖𝑐 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
𝑖𝑠
𝒇(𝒙) = 𝒙𝟐 − 𝟑𝒙 + 𝟔 𝒐𝒓 𝒚 = 𝒙𝟐 − 𝟑𝒙 + 𝟔

7. Example 3:
Find the quadratic function represented by table of values
below.
X 1 2 3 4 5 6 7
y 5 11 19 29 41 55 71
First difference 6 8 10 12 14 16
2
2
2 2
2
Second difference
In Quadratic Function 𝒚 = 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄,
𝒂 = 𝒔𝒆𝒄𝒐𝒏𝒅 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒄𝒆
2

8. Solve for 𝐛 𝐚𝐧𝐝 𝐜: Standard form of a quadratic function was 𝒚 = 𝒂
𝒙𝟐 + 𝒃𝒙 + 𝒄.
5 = (1)(1)2 + 𝑏(1) + 𝑐
5 = 1 + 𝑏 + 𝑐 𝟒 = 𝒃 + 𝒄 eq. 1
11 = (1)(2)2 + 𝑏(2) + 𝑐
11 = 4 + 2𝑏 + 𝑐 𝟕 = 𝟐𝒃 + 𝒄 eq. 2
Use eq. 1 and eq. 2: 𝑏 + 𝑐 = 4, 𝑎𝑛𝑑 2𝑏 + 𝑐 = 7, 𝑏𝑦 𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛

9. eq. 1: 𝑏 + 𝑐 = 4 . . . multi by -1 . . . −𝑏 − 𝑐 = −4
eq. 2: 2𝑏 + 𝑐 = 7 . . . . . . . . . . . 2𝑏 + 𝑐 = 7
𝑏 = 3 𝑏 = 3
Use 𝑒𝑞. 1 𝑡𝑜 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑐: 𝑏 + 𝑐 = 4 . . . . 3 + 𝑐 = 4 𝑐 = 1
Thus, 𝒂 = 𝟏, 𝒃 = 𝟑𝒂𝒏𝒅 𝒄 = 𝟏, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡ℎ𝑒 𝑞𝑢𝑎𝑑𝑟𝑎𝑡𝑖𝑐 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑖𝑠
𝒇(𝒙) = 𝒙𝟐 + 𝟑𝒙 + 𝟏 𝒐𝒓 𝒚 = 𝒙𝟐 + 𝟑𝒙 + 𝟏
Example 4: Given the graph
Use the 𝑉(ℎ, 𝑘) form of a quadratic
function 𝑦 = 𝑎(𝑥 − ℎ)2 + 𝑘.

10. Substitute ℎ = 2 𝑎𝑛𝑑 𝑘 = −3, 𝑦 = 0 𝑎𝑛𝑑 𝑥 = 5 from
𝑦 = 𝑎(𝑥 − ℎ)2 + 𝑘.
0 = 𝑎(5 − 2)2 − 3.
0 = 𝑎(3)2 − 3.
0 = 9𝑎 − 3.
1
𝑎 =
3
Thus, 𝑎 = 1 , ℎ = 2, 𝑘 = −3, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡ℎ𝑒 𝑞𝑢𝑎𝑑𝑟𝑎𝑡𝑖𝑐
3
𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝒚 = 𝟏 (𝒙 − 𝟐)𝟐 − 𝟑
3

11. Lesson 2
Solve Problems Involving
Quadratic Functions

12. A length of rectangular field is 20 m greater than the
width. Its area is 2400 m2. Find the length and the
width.
L and W rectangular field:
x
x + 20
width: x
length: x + 20
Conditions Presentation
The area is 2400 m2 x(x + 20) = 2400
Use equation 1: 𝑥(𝑥 + 20) = 2400
Multiply 𝑥2 + 20𝑥 = 2400
Transform into 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 𝑥2 + 20𝑥 − 2400 = 0
Solve for x 𝑥 − 40 = 0 , 𝑥 + 60 = 0
𝑥 = 40, 𝑥 = −60

13. Since we are looking for the length and the width, -60 is not possible
length.
The width is 40 m. Since the area is 2400, then
L = x + 20 = 40 + 20 = 60. L = 60 m and the W = 40 m.
Check: The area is 2400 m2 so, (40)(60) = 2400 m2.
The use of the quadratic function can be seen in many different
fields like physics, industry, business and in variety of mathematical
problems.
Example 1:
The sum of two numbers is 30. Find the numbers so that the
product is to be maximum.
Solution: Let 𝑛 be the number
The product is to be maximum.
30 − 𝑛 is the other number

14. 𝑦 = 𝑛(30 − 𝑛)
𝑦 = 30𝑛 − 𝑛2
𝑦 = −(𝑛2 − 30𝑛)
𝑦 = −(𝑛2 − 30𝑛 + 225) + 225
𝑦 = −(𝑛 − 15)2 + 225
The numbers are 15 and 15 and the maximum product is 225