The document summarizes a training session for teachers on quadratic functions. It defines quadratic functions, discusses their various forms, characteristics, and how to determine their domain and range. It provides examples of solving problems involving quadratic functions, such as finding the maximum value of a function. The objectives are for participants to be able to define quadratic functions, determine their domain and range, and solve problems involving them.

1. DIVISION TRAINING OF TEACHERS ON LEAST
LEARNED COMPETENCIES/SKILLS IN
GRADE 9 MATHEMATICS
January 28-29, 2021

2. Quadratic Functions
EMETERIO J. FLORESCA JR. T-III
Mathematics Trainer

3. At the end of the session, the
participants should be able to:
OBJECTIVES
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
1. Define quadratic function
2. Determine the domain and range of
the quadratic function
3. Solve problems involving quadratic
function.

4. 1. Determine the range of
y = x2 + 3x +1
A. y > 1 C. y ≥ −
𝟓
𝟒
B. y < 1 D. y ≤
𝟓
𝟒
ACTIVITY
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
ans. C

5. 2. The sum of two numbers is 38.
What is the maximum product of the
two numbers?
A. 72 C. 360
B. 357 D. 361
ACTIVITY
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
ans. D

6. 3. A ball is thrown into the air. Its height above
the ground, h (measured in feet), at any given
time after the ball is thrown, t (measured in
seconds), can be modeled using the quadratic
function h(t) = -16t2 + 16t + 32. When did the ball
hit the ground?
A. 1 second C. 2 seconds
B. 1.5 second D. 3 seconds
ACTIVITY
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
ans. C

7. •A quadratic function is any
equation/ function with a degree of
2 that can be written in the
form y/f(x) = ax2 + bx + c, where a, b,
and c are real numbers, and a does
not equal 0.
QUADRATIC FUNCTIONS DEFINED
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
DEFINITION:

8. Standard Form:
FORMS OF QUADRATIC FUNCTIONS
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
y/f(x) = ax2 + bx + c

9. Factored Form:
FORMS OF QUADRATIC FUNCTIONS
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
y/f(x) = (ax + c)(bx + d)

10. Vertex Form:
FORMS OF QUADRATIC FUNCTIONS
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
y/f(x) = a(x + b)2 + c

11.  The graph of a quadratic
function is always a parabola.
 Upward when a is positive.
 Downward when a is negative.
CHARACTERISTICS OF QUADRATIC FUNCTIONS
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO

12.  The domain of a
quadratic function are
all real numbers.
CHARACTERISTICS OF QUADRATIC FUNCTIONS
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO

13.  The range of a
quadratic function is the
set of all real values of y.
CHARACTERISTICS OF QUADRATIC FUNCTIONS
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO

14.  The vertex is the lowest
point when the parabola
opens upward.
CHARACTERISTICS OF QUADRATIC FUNCTIONS
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO

15.  The vertex is the
maximum point when the
parabola opens downward.
CHARACTERISTICS OF QUADRATIC FUNCTIONS
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO

16. Example 1. y = x2
Domain: All real numbers.
Range: y ≥ 0
Example 2. y = -x2 + 5
Domain: All real numbers.
Range: y ≤ 5
DETERMINING THE DOMAIN & RANGE OF QUADRATIC FUNCTIONS
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO

17. Example 3. y = 25x2 + 4
Domain: All real numbers.
Range: y ≥ 4
Example 4. y = x2 + 3x + 1
Domain: All real numbers.
Range: y ≥ -1.25
DETERMINING THE DOMAIN & RANGE OF QUADRATIC FUNCTIONS
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO

18. Example 4. y = x2 + 3x + 1
Solution: x2 + 3x + 1 – y = 0
a = 1; b = - 3; c = 1 – y
Using 𝒃𝟐 − 𝟒𝒂𝒄 ≥ 0, −𝟑 𝟐 − 𝟒(𝟏)(𝟏 − 𝒚) ≥ 0;
𝟗 − 𝟒 + 𝟒𝒚 ≥ 𝟎; 𝟓 + 𝟒𝒚 ≥ 𝟎; 𝒚 ≥ −𝟏. 𝟐𝟓
Thus, the range of the function, y ≥ - 1.25
DETERMINING THE DOMAIN & RANGE OF QUADRATIC FUNCTIONS
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO

19. In any quadratic function,
y = ax2 + bx + c, the domain is the
set of all real numbers while the range of a
quadratic function depends on the value
of a, where a is the numerical coefficient
of x2.
DETERMINING THE DOMAIN & RANGE OF QUADRATIC FUNCTIONS
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO

20. In any quadratic function,
y = ax2 + bx + c, the domain is the set of all real
numbers while the range of a quadratic function depends on
the value of a, where a is the numerical coefficient of x2.
If a > 0,
range: 𝒚 ≥ 𝒄 −
𝒃𝟐
𝟒𝒂
or 𝒄 −
𝒃𝟐
𝟒𝒂
, ∞
If a < 0,
range: 𝒚 ≤ 𝒄 −
𝒃𝟐
𝟒𝒂
or −∞, −
𝒃𝟐
𝟒𝒂
DETERMINING THE DOMAIN & RANGE OF QUADRATIC FUNCTIONS
DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO

21. DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
APPLICATIONS
PROBLEM 1: NUMBER PROBLEM
The sum of two numbers is 14. Determine the maximum product of two
numbers.
Let x = first number
14 – x = second number
y = maximum product of two numbers
y = (14 – x)(x) 14𝑥 − 𝑥2; a = - 1; b = 14; c = 0
Solution: Solve for k, where k is the maximum value of the two numbers.
k = 𝑐 −
𝑏2
4𝑎
0 -
142
4(−1)
−196
−4
= 49
Thus, the maximum product of two numbers is 49.

22. DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
APPLICATIONS
PROBLEM 2: GEOMETRY PROBLEM
What are the dimensions of the largest rectangular field that can be
enclosed by 60 m fencing wire?
Solution: Let l and w be the length and width of the rectangle respectively.
Then the perimeter of the rectangle is P = 2L + 2w
Since P = 60, then we have 2L + 2w = 60; l + w = 30; l = 30 – w
Substituting in the formula for area A of a rectangle.
Area = length x width
Area = (30 – w)(w) 30w – w2-; a = -1; b = 30,
Since the area is a function of w, so we have, 𝑤 =
−𝑏
2𝑎
; 𝑤 = 15 𝑚; 𝑙 = 30 − 𝑤 = 15 𝑚
Thus, the length should be 15 m and width should be 15 m. The rectangle that gives the
maximum area is a square with an area of 225 sq. m.

23. DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
APPLICATIONS
PROBLEM 3: PROJECTILE PROBLEM
• A ball is thrown into the air. Its height above the ground, h
(measured in feet), at any given time after the ball is
thrown, t (measured in seconds), can be modeled using the
quadratic function h(t) = -16t2 + 16t + 32.
a) From what height above the ground was the ball thrown?
b) How many seconds after the ball was thrown did it reach
its maximum height?
c) What was the ball's maximum height?
d) When did the ball hit the ground?

24. DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
APPLICATIONS
PROBLEM 3: PROJECTILE PROBLEM
• A ball is thrown into the air. Its height above the ground, h
(measured in feet), at any given time after the ball is thrown,
t (measured in seconds), can be modeled using the quadratic
function h(t) = -16t2 + 16t + 32.
a) From what height above the ground was the ball thrown?
Solution:
Assume that t = 0, h(0) = -16(0)2 + 16(0) + 32 = 32
Thus, the ball was thrown 32 ft above the ground.

25. DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
APPLICATIONS
PROBLEM 3: PROJECTILE PROBLEM
• A ball is thrown into the air. Its height above the
ground, h (measured in feet), at any given time after
the ball is thrown, t (measured in seconds), can be
modeled using the quadratic function h(t) = -16t2 +
16t + 32.
b) How many seconds after the ball was thrown did it reach its
maximum height?
Solution: Use t = -b/2a
t = -16/-2(-16)
t = 0.5 seconds
Thus, the ball reaches its maximum height one-half
second after it is thrown into the air.

26. DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
APPLICATIONS
PROBLEM 3: PROJECTILE PROBLEM
A ball is thrown into the air. Its height above the ground, h
(measured in feet), at any given time after the ball is
thrown, t (measured in seconds), can be modeled using the
quadratic function h(t) = -16t2 + 16t + 32.
c) What was the ball's maximum height?
Solution:
h(0.5) = -16(-0.5)2 + 16(0.5) + 32
h = 36
Thus, the maximum height of the ball reached was
36 ft. above the ground.

27. DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
APPLICATIONS
PROBLEM 3: PROJECTILE PROBLEM
• A ball is thrown into the air. Its height above the ground, h
(measured in feet), at any given time after the ball is
thrown, t (measured in seconds), can be modeled using the
quadratic function h(t) = -16t2 + 16t + 32.
d) When did the ball hit the ground?
0 = -16t2 + 16t + 32
Solution:
0 = -16(t2 – t – 2)
0 = t2 – t – 2
0 = t – 2; 0 = t + 1
0 = (t – 2)(t + 1)
Thus, the ball hit the ground after 2 seconds.

28. DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
APPLICATIONS
PROBLEM 4: BUSINESS PROBLEM
• The profit (in thousands of pesos) of a company is given by.
P(x) = 5000 + 1000 x - 5x2
where x is the amount ( in thousands of pesos) the company
spends on advertising.
a) Find the amount, x, that the company has to spend to
maximize its profit.
b) Find the maximum profit P(max).

29. DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
APPLICATIONS
PROBLEM 4: BUSINESS PROBLEM
• The profit (in thousands of dollars) of a company is given by.
P(x) = 5000 + 1000x - 5x2
where x is the amount ( in thousands of dollars) the company spends on
advertising.
a) Find the amount, x, that the company has to spend
Solution: Function P that gives the profit is a quadratic function with the
leading coefficient a = - 5.
This function (profit) has a maximum value at
x = h = - b / (2a)
x = h = -1000 / (2(-5)) = 100
Thus, the company has to spend 100 thousand pesos
to maximize its profit.

30. DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
APPLICATIONS
PROBLEM 4: BUSINESS PROBLEM
• The profit (in thousands of dollars) of a company is given by.
P(x) = 5000 + 1000 x - 5 x2
where x is the amount ( in thousands of dollars) the company
spends on advertising.
b) Find the maximum profit Pmax.
Solution:
P(100) = 5000 + 1000 (100) – 5(100)2
P(100) = 55000
Thus, the company’s maximum profit is ₱55,000.

31. DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
NON-ROUTINE PROBLEMS IN QUADRATIC FUNCTIONS
If y = a(x – 2)2 + c and y=(2x – 5)(x – b)
represent the same quadratic
function, determine the numerical
value of b?
PROBLEM 5: NON-ROUTINE PROBLEM

32. DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
NON-ROUTINE PROBLEMS IN QUADRATIC FUNCTIONS
Solutions:
PROBLEM 5: NON-ROUTINE PROBLEM
We expand the two functions.
First, y = a(x - 2)2 + c
= ax2 – 4ax + (4a + c)
Second, y = (2x – 5)(x – b) = 2x2 – (5 + 2b)x + 5b
Equate the coefficients,
ax2 – 4ax + 4a + c = 2x2 – (5 + 2b)x + 5b
So, a = 2.
From the coefficient of x, 4a = 5 + 2b; since a = 2, then b =
3
2
.
Thus, the numerical value of b =
3
2
.

33. DEPARTMENT OF EDUCATION-SCHOOLS DIVISION OF SOUTH COTABATO
Thank you for listening!!!