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PROBLEM SOLVING 
INVOLVING QUADRATIC 
Reynaldo B. Pantino FUNCTION
Objectives: 
Determine the 
maximum or the 
minimum value 
of 
f(x) =ax2 + bx 
+ c. 
Write the given 
and the unknown 
quantity of the 
problem. 
Solve and show 
the solution of 
solving the 
problem.
A professional pyro-technician shoots fireworks vertically into 
the air off of a building that is 80 feet tall. The initial 80 feet velocity tall 
of 
the firework initial velocity 
is 64 feet per second. 
64 feet per second 
When 
will the fireworks reach their maximum height? 
What is the maximum height reached by the fireworks? 
maximum height 
h = –16t2 + v0t + h0 
– b 
2a 
t =
WHAT IS PARABOLA? 
A parabola is the 
graph of a 
quadratic function. 
Unlocking of Difficulties
WHAT IS VERTEX? 
Vertex 
Vertex 
The vertex of a parabola is either the 
lowest point on the graph or the highest 
point on the graph. 
Unlocking of Difficulties
WHAT IS AXIS OF 
SYMMETRY? 
The axis of symmetry is the line that 
divides the parabola into two equal 
parts. 
Unlocking of Difficulties 
Axis of symmetry
WHAT IS MAXIMUM OR 
MINIMUM VALUE? 
When a parabola opens up and the vertex 
is the lowest point the y-coordinate of the 
vertex is the minimum. 
When a parabola opens down and the 
vertex is the highest point the y-coordinate 
of the vertex is the maximum. 
Unlocking of Difficulties
WHAT IS MAXIMUM OR 
MINIMUM VALUE? 
Unlocking of Difficulties 
minimum 
maximum
WHAT IS ZERO OF A 
FUNCTION? 
Zero of a function is a number x such 
that f(x) = 0. 
Example. F(x) = x2 – 4x + 4. 
What is the value of x such that f(x) will 
be zero? 
Unlocking of Difficulties
Illustrative Example 1. 
1. Find two real numbers whose sum is 8 and whose product is 16. 
Solution: 
Let x be one of the numbers. Then 8 – x is the other number. If the 
product of the two numbers is 16, then, 
x(8 – x) = 16 
8x – x2 = 16 
- x2 + 8x – 16 = 0 by factoring we have (x - 4)(-x + 4) = 0 
x = 4 ( one of the number) 
What is 8 – x ? 
Since x = 4 then 8 – (4) = 4 (other number)
Illustrative Example 1. 
The table of values and the graph of the 
function, f(x) = - x2 + 8x – 16 are as follows; 
x - x2 + 8x – 16 f(x) 
1 - (1)2 + 8(1) – 16 -9 
2 - (2)2 + 8(2) – 16 -4 
3 - (3)2 + 8(3) – 16 -1 
4 - (4)2 + 8(4) – 16 0 
5 - (5)2 + 8(5) – 16 -1 
6 - (6)2 + 8(6) – 16 -4 
7 - (7)2 + 8(7) – 16 -9 
Notice that the maximum value of the function occurs at (4,0) representing the 
vertex of the parabola. The point of intersection is at x –axis. Therefore, if x = 4 is 
one of the number, then the other number, 8 – x is 4.
Illustrative Example 2. 
A projectile is launched from a point 40 meters above the 
ground. The height at ground level h in meters after t 
seconds is given by the function, h(t) = 40t – 5t2. How 
many seconds after the launch will the projectile hit the 
ground? 
Solution: 
At the ground level, the value of h will be 0. Thus, 
the quadratic equation is; 0 = 40t – 5t2 
0 = 40t – 5t2 
0 = 5t(8 – t) by common factor 
5t = 0 solve for the value of t 
t = 0 
8 – t = 0 
t = 8
Illustrative Example 2. 
Since t represents the time in seconds 
after the launch, t ≠ 0. The projectile will 
hit the ground 8 seconds after launch. 
You can always solve this equation by 
looking for the zeros of the related 
quadratic function. 
Axis of symmetry: Vertex 
t = - b /2a h(t) = 40t – 5t2 
= - 40/2(-5) h(4) = 40(4) – 5(4)2 
= 4 = 80
2 4 6 8 
120 
100 
Axis of symmetry 
80 
60 
40 
20 
0 
Illustrative Example 2. 
Thus, the vertex of the graph is the 
point (4,80) where the maximum 
value occurs. The graph and table of 
values is given below; 
x 40t – 5t2 f(x) 
0 40(0) – 5(0)2 0 
1 40(1) – 5(1)2 35 
2 40(2) – 5(2)2 60 
3 40(3) – 5(3)2 75 
4 40(4) – 5(4)2 80 
5 40(5) – 5(5)2 75 
6 40(6) – 5(6)2 60 
7 40(7) – 5(7)2 35 
8 40(8) – 5(8)2 0 
Maximum value 
vertex 
(4,80)
Illustrative Example 3. 
One hundred meters of fencing is used to 
fence a rectangular region. What 
dimensions of the rectangle will maximize 
the area of the region? 
Solution: 
The formula for the perimeter of a 
rectangle is P = 2(l + w) 
Since P = 100 meters, the length plus the 
width is 50 meters. 
Thus, let x be the length of the rectangle 
50 – x is the width of the 
rectangle 
Since A = lw for the area of the rectangle, 
A (x) = x(50 – x) 
= 50x – x2 
The area is a 
quadratic function 
of the length. 
The maximum value 
occurs when; 
x = -b /2a 
Thus; x = -50/2(-1) 
= 25 
The length should be 
25 m and the width 
should be 25 m. The 
rectangle that gives 
the maximum area is 
a square with an 
area of 625 m2.
Activity 1. Find the maximum or the minimum value of 
each quadratic function 
1.) f(x) = x2 – 5x + 6 6.) f(x) = -x2 + 4x – 3 
2.) f(x) = x2 – 10x + 21 7.) f(x) = -x2 – 5x – 3 
3.) f(x) = x2 – 6x – 6 8.) f(x) = -2x2 + 2x - 5 
4.) f(x) = x2 – 2x + 4 9.) f(x) = 2x2 + 2x – 4 
5.) f(x) = 2x2 – x + 6 10.) f(x) = 6x2 – x + 2
Activity 2. Solve at least 2 of the following problems. 
1.) Find the two real numbers whose sum is 4 and whose product is a 
maximum. 
2.) The perimeter of a rectangle is 38 meters. Find the dimensions of 
the rectangle that will contain the greatest area. 
3.) Find the number of computer units x that needs to be sold to 
produce a maximum profit P (in pesos) if P = 900x – 0.1x2. 
4.) The height in feet of a ball t seconds after being tossed upwards is 
given by the function h(t) = 84t – 16t2. 
a. After how many seconds will it hit the ground? 
b. What is its maximum height? 
5.) The sum of two numbers is 22 and the sum of their squares is 250. 
Find the numbers.
Quiz. (1 whole sheet of paper) 
1. The sum of two numbers is 16 and the 
sum of their squares is 146. Find the 
numbers. 
2. The perimeter of a rectangle is 138 
meters and the area is 1080 square 
meters. Find the length and the width of 
the rectangle.

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SOLVE QUADRATIC PROBLEMS AND MAXIMIZE AREAS

  • 1. PROBLEM SOLVING INVOLVING QUADRATIC Reynaldo B. Pantino FUNCTION
  • 2. Objectives: Determine the maximum or the minimum value of f(x) =ax2 + bx + c. Write the given and the unknown quantity of the problem. Solve and show the solution of solving the problem.
  • 3. A professional pyro-technician shoots fireworks vertically into the air off of a building that is 80 feet tall. The initial 80 feet velocity tall of the firework initial velocity is 64 feet per second. 64 feet per second When will the fireworks reach their maximum height? What is the maximum height reached by the fireworks? maximum height h = –16t2 + v0t + h0 – b 2a t =
  • 4. WHAT IS PARABOLA? A parabola is the graph of a quadratic function. Unlocking of Difficulties
  • 5. WHAT IS VERTEX? Vertex Vertex The vertex of a parabola is either the lowest point on the graph or the highest point on the graph. Unlocking of Difficulties
  • 6. WHAT IS AXIS OF SYMMETRY? The axis of symmetry is the line that divides the parabola into two equal parts. Unlocking of Difficulties Axis of symmetry
  • 7. WHAT IS MAXIMUM OR MINIMUM VALUE? When a parabola opens up and the vertex is the lowest point the y-coordinate of the vertex is the minimum. When a parabola opens down and the vertex is the highest point the y-coordinate of the vertex is the maximum. Unlocking of Difficulties
  • 8. WHAT IS MAXIMUM OR MINIMUM VALUE? Unlocking of Difficulties minimum maximum
  • 9. WHAT IS ZERO OF A FUNCTION? Zero of a function is a number x such that f(x) = 0. Example. F(x) = x2 – 4x + 4. What is the value of x such that f(x) will be zero? Unlocking of Difficulties
  • 10. Illustrative Example 1. 1. Find two real numbers whose sum is 8 and whose product is 16. Solution: Let x be one of the numbers. Then 8 – x is the other number. If the product of the two numbers is 16, then, x(8 – x) = 16 8x – x2 = 16 - x2 + 8x – 16 = 0 by factoring we have (x - 4)(-x + 4) = 0 x = 4 ( one of the number) What is 8 – x ? Since x = 4 then 8 – (4) = 4 (other number)
  • 11. Illustrative Example 1. The table of values and the graph of the function, f(x) = - x2 + 8x – 16 are as follows; x - x2 + 8x – 16 f(x) 1 - (1)2 + 8(1) – 16 -9 2 - (2)2 + 8(2) – 16 -4 3 - (3)2 + 8(3) – 16 -1 4 - (4)2 + 8(4) – 16 0 5 - (5)2 + 8(5) – 16 -1 6 - (6)2 + 8(6) – 16 -4 7 - (7)2 + 8(7) – 16 -9 Notice that the maximum value of the function occurs at (4,0) representing the vertex of the parabola. The point of intersection is at x –axis. Therefore, if x = 4 is one of the number, then the other number, 8 – x is 4.
  • 12. Illustrative Example 2. A projectile is launched from a point 40 meters above the ground. The height at ground level h in meters after t seconds is given by the function, h(t) = 40t – 5t2. How many seconds after the launch will the projectile hit the ground? Solution: At the ground level, the value of h will be 0. Thus, the quadratic equation is; 0 = 40t – 5t2 0 = 40t – 5t2 0 = 5t(8 – t) by common factor 5t = 0 solve for the value of t t = 0 8 – t = 0 t = 8
  • 13. Illustrative Example 2. Since t represents the time in seconds after the launch, t ≠ 0. The projectile will hit the ground 8 seconds after launch. You can always solve this equation by looking for the zeros of the related quadratic function. Axis of symmetry: Vertex t = - b /2a h(t) = 40t – 5t2 = - 40/2(-5) h(4) = 40(4) – 5(4)2 = 4 = 80
  • 14. 2 4 6 8 120 100 Axis of symmetry 80 60 40 20 0 Illustrative Example 2. Thus, the vertex of the graph is the point (4,80) where the maximum value occurs. The graph and table of values is given below; x 40t – 5t2 f(x) 0 40(0) – 5(0)2 0 1 40(1) – 5(1)2 35 2 40(2) – 5(2)2 60 3 40(3) – 5(3)2 75 4 40(4) – 5(4)2 80 5 40(5) – 5(5)2 75 6 40(6) – 5(6)2 60 7 40(7) – 5(7)2 35 8 40(8) – 5(8)2 0 Maximum value vertex (4,80)
  • 15. Illustrative Example 3. One hundred meters of fencing is used to fence a rectangular region. What dimensions of the rectangle will maximize the area of the region? Solution: The formula for the perimeter of a rectangle is P = 2(l + w) Since P = 100 meters, the length plus the width is 50 meters. Thus, let x be the length of the rectangle 50 – x is the width of the rectangle Since A = lw for the area of the rectangle, A (x) = x(50 – x) = 50x – x2 The area is a quadratic function of the length. The maximum value occurs when; x = -b /2a Thus; x = -50/2(-1) = 25 The length should be 25 m and the width should be 25 m. The rectangle that gives the maximum area is a square with an area of 625 m2.
  • 16. Activity 1. Find the maximum or the minimum value of each quadratic function 1.) f(x) = x2 – 5x + 6 6.) f(x) = -x2 + 4x – 3 2.) f(x) = x2 – 10x + 21 7.) f(x) = -x2 – 5x – 3 3.) f(x) = x2 – 6x – 6 8.) f(x) = -2x2 + 2x - 5 4.) f(x) = x2 – 2x + 4 9.) f(x) = 2x2 + 2x – 4 5.) f(x) = 2x2 – x + 6 10.) f(x) = 6x2 – x + 2
  • 17. Activity 2. Solve at least 2 of the following problems. 1.) Find the two real numbers whose sum is 4 and whose product is a maximum. 2.) The perimeter of a rectangle is 38 meters. Find the dimensions of the rectangle that will contain the greatest area. 3.) Find the number of computer units x that needs to be sold to produce a maximum profit P (in pesos) if P = 900x – 0.1x2. 4.) The height in feet of a ball t seconds after being tossed upwards is given by the function h(t) = 84t – 16t2. a. After how many seconds will it hit the ground? b. What is its maximum height? 5.) The sum of two numbers is 22 and the sum of their squares is 250. Find the numbers.
  • 18. Quiz. (1 whole sheet of paper) 1. The sum of two numbers is 16 and the sum of their squares is 146. Find the numbers. 2. The perimeter of a rectangle is 138 meters and the area is 1080 square meters. Find the length and the width of the rectangle.

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