Q1 I.

Inequalities
W4
At the end of the lesson, you are expected to:
 graph the solution set of inequalities What is a number line?
 write the set notation and interval of inequality
A straight line with numbers placed at equal
intervals or segments along its length.

Statements like “Only ages 18 and above are

allowed to go outside their houses in this
pandemic and “I spent less than P500 in
shopping” when translated into algebraic
expressions can have many solutions.

Ages 18 and above can mean ages like 18,

18.5, 19, 20, …

Spent less than P500 refers to amounts like

P499.5, P499, P498, …
To be guided in graphing the linear inequality in one
These statements are examples of inequalities. variable, follow the table below:

Inequality Tail of the Extension of the

Definition: symbol arrow arrow
Inequality - is a statement that shows two empty circle extended to the left of
numbers or expressions that are not equal. < “o” the number line
solid circle extended to the left of
Solution of an inequality - is a value/s of the
≤ “●” the number line
empty circle extended to the right
variable that makes the inequality true. > “o” of the number line
Example: x ≤−3 is a solution to the inequality Solid circle extended to the right
2 x ≤−6 ≥ “●” of the number line
Solution set - is the set of all solutions.
Example: {−3 ,−4 ,−5 ,−6 ,−7 , … } this set is
a solution set of 2 x ≤−6
- may be written in set notation using braces,
{ }.
Example: {x : x ≤−3 } is a solution set express
in set notation of the set
{−3 ,−4 ,−5 ,−6 ,−7 , … }

Illustrative Example:
a. Graph the solution set {x : x< 3 }

Solution:
Steps:
1. Locate 3 on the number line
How to read a set notation of an inequality? 2. Draw an empty circle (o), since the inequality
symbol is < then extend the arrow to the left.

{ x | x < 8}

The set of all x x is less than 8

such that
Solutions set of an inequality can be
illustrated too on a number line
b. Graph the solution set {x : x>−1}

1
 Steps: Illustrative Example:
1. Locate -1 on the number line Write the interval notation of the following set notation.
2. Draw an empty circle (o), since the inequality 1. {x : x>−1}
symbol is > then extend the arrow to the right. 2. {x : x< 3 }
3. {x  2}
4. {x  -4}
5. 2 < x < 8
6. 2 ≤ x < 8

c. Graph the solution set {x  2} Solution:

Steps: 1. “x is greater than -1” {x : x>−1}
1. Locate 2 on the number line
2. Draw a solid circle (●), since the inequality
symbol is  then extend the arrow to the left.

Corresponding interval: (-1, +∞) Use ( ) for -1

since we are using >.
Notice that the
interval notation is
written from left to
d.Graph the solution set {x  -4}
right of the number
Steps: line.
1. Locate -4 on the number line
2. Draw a solid circle (●), since the inequality 2. “x is less than 3” {x : x< 3 }
symbol is  then extend the arrow to the right.

Corresponding interval: (-∞, 3) Use ( ) for 3 since

we are using <.
Notice that the
interval notation is
written from left to
Inequalities can also be represented on a right of the number
number line to show a range of line.
solutions with an upper and lower limit.
Example: 3. “x is less than or equal to 2” {x  2}

Corresponding interval: (-∞, 2] Use [ ] for 2 since

we are using ≤.

4. “x is greater than or equal to -4” {x  -4}

Set notation of Inequalities and Interval
notation.
Set notation of inequalities can also be written into
"Interval Notation" where we just write the beginning
and ending numbers of the interval, and use: Corresponding interval: [-4, +∞) Use [ ] for -4
since we are using
 [ ] a square bracket when we want to include the ≥.
end value, or when we are using the symbols {≤ or
≥} or when using solid circle. 5. “2 is less than x and x is less than 8” 2<x<8
 ( ) a round bracket when we don't include the end
value or when we use symbols {< or >} or when
using empty circle.
 The symbol (+∞) is read as positive infinity and
indicates that arrow extends to the right on a number Corresponding interval: (2, 8)
line
 The symbol (-∞) is read as negative infinity and
indicates that arrow extends to the left on a number
line.
 The infinity uses the ( )
2
 6. “2 is less than or equal to x and x is less than 8”
2≤x<8

Corresponding interval: [2, 8) we used [ ] for 2 Illustrative Example:

and () for 8
Find the solution set of the quadratic inequality
Illustrative Example:
1. x2 + 3x – 4 > 0
Write the set notation and interval notation for each
2. x2 + 2x – 8 ≥ 0
graph.
3. x2 + 8x + 15 < 0
4. x2 - 5x+ 4 ≤ 0
1.
Solution:

1. x2 + 3x – 4 >0
2.
x2 + 3x – 4 = 0 rewrite into quadratic equation

Solve for the Roots:

3.
x2 + 3x – 4 = 0 where a = 1, b = 3 and c = -4

Product: a ∙ c = 1 ∙ -4 = -4
4.
Sum : b = 3
Find 2 numbers
whose Product is -
-4
4 and whose sum
Solution: is 3
1. Set Notation: {x|x < 4}
Interval Notation: (-∞, 4) 4 -1 The 2 numbers
2. Set Notation: {x|x ≥ 4} are 4 and -1.
Since 4 ∙ -1 = -4
Interval Notation: [4, +∞)
3. Set Notation: {x| 0 ≤x ≤ 4 }
3 and
4+ -1 = 3
Interval Notation: [0, 4]
4. Set Notation: {x| 0 < x < 4 }
Divide the 2 numbers by the value of a which is
Interval Notation: (0, 4)
a = 1. Do not forget this part, otherwise you will get
wrong roots.
II.Quadratic Inequalities
4 -1
At the end of the lesson, you are expected to:
,
 solve quadratic inequalities using solution to 1 1 a=1
quadratic equation and number line
and -1
Definition:
= 4
A quadratic inequality in one variable is an
inequality that contains polynomial whose highest = -4 and 1 Lastly, change the signs to
exponent is 2. The general forms are the opposites.
following, where a, b, and c are real numbers with
a > 0 and a ≠ 0. Therefore the roots are -4 and 1 .

ax2 + bx + c > 0 ax2 + bx + c < 0

Plot the roots on a number line using empty circles
ax2+ bx + c ≤ 0 ax2+ bx + c ≥ 0
since the inequality is >. Look at the 3 partitions made
In solving quadratic inequalities, we need to find by the roots. Write the 3 intervals.
its solution set. The solution set of a quadratic
inequality can be written as a set and can be x < -4 -4 < x < 1 x>1
illustrated through a number line. or or or
(-∞, -4) (-4, 1) (1, +∞)
Steps in solving quadratic inequalities:
1. Rewrite the inequality into its corresponding
equality.
2. Solve the roots of the quadratic equation. 3
3. Plot the roots on a number line.
4. Write the three intervals.
5. Test a number from each interval against the
inequality.
 x2 + 8x + 15 = 0 rewrite into quadratic equation

The roots are -5 and -3 .

Plot the roots on a number line using empty circles

since the inequality is < and write the 3 intervals.
Test a number from each interval against the inequality
x2 + 3x – 4 >0 x < -5 -5 < x < -3 x > -3
or or or
For (-∞, -4), For (-4, 1), For (1, +∞), use
(-∞, -5) (-5, -3) (-3, +∞)
use x = -6 use x= -2 x=4
(it is up to you what (it is up to you what (it is up to you what
number you will choose number you will choose number you will choose
as long as it is an as long as it is an as long as it is an
element of (-∞, -4). element of (-4, 1). element of (1, +∞).
Notice that -4 is not Notice that -4 and 1 Notice that 1 is not
included. should not be included included.

x2 + 3x – 4 >0 x2 + 3x – 4 >0 x2 + 3x – 4 >0 Test a number from each interval against the inequality
(-6)2 + 3(-6) - 4 > 0 (-2)2 + 3(-2) - 4 > 0 (4)2 + 3(4) - 4 > 0
36 + -18 – 4 > 0 x2 + 8x + 15 < 0
4 + -6 – 4 > 0 16 + 12 – 4 > 0
18 – 4 > 0 -2 – 4 > 0 28 – 4 > 0
14 > 0, true -6 > 0, false 24 > 0, true For (-∞, -5) For (-5, -3) For (-3, +∞)
use x = -6 use x= -4 use x = 0
Therefore, intervals solutions are
(-∞, -4) or (1, +∞). In other words, the solution x2 + 8x + 15 < 0 x2 + 8x + 15 < 0 x2 + 8x + 15 < 0
(-6)2 + 8(-6) +15 < 0
set of the inequality is {x| x < -4 or x > 1}. (-4)2 + 8(-4) +15 < 0 (0)2 + 8(0) +15 < 0
36 - 48 + 15 < 0 16 - 32 + 15 < 0 0 + 0 + 15 < 0
3 < 0, false -16 + 15 < 0 15 < 0, false
-1 < 0, true
2. x2 + 2x – 8 ≥ 0
Therefore, interval solution is (-5, -3) . In other
x2 + 2x – 8 = 0 rewrite into quadratic equation words, the solution set of the inequality is
{x| -5< x ≤ -3}.

The roots are -4 and 2 . 4. x2 - 5x+ 4 ≤ 0

Plot the roots on a number line using solid circles since x2 - 5x+ 4 = 0 rewrite into quadratic equation
the inequality is ≥ and write the 3 intervals.
The roots are 4 and 1 .
x ≤ -4 -4 ≤ x ≤ 2 x≥2
or or or Plot the roots on a number line using solid circles since
(-∞, -4] [-4, 2] [2, +∞) the inequality is ≤ and write the 3 intervals.

x≤1 1≤x≤4 x≥4

or or or
(-∞, 1] [1, 4] [4, +∞)

Test a number from each interval against the inequality

x2 + 2x – 8 ≥ 0

For (-∞, -4] For [-4, 2], For [2, +∞)

use x = -5 use x= 0 use x = 3
(it is up to you what (it is up to you what
Test a number from each interval against the inequality
(it is up to you what
number you will choose number you will choose number you will choose x2 - 5x+ 4 ≤ 0
as long as it is an as long as it is an as long as it is an
element of (-∞, -4]. element of [-4, 2]. element of [2, +∞).
Notice that -4 is Notice that -4 and 2 are Notice that 2 is included. For (-∞, 1] For [1, 4] For [4, +∞)
included. included use x= 1 use x = 5
use x = 0
x2 + 2x – 8 ≥ 0 x2 + 2x – 8 ≥ 0 x2 + 2x – 8 ≥ 0
(-5)2 + 2(-5) – 8 ≥ 0 (0)2 + 2(0) – 8 ≥ 0 (3)2 + 2(3) – 8 ≥ 0 x2 - 5x + 4 ≤ 0 x2 - 5x + 4 ≤ 0 x2 - 5x + 4 ≤ 0
25 + -10 – 8 ≥ 0 0+0–8≥0 9+6–8≥0 (0)2 - 5(0) + 4 ≤ 0 (1)2 - 5(1) + 4 ≤ 0 (5)2 - 5(5) + 4 ≤ 0
15 – 8 ≥ 0 – 8 ≥ 0, false 15 – 8 ≥ 0 0-0+4≤0 1-5+4≤0 25 - 25 + 4 ≤ 0
7 ≥ 0, true 7 ≥ 0, true 4 ≤ 0, false -4+4 ≤ 0 4 ≤ 0, false
0 ≤ 0, true
Therefore, interval solutions are
(-∞, -4] or [2, +∞). In other words, the solution Therefore, the interval solution is [1, 4] . In other
set of the inequality is {x| x ≤ -4 or x ≥ 2}. words, the solution set of the inequality is
{x| 1 ≤ x ≤ 4}.
3. x2 + 8x + 15 < 0
4
 Illustrative Example:
1. Plot the point (-5, 3)

Solution:

(-5, 3) ----- means 5 Left and 3 Up

Find the solution set of the following quadratic
inequalities then graph.
1. x2 – 4 > 0
2. x2 – 4 < 0
3. x2 – 4  0
4. x2 – 4  0

Answer key: How many did you get right?

2. Plot (1 , -4)
Solution:
(1, -4) ----- means 1 Right and 4 Down

III. Cartesian Coordinate System

At the end of the lesson, you are expected to:
 Plot points on a Cartesian Plane

Plot the following on a Cartesian plane

 A Cartesian plane (named after French
mathematician Rene Descartes, who a. (2, 3)
formalized its use in mathematics) is defined b. (-3, 1)
by two perpendicular number lines: c. (0,0)
 the x-axis, which is horizontal and the y- d. (-1.5, -2.5)
axis, which is vertical. A funny mnemonics
Answer key: How many did you get right?
would be, Horizontal line for (“higa”) and
Vertical for (“varog”).
 The ordered pair ( x, y) where x is the abscissa/x-
coordinate and y is the ordinate/y-coordinate.

How do we plot points on Cartesian plane?

 The movement of x and y is represented below:

(x, y)
Left/Right Up/Down
 The sign of the number tells us the direction.
 Positive x means move Right, Negative x
means move Left
 Positive y means move Up, Negative y means
move Down

5
 4: y = (x – 2) (x + 4) --- is a quadratic function its
standard form is
IV. Quadratic Functions y = x2 + 2x – 8

Illustrative Example:
At the end of the lesson, you are expected to:
For each values of x in the table, find the corresponding
 recognize quadratic functions and differentiate value of y by evaluating y = x2 – 4
them from quadratic equations .
 represent and describe quadratic functions using y = x2 – 4
(a) table of values (b)graph and (c)equations
x -3 -2 -1 0 1 2 3
y
Definition:

A quadratic function is a second degree polynomial Solution:

represented as f(x) = ax2 + bx + c or y = ax2 + bx + c, If x = -3 then y = (-3)2 – 4 = 9 – 4 = 5
a ≠ 0 where a, b, and c are real numbers. If x = -2 then y = (-2)2 – 4 = 4 – 4 = 0
The form ax2+bx + c = 0 is a quadratic equation, while y =
ax2+bx + c is a QUADRATIC FUNCTION. Quadratic x -3 -2 -1 0 1 2 3
Equations and Quadratic Functions are almost the same in
y 5 0 -3 -4 -3 0 5
terms in the role of the independent variable or the x.
Notice that 0 from the quadratic equation was replaced by a
variable y in the quadratic function.
The graph of a quadratic function is a Parabola.

STANDARD FORM OF A QUADRATIC

FUNCTION
If a function is quadratic the second
2
y = ax +bx + c 2
OR f(x) = ax +bx + c difference of the value of y will be the
same.
 y can also be written in a function f (x) which  Let us consider y = x2 – 4
reads as “f of x”
 y is considered as the dependent variable in the
equation, meaning its value depends on x
 x is considered as the independent variable and
its highest exponent should be 2
 The right side of the equation is arrange in
decreasing order.
 Unlike the quadratic equation, sometimes the
standard quadratic function may have a to be Tips on identifying quadratic functions if the
negative. given is:
 Remember that a  0. If a = 0, the quadratic
function is no longer quadratic, instead we  a table of values - we can know if it
would have a linear function y = bx + c. represents a quadratic function if the
second differences of y are equal.
 a graph we can know if it represents
Illustrative Example: a quadratic function if it is a parabola.
Determine the values of a, b and c in the quadratic  an equation we can know if it
functions below: represents a quadratic function the highest
Solution: exponent of the dependent variable must
1. y = 3x2 + 4x + 7 a=3 b=4 c=7 be 2
2. y = x2 - 5x a = 1 b = -5 c = 0
3. f(x) = -6x2 + 8 a = -6 b = 0 c = 8
4. f(x) = x 2
a=1 b=0 c=0

Illustrative Example:
Identify whether the equation is a quadratic function or
not.
Solution: Identify which of the following represent quadratic
1. y = 2x -10 ---- is not a quadratic function Function. Write QF on the blank, if quadratic function
because the highest and NQF if not.
exponent of x is not 2.

2: y = 4x3 +3x + 7 --- is not a quadratic function

because the highest
exponent of x is again not
2.

3: f(x) = 5x – x2 --- is a quadratic function and its

standard form is f(x) = – x2 + 5x

6
 When Parabola opens upward or Parabola
opens downward
To determine if the parabola opens upward or
downward is to first identify the value of a.

 Parabola opens up if a is positive. Since

parabola opens upward then the vertex is the
minimum point or the lowest point of the
parabola.
 Parabola opens downward if a is negative.
Since parabola opens downward then the
vertex is the maximum point or the highest
point of the parabola.

Answer key: How many did you get right?

1. QF 2. QF 3. QF 4.NQF

Illustrative Example:
V. Graphing Quadratic Functions State whether the parabola opens up or down and
At the end of the lesson, you are expected to: whether the vertex is a maximum or a minimum point.
 graphs a quadratic function: (a) domain; (b)
1. y  2 x  4 x  5
2
range; (c) intercepts; (d) axis of symmetry; (e) --- a = 2 , since it is
vertex; (f) direction of the opening of the positive then the parabola
parabola. opens upward. The vertex
is the minimum point.
Graph of Quadratic Functions 2.
The graph of a quadratic function is a smooth --- The standard form is
u-shaped curve called parabola , where a = -3 ,
since it is negative then the
Characteristics/Properties of its graph are the parabola opens downward. The
following vertex is the maximum
1. Opening of the parabola – the direction of point.
the parabola whether it is upward or
downward.
Finding the Vertex and the Axis of Symmetry
2. Axis of symmetry - the line that divides the
and the maximum or minimum value of the
parabola into two parts that are mirror images
quadratic function y = ax2 +bx + c
of each other
3. Vertex - is the point of intersection of the axis
 Vertex is a point represented by (h, k)
of symmetry with the parabola. The vertex has
the coordinates (h, k). It is also called the
minimum point (if the graph opens upward) or  The h value can be found using h =
the maximum point (if the graph opens  To find the k value, we can replace y with k
downward) and replace the x with h.
4. Minimum value/Maximum Value – the
value of k in the vertex (h, k)
 Axis of Symmetry:
5. Domain – the set of all possible values of x in
 Maximum value if the parabola opens
a given function which is the set of real
downward
numbers
Maximum value = k
6. Range – the set of all possible values of y in a
 Minimum value if the parabola opens
given function
upward
7. Intercepts – for the y intercept it a point
Minimum value = k
where the graph intersects the y-axis and this
Obviously both maximum and minimum value
the point (0, c) where c is the constant in the
always refers to the value of k
quadratic function. The x-intercepts also serve
as the zeroes of the functions and to find them
we let y = 0 and then do the same process is
Illustrative Example:
as finding the roots of quadratic equation.
Determine if the parabola opens up or down then
find the vertex , the axis of symmetry and the
The Parabola minimum/maximum value of the quadratic function
7
 y – intercept is (0, 4)
1: y =2x2 +4x + 5
2. y =-3x2 +4 3. y =2x2 Since c = 0, the
y – intercept is (0, 0)

Solution:
1. y =2x2 +4x + 5 In the standard form Zeroes of the Function
where a = 2, b = 4 and
c=5. To find the zeroes of the function y =ax2 +bx + c
we let y = 0 that becomes ax2 +bx + c = 0
 Parabola opens upward since a = 2, which is and then do the same process as finding the roots of
positive quadratic equation.

Find Vertex (h, k)

Illustrative Example:
Find the zeroes of the following quadratic function.
h= =
1. y = x2 -3x + 2
To find the value of k, we will replace y with k and 2. y = x2 + 4x – 2
substitute h = -1 to the x variables in the quadratic
function Solution:
y = 2x2 +4x + 5 1. y = x2 -3x + 2
k = 2(-1)2 + 4(-1) + 5
k = 2(1) + -4 + 5 Let y = 0
k = 2 + -4 + 5 0 = x2 -3x + 2 or x2 -3x + 2 = 0 where a = 1, b = -3
k = -2 + 5 and c = 2
k=3
 Therefore the vertex is (-1, 3) Product: a ∙ c = 1 ∙ 2 = 2 Sum : b = -3
 The axis of symmetry is x = h which is x = -1
 The minimum value is k = 3 .

2. y =-3x2 +4 2
The 2 numbers
y =-3x2 +4 In the standard form where
are -2 and -1.
a = -3, b = 0 and c = 4
-2 -1 Since -2∙ -1= 2
 Parabola opens downward since a = -3, which is and -2+ -1 = -3
negative.
-3
Find Vertex (h, k)
Divide the 2 numbers by the value of a which is
a = 1.
h= =-
-2 -1
y =-3x2 +4 ,
k = -3(0)2 + 4 1 1 a=1
k = -3(0) + 4 = -2 and -1
k=0+4 = 2 and 1 Lastly, change the signs to
k=4 opposites.
 Therefore the vertex is (0, 4) Therefore, the zeroes of the function
 The axis of symmetry is x = h so that is x = 0 y = x2 -3x + 2 are 2 and 1.
 The maximum value is k = 4

2. y = x2 + 4x – 2
y-intercept of the Quadratic Function
Let y = 0
y-intercept = (0, c) where c is the constant c in the
x2 + 4x - 2 = 0 where a = 1, b = 4 and
y =ax2 +bx + c
c = -2

Since it is non-factorable we will use the quadratic

Illustrative Example: formula.
Find the y- intercept of the quadratic function below:

1. y =2x2 +4x + 5 Since c = 5, the

y – intercept is (0, 5) x=

2. y =-3x2 +4 Since c = 4 , the

8
x=

x=

x=

x=

Zeros of the function Replace y = 0 then find

x= divide each term by 2 the roots of the quadratic
x= equation
Domain All Real Numbers
Therefore the zeroes of y = x2 + 4x – 2 are
Range The range are the values
which are less than or
and
equal to k or y ≤ k
Maximum value k

Characteristics of the a>0

parabola when it is
facing upward
Opens Upward
Vertex (h, k)
Axis of Symmetry x=h To Graph a Quadratic Function

/a/<1 The parabola is wider 1. Determine the vertex

compared to its graph 2. Prepare the table of values. Basically, we only
when a= 1 need 3 points to graph a parabola. (include the
/a/>1 The parabola is narrower vertex, y-intecept and x-intecepts/zeroes)
compared to its graph 3. Connect the points with a smooth curve.
when a= 1 4. Draw a straight line for the axis of symmetry.
y-intercept of the graph (0, c)
Zeros of the function Replace y = 0 then find
the roots of the quadratic Illustrative Example:
equation
Determine the characteristics of the quadratic function
Domain All Real Numbers then graph it.
Range The range are the values
which are greater than or 1. y = 2x2 – 2
equal to k or { y ≥ k}
Minimum value k Solution:
y = 2x2 – 2 In standard form where a = 2 ,
Characteristics of the a<0 b = 0 and c = -2
parabola when it is
facing downward
Opens Downward Opens Upward
Vertex (h, k) Vertex (0, -2)
Axis of Symmetry x=h
Axis of Symmetry x=0
/a/<1 The parabola is wider
/a/ =/2/ = 2 > 1 Since 2 > 1, The parabola
compared to its graph
is narrower compared to
when a= 1
its graph when a= 1
/a/>1 The parabola is narrower
y-intercept of the (0, -2)
compared to its graph
graph
when a= 1
Zeros of the -1 and 1
y-intercept of the graph (0, c)
function

9
 Domain All Real Numbers y = -12
Range The range are the values
x -1 0 1
which are greater than or
y -12 -3 0
equal to -2 or { y ≥ -2}
Minimum value -2
Graph: y = -3x2 +6x - 3

Graph: y = 2x2 – 2
Opens Downward
Vertex (1, 0)
Axis of Symmetry x=1
/a/ = /-3/ = 3 >1 Since 3>1, the parabola
To find k, replace y
is narrower compared to
its graph when the a= 1 to k and -3 to x in
y-intercept of the graph (0, -3) the original equation
VI. Transform the quadratic
Zeroes of the function 1 function defined by y = ax2 + bx + c
into the form y = a(x – h)2 + k.
Domain All Real Numbers At the end of the lesson you are expected to:
 transform the quadratic function defined by
Range The range are the values y = ax2 + bx + c into the form y = a(x – h)2 + k.
which are less than or  analyze the effects of changing the values of a, h and
equal to 0 or y ≤ 0 k in the equation y = a(x – h)2 + k of a quadratic
Maximum value 0 function on its graph.

Definition
 Quadrati
the form
VERTEX
 To do th
k, then s

Illustrative Example:
Rewrite the following into the vertex form
2. y = -3x2 +6x - 3 1. y = = x2 +6x + 5

y = -3x2 +6x – 3 In standard form where a = -3 , Solution:

b = 6 and c = -3
Determine the value of a, b, h and k
a=1, b=6,h=?, k=?
Obviously we need 3 points to graph a parabola so we
have to add another point to form u shaped parabola.
So letting x = -1
h=
y = -3x +6x – 3
2
y = x2 +6x + 5
y = -3(-1)2 +6(-1) – 3
k = (-3)2 +6(-3) + 5
y = -3(1) -6 -3 k = 9 + -18 + 5
y = -3 – 6 – 3 k = -9 + 5
10
 k = -4 y = a(x-h)2 + k
Thus , a = 2, h = 4 and k = 0
We will substitute a = 1, h = -3 and k = -4 to
y = a(x-h)2 + k 2. y = 3x2 – 1

y = 1(x – (-3))2 + -4 Since -(-3) = +3 y = 3(x – 0)2 + -1 following the form

y = (x + 3) 2 – 4 y = a(x-h)2 + k
Thus , a = 3, h = 0 and k = -1
Therefore, y = x +6x + 5 in vertex form is
2

y = (x + 3) 2 – 4 where a = 1, h = -3 and k = -4. 3. y = (x + 3)2 + 3

y = (x –(-3))2 + 3
2. y = 2x2 +12x + 14
Thus , a = 1, h = -3 and k = 3
Solution:
Determine the value of a, b, h and k 4. y = 4(x - 2)2 – 7
a = 2, b = 12, h = ? , k = ? y = 4(x – (+2))2 + -7

Thus , a = 4, h = 2 and k = -7

h=
THE EFFECTS OF CHANGING THE
y = 2x2 +12x + 14 To find k, replace y
k = 2(-3)2 +12(-3) + 14 VALUES OF a in y = a(x-h)2 + k
to k and -3 to x in
k = 2(9) + -36 + 14
the original equation
k = 18 + -36+14
k = -18 + 14
Cases on The effects of changing the values of h
k = -4
and k in y = a(x-h)2 + k:
We will substitute a = 2, h = -3 and k = -4 to
Case 1: To graph y = a(x-h)2 , when h is positive, is
y = a(x-h)2 + k
to slide the graph of y = ax2 TO RIGHT by h units but
if h is negative slide the graph of y = ax2 TO LEFT by
y = 2(x – (-3))2 + -4 h units. In other words the sign of h tells us whether to
y = 2(x + 3) 2 – 4 Since -(-3) = +3 move right or left.
Therefore, y = x2 +6x + 5 in vertex form is
y = 2(x + 3) 2 – 4 where a = 2, h = -3 and k = -4. Illustrative Example:
1. Graph y = 2(x-4)2
Solution: a = 2, h = 4 , k = 0
How about rewriting y = a(x-h)2 + k to the standard
Let usLook
first at the ygraph
graph = ax2above,
which notice that
is y=2x 2
for
form y = ax2 +bx + c?
whenever the /a/ gets smaller, the graph
Illustrative Example: Graphbecomes
of y=2x“wider”,
2
but as it gets bigger the
Rewrite the equation into y = ax2 +bx + c graph becomes “narrower” or “skinnier”

1: y = 3(x-2)2 + 4

Solution:
y = 3(x - 2)2 + 4
y = 3(x2 -4x + 4) + 4 Since (x - 2)2 = x2 -4x + 4
y = 3x2 -12x + 12 + 4 Distribute 3 to x2 -4x + 4
y = 3x2 -12x + 16 Since 12 + 4 = 16

Therefore y = 3(x-2)2 + 4 in the standard form is

y = 3x2 -12x + 16

Illustrative Example:
Identify the value of a, h and k of the following
quadratic functions: Since h = 4 and it is positive we will slide 4 units to the
1. y = 2(x - 4)2 right

11
 Solution: a = 2, h = 0 , k = 4
Let us first graph y = ax2 which is y=2x2 and since
k = 4 and its positive we will move 4 units upward.

2: Graph y = 3(x + 2)2

Solution: a = 3, h = -2 , k = 0 2: Graph y = 3x2 – 2

Solution: a = 3, h = 0 , k = -2
Let us first graph y = ax2 which is y=3x2 and since
h = -2 and its negative we will move 2 units to the Let us first graph y = ax2 which is y=3x2 and since
left. k = -2 and its negative we will move 2 units
downward.

Case 2: To graph y = ax2 + k , when k is positive then

Case 3: To graph y = a(x-h)2 + k , then slide the graph
slide the graph of y = ax2 UPWARD by k units but if k
of y = ax2 RIGHT OR LEFT by h units and UPWARD
is negative slide the graph of y = ax2 DOWNWARD
or DOWNWARD by k units. The sign of h and k
by k units. In other words the sign of k tells us to move
will tell us which direction to move.
upward or downward.
Illustrative Example:
Illustrative Example:
1. Graph y = (x + 3) 2 + 4
1.Graph y = 2x2 + 4
12
 Solution: a = 1, h = -3 , k = 4
Let us first graph y = ax2 which is y=x2 and move to WRITING A QUADRATIC FUNCTION GIVEN
the left by 3 units and upward by 4 units. ITS VERTEX AND A POINT.

Procedure:
To write a quadratic function given its vertex (h,
k ) and another point (x, y)
a. Substitute the values of the vertex (h, k)
and the values of (x, y) in the equation
y = a(x-h)2 + k
b. Solve for a.
c. Simplify the equation.

Illustrative Example:
Find the equation of the parabola whose vertex is (1, 5)
and which passes through the point (2, 7)

Solution:
Given: Vertex = (1, 5)
Another point = (2, 7)
which means that h = 1, k = 5 , x = 2 and y = 7

a. Substitute h = 1, k = 5 , x = 2 and y = 7 to
y = a(x-h)2 + k
7 = a (2 – 1)2 + 5

b. Solve for a.
7 = a (2 – 1)2 + 5
7 = a(1)2 + 5
7 = a (1) + 5
7=a+5
7–5=a
2 = a or a = 2

2: Graph y = 2(x - 3) 2 – 5 c. Simplify the equation by substituting a = 2

and h = 1, k = 5 to
Solution: a = 2, h = 3 , k = -5 y = a(x-h)2 + k
Let us first graph y = ax2 which is y=2x2 and move to y = 2 (x-1)2 + 5
the right by 3 units and downward by 5 units. y = 2 (x2 -2x + 1) + 5
y = 2x2 + 2-2x + 21 + 5
y = 2x2 -4x + 2 + 5

y = 2x2 -4x + 7

2. Find the equation of the parabola whose vertex is

(2, 3) and which passes through the point (0, 0)

Solution:
Given: Vertex = (2, 3)
Another point = (0, 0)
which means that h = 2, k = 3 , x = 0 and y = 0

a. Substitute h = 2, k = 3 , x = 0 and y = 0 to
y = a(x-h)2 + k
0 = a (0 – 2)2 + 3

b. Solve for a.
0 = a (0 – 2)2 + 3
0 = a(– 2)2 + 3
0 = a (4) + 3
VII. FINDING THE EQUATION OF 0 = 4a + 3
QUADRATIC FUNCTIONS 0 – 3 = 4a
At the end of the lesson you are expected to: -3 = 4a
 determines the equation of a quadratic function given:
(a) a table of values; (b) graph; (c) zeros.

13
 -2 = a or a = -2

or c. Simplify the equation by substituting a = -2

and h = 1, k = 3 to
c. Simplify the equation by substituting y = a(x-h)2 + k
y = -2 (x-1)2 + 3
and h = 2, k = 3 to y = -2 (x2 -2x + 1) + 3
y = a(x-h)2 + k y = -2x2 + -2-2x + -21 + 3
y = -2x2 + 4x - 2 + 3

y=
y = -2x2 + 4x + 1 equation of the
parabola
y=

WRITING A QUADRATIC FUNCTION given the

y= TABLE OF VALUES

Procedure:
y= To write a quadratic function given the TABLE
OF VALUES
a. Write 3 equations by substituting the points in
y= the equation y = ax2 + bx + c
b. Simplify the 3 equations
c. Substitute the value determined for c into the
y= other two equations.
d. Solve the resulting system of two equations in
two unknowns.
e. Substitute the values of a, b and c in the
equation y = ax2 + bx + c

3. Find the equation of the quadratic function

determined from the GRAPH below Illustrative Example:
1. Determine the equation of the quadratic
function represented by the table of values
below
x -1 0 3
y 0 3 0

Solution:
a. Determine the 3 equations by substituting the
points in the equation y = ax2 + bx + c:

Using (-1, 0)
0 = a(-1)2 + b(-1) + c:
0=a–b+c --- Equation 1

Using (0, 3)
3 = a(0)2 + b(0) + c:
Solution: 3= c --- Equation 2
We only need the vertex and a given point.
Given: Vertex = (1, 3) Using (3, 0)
Another point = (0, 1) 0 = a(3)2 + b(3) + c:
which means that h = 1, k = 3 , x = 0 and y = 1 0 = 9a + 3b + c --- Equation 3
b. Simplify the 3 equations
Equation 1: a – b + c = 0
a. Substitute h = 1, k = 3 , x = 0 and y = 1 to
Equation 2: c = 3
y = a(x-h)2 + k Equation 3: 9a + 3b + c = 0
1 = a (0 – 1)2 + 3
c. Since c = 3 already we will substitute c = 3 to
b. Solve for a. the Equations 1 and 3.
1 = a (0 – 1)2 + 3 Equation 1: a – b + 3 = 0
1 = a(-1)2 + 3 Equation 3: 9a + 3b + 3 = 0
1 = a (1) + 3
1=a+3 d. Do elimination method in Equation 1 and 3 in
letter c.
1–3=a
14
 Equation 1: a – b + 3 = 0 y = 2x2 - 6x – 8
Equation 3: 9a + 3b + 3 = 0

The quadratic functions y = x2 - 3x – 4 and

To do elimination we must multiply -9 to the
Equation 1 since -9 is the opposite of the y = 2x2 - 6x – 8 were not unique since a can be any
numerical coefficient of 9a. nonzero number, thus the answer is:

Equation 1: -9(a – b + 3) = -9(0) y = a (x2 - 3x – 4) where a is any nonzero

-9a + 9b -27 = 0 constant.
Add -9a + 9b -27 = 0 to equation 3: 9a + 3b + 3 = 0
VIII. APPLICATIONS OF
-9a + 9b -27 = 0 QUADRATIC FUNCTIONS
9a + 3b + 3 = 0
12b – 24 = 0 (notice that -9a + 9a
At the end of the lesson you are expected to:
was eliminated)  solves problems involving quadratic functions

Simplify: 12b -24 = 0 Quadratic functions have many applications.

12b = 24 Analyze the graph below representing the
relationship between the time in seconds when a
projectile is launched vertically into the air, and its
height at that time. Answer the questions that follow.

b=2
Using Equation 1: a – b + 3 = 0 to find a.
Substitute b = 2
a–2+3=0
a+1=0
a = -1

Therefore, a = -1, b = 2 and c = 3. Substitute these

values in the quadratic function y = ax2 + bx + c

y = -1x2 + 2x + 3 or

y = -x2 + 2x + 3

WRITING A QUADRATIC FUNCTION given the

zeroes

Procedure:

a. Substitute roots r1 and r2 to the equation

y = a(x - r1 )(x – r2 ) where a is any nonzero
constant.
b. Simplify y = a(x - r1 )(x – r2 ) by multiplying
(x - r1 )(x – r2 )

Illustrative Example:
Find a quadratic function whose zeroes are -1 and 4

Solution: Given r1 = -1 and r2 = 4

a. Substitute r1 = -1 and r2 = 4 to
y = a(x - r1 )(x – r2 )
y = a (x - -1) (x – 4)
y = a (x + 1) (x – 4)

b. Simplify y = a (x + 1) (x – 4)

y = a (x2 + 1x - 4x – 4)
y = a (x2 - 3x – 4) where a is any nonzero
constant.

Possible quadratic functions:

If a = 1 then y = 1 (x2 - 3x – 4)
y = x2 - 3x – 4
If a = 2 then y = 2 (x2 - 3x – 4)
15
 Problem: What are the dimensions of the largest
rectangular field that can be enclosed by 60m of fencing
wire?

Solution:
Representation:
Let l be the length
Let w be the width
Perimeter P of a rectangle is P = 2l + 2w.

w
l = 30 -w
Questions:
Since P = 60m, then we have
1. What is the maximum height reached by a
2l +2w = 60
projectile?___________
l +w = 30
2. How long does it take for a projectile to reach its
l = 30-w (Expressing the
maximum height?___________
length as a function of w)
3. How long does it take for a projectile to return to
the ground?_____________
Substituting in the formula for the area A of a rectangle
4. What is the axis of symmetry of the graph?
A(w) = lw
________________
A(w) = (30-w)w
5. What is the vertex of the graph?________
A(w) = 30 –w2
Now you will explore situations and real-life problems A(w) = -w2 +30w
that can be modeled and can be solved using the
concepts of quadratic functions. Solve for h (in this case h will be of the same value
with w)
a = -1 and b = 30
Illustrative Example:
A. Number Problem
Problem: Find two real numbers whose sum is 14 and =
whose product is a maximum. h = 15 or w = 15

Solution: Solve for L

Representation: L = 30- w
Let x be the number. L = 30 – 15
14 - x is the other number. L = 15
P(x) is the product
Substituting this to A(w) = -w2 + 30w
Now, If the product of the two numbers is a maximum, = -152 + 30(15)
then its equivalent quadratic function = -225 + 450
Product = x(14 - x) = 225
P(x) = 14x – x2 Therefore, the length and width of the
P(x) = - x2 + 14x (attains its maximum rectangle should be 15m which gives the
value at its vertex (h, k)) maximum area of a square. The maximum area is
225m2.
Solve for h (in this case h will be of the same value
with x) C. Free Falling Bodies
a = -1 and b = 14 Problem: A ball is thrown vertically upward with a
velocity of 40m/sec. The height S(t) of the ball above
the building after t seconds is given by the function
= S(t) = 40t - 5t2 .
h=7
a. How long will it take the object to reach the
Since h = 7, then the first number x is 7 maximum height?
Second number = 14 – x = 14- 7 = 7
Solution:
B. Geometry
By using the abscissa of the vertex of the parabola or
the value of h
16
Solve for h: (in this case h will be of the same value
with t)
a = -5 b = 40

h = 4 or t = 4

Thus, the object is at its maximum height after 4

seconds.

b. What maximum height will the object reach? Mathematics A

Q1
Substitute the value of t = 4 or the abscissa to the given W4 Name: ___________________________ Sectio
function. Subject Teacher: Mrs. Justin P. Galinato
S(t) = 40t - 5t2
= 40(4) – 5 (4)2 A. Multiple Choice. Directions: Choose the letter of the
= 160 - 80 correct answer. Write ONLY the letter of your answer
= 80 on the blank (2 pts each)
Therefore, the maximum height that the object
will reach is 80 m. _____1. Which of the following is a solution set to the
inequality x 2+ 9 x+14 ≥0
c. Find the time at which the object is on the A. {x/x ≤ -7 or x ≥ -2} B. {x/x ≤ -2 or x ≥ -7}
ground. C. {x/x 2 ≤ x ≤ 7} D. {x/x ≤ -7 or x ≥ 2}
_____2. “The floor of a conference hall can be covered
At ground level, the value of S(t) = 0. Thus the completely with tiles. Its length is 36 ft. longer than
quadratic equation is its width. The area of the floor is less than 2,040
square feet.” What mathematical sentence could
S(t) = 40t - 5t2 represent the given situation?
0 = 40t – 5t2 A. 2
w + 36 < 2040 B. w 2 + 36 > 2040
0 = 5t(8 - t) factoring C. w 2 + 36w > 2040 D. w 2 + 36w < 2040
5t = 0 or 8 – t = 0
_____3. Which of the following equations represents a
t=0 t=8
quadratic function?
Since t represents the time in seconds and t must
A. g(x) = 3 + 2x2 B. f(x) = 3x - 22
not be equal to 0.
C. x = 2y2 + 3 D. y = 2x – 3
_____4. Which of the following parabolas has a vertex
Thus, the object will hit the ground 8 seconds (-1, 1)?
after launch. A. f(x) = (x + 1)2 + 1 B. f(x) = (x + 1)2 - 2
C. f(x) = (x + 1)2 + 2 D. f(x) = (x + 1)2 - 1
_____5. Which of the following are the zeros of
y = 2x2 - x – 10?

A. B.-2, 5 C. D.
For numbers 6-8

17
 _____6. What is the vertex of the parabola at the right?
A. (4, 0) B. (-1, 4) C. (4, -1) D. (0, 4)

_____7. What is the Range of the parabola?

A. {y/y ≤ 4} B.{y/y ≥ 4} C. {y/y ≤ -1} D. {y/y ≥ -1}

____8. Find the equation of the parabola.

2 2
A. f ( x )=x + 4 x−8 B . f ( x ) =3 x −12 x
2 2
C . f ( x )=x + 4 x+1 D . f ( x )=−2 x −4 x +2

Parent’s Name & Signature

B. Write the letter of the correct answer then

show your solution. (You may use the back page for
your solution) 5 pts each

_____9. Pete has 120 m of fencing materials. What is

the maximum area of a rectangular garden that he can
fence?
A. 3 600 m 2 B. 500 m 2 C. 600 m 2 D. 900 m 2
Solution:

____10. Homer wishes to fence three sides of a

rectangular yard for his carabao. The fourth side of the
yard will be the side of his house. He has 80 meters of
fencing available. Find the maximum area that can be
fence.
A. 800 m 2 B. 400 m 2 C. 500 m 2 D. 1000 m 2
Solution:

C. Graph the quadratic function y = -x2 +4x -1 on

the Cartesian plane, including its axis of symmetry.
(4 pts.)

18