2. 2
Let a, b, and c be real numbers a ¹ 0. The function
f (x) = ax2 + bx + c
is called a quadratic function.
The graph of a quadratic function is a
Epavrearyb oplaar.abola is symmetrical about a line called the axis
(of symmetry).
The intersection point of the
parabola and the axis is
called the vertex of the
parabola.
f (x) = ax2 + bx + c
x
y
vertex
axis
3. 3
The leading coefficient of ax2 + bx + c is a.
When the leading coefficient
is positive, the parabola
opens upward and the
vertex is a minimum.
vertex
maximum
When the leading
coefficient is negative,
the parabola opens downward
and the vertex is a maximum.
x
y
f(x) = ax2 + bx + c
a > 0
opens
upward
vertex
minimum
x
y
f(x) = -ax2 + bx + c
a < 0
opens
downward
4. 4
The simplest quadratic functions are of the form
f (x) = ax2 (a ¹ 0)
These are most easily graphed by comparing them with the
graph of y = x2.
Example: Compare the graphs of
f (x) = 1 x g(x) = 2x2
, and y = x2 2
2
5
y
y = x2
g(x) = 2x2
x
f (x) = 1 x
2
2
-5 5
5. 5
Example: Graph f (x) = (x – 3)2 + 2 and find the vertex and axis.
f (x) = (x – 3)2 + 2 is the same shape as the graph of
g (x) = (x – 3)2 shifted upwards two units.
g (x) = (x – 3)2 is the same shape as y = x2 shifted to the right
three units.
f (x) = (x – 3)2 + 2
y = x g (x) = (x – 3)2 2
- 4
(3, 2)
x
y
4
4
vertex
6. 6
The standard form for the equation of a
quadratic function is:
f (x) = a(x – h)2 + k (a ¹ 0)
The graph is a parabola opening upward if
a > 0 and opening downward if a < 0. The
axis is x = h, and the vertex is (h, k).
7. 7
Example: Graph and find the vertex, line
of symmetry and x- and y-intercepts of
f (x) = –x2 + 6x + 7.
f (x) = – x2 + 6x + 7 original equation
f (x) - 7 = – ( x2 – 6x) isolate 7 and factor out –1
f (x) -7 - 9= – ( x2 – 6x + 9) complete the square add -9 to both sides
f (x) – 16 = – ( x - 3)2 factor out
f (x) = – ( x – 3)2 + 16 add 16 to both sides
a < 0 ® parabola opens downward.
h = 3, k = 16 ® axis x = 3, vertex (3, 16).
8. 8
x
y
4
(0, 7)
(3, 16)
(–1, 0) (7, 0)
4
Example: Graph and find the vertex, line of symmetry
and x- and y-intercepts of f (x) = –x2 + 6x + 7.
a < 0 ® parabola opens downward.
h = 3, k = 16 ® axis x = 3, vertex (3, 16).
Find the x-intercepts by letting y = 0
and solving for x
–x2 + 6x + 7 = 0.
(x - 7 )( x + 1) = 0 factor
x = 7, x = –1
x-intercepts are 7 and -1
x = 3
f(x) = –x2 + 6x + 7
x2 - 6x - 7 = 0
Find the y-intercept let x = 0
and solve for y.
f(x)=- (0)2 + 6(0) + 7 = 7.
9. 9
Example: Graph the parabola f (x) = 2x2 + 4x – 1 and find
the axis and vertex and the x-intercepts.
f (x) = 2x2 + 4x – 1 original equation
f (x) + 1 = 2( x2 + 2x) add 1 both sides and factor out
2
f (x) + 1 + 2 = 2( x2 + 2x + 1) complete the square add 2 both sides
f (x) + 3 = 2( x + 1)2 factor out
f (x) = 2( x + 1)2 – 3 add -3 both sides
a > 0 ® parabola opens upward like y = 2x2.
h = –1, k = –3 ® axis x = –1, vertex (–1, –3).
10. x
10
Example: Graph the parabola f (x) = 2x2 + 4x – 1 and find
the axis and vertex and the x- and y-intercepts.
a > 0 ® parabola opens upward like y = 2x2.
h = –1, k = –3 ® axis x = –1, vertex (–1, –3).
y
-2+ 6
x = –1
f (x) = 2x2 + 4x – 1
(–1, –3)
x- intercepts are:
x=-2+ 6
2
x=-2- 6
2
y- intercept:
y = -1
( ,0)
2
-2- 6
( ,0)
2
(0,-1)
11. 11
Example: Write each function in the form f (x) = a(x-h)2 +k
and find direction of the opening of the graph, the axis of
symmetry and vertex , the x- and y-intercepts.
1.f(x) = x2 + 2x + 3
2.f(x) = x2 -6x - 2
3.f(x) = 4x2 + 8x - 5
4.f(x) = -3x2 +12x +1
12. 12
Vertex of a Parabola
The vertex of the graph of f (x) = ax2 + bx + c (a ¹ 0)
b f b
a a
æ- æ- öö ç çè ÷ø÷ è ø
is ,
2 2
Example: Find the vertex of the graph of f (x) = x2 – 10x + 22.
f (x) = x2 – 10x + 22 original equation
a = 1, b = –10, c = 22
10
10
= - = - ( - )
= =
At the vertex, 5
2
2 1
2
( )
a b
x
f b
æ - f
a
ö çè
2 - = + - = = ÷ø
(5) 5 10(5) 22 3
2
So, the vertex is (5, -3).
13. 13
Vertex of a Parabola
The vertex of the graph of f (x) = ax2 + bx + c (a ¹ 0)
b f b
a a
æ- æ- öö ç çè ÷ø÷ è ø
is ,
2 2
1. Find the vertex of the graph of f (x) = 7x2 – 12x + 5.
2. Find the vertex of the graph of f (x) = 2 x2 + x -10.
3. Find the vertex of the graph of f (x) = -2x2 + 11 x -14.
4. Find the vertex of the graph of f (x) = -3x2 + 10 x +.13
14. Zeros of a Quadratic Function
• The roots of a quadratic equation are called
the zeros of the function.
• A root or solution of an equation is a value
of the variable that satisfies the equation.
• The zeros of a function are points on the
graph of the function where it intersects the
x-axis. These are the x-intercepts.
14
15. Zeros of a Quadratic Function
15
x
y
4
(–1, 0) (7, 0)
4
(0, 7)
(3, 16)
Ans. -1 and 7
16. Zeros of a Quadratic Function
16
Ans. -2 and 4
(1, 9)
(-2, 0) (4, 0)
17. Determine the zeros of the given
quadratic function
1.f(x) = x2 – 2x – 24
2. f(x) = x2 + 8x -20
3. f(x) = 2x2 + 11x -21
17
18. Deriving Quadratic Functions from Zeros of the
functions
Example. If -2 and 3 are zeros of a quadratic
function, find the quadratic function.
x = -2 or x = 3
It follows that x + 2 = 0 or x – 3 = 0, then
y = (x + 2)(x – 3)
y = x2 – x – 6
thus, the ans. is y = a(x2 – x – 6) where a is any
nonzero constant.
18
19. Deriving Quadratic Functions from Zeros of the
functions
Example. If - 1 and 4 are zeros of a quadratic
function, find the quadratic function.
x = -1 or x = 4
It follows that x + 1 = 0 or x – 4 = 0, then
y = (x + 1)(x – 4)
y = x2 – 3x – 4
thus, the ans. is y = a(x2 – 3x – 4) where a is any
nonzero constant.
19
20. Deriving Quadratic Functions from Zeros of the
functions
3± 2
Example. If are zeros of a
quadratic function, find the quadratic
function.
20
3
21. Deriving Quadratic Functions
from Zeros of the functions
Example. If the x-intercepts of a quadratic
function are and -5, find the quadratic
function.
21
3
2
22. Deriving Quadratic Functions
from a vertex and a point of the
functions
22
Example: Find an equation for the
parabola with vertex (2, –1)
passing through the point (0, 1).
23. 23
Example: Find an equation for the parabola with vertex (2, –1)
y
x
passing through the point (0, 1).
f (x) = a(x – h)2 + k standard form
f (x) = a(x – 2)2 + (–1) vertex (2, –1) = (h, k)
y = f(x)
f (x) 1 x 2 2 1
= ( - 2) -1®
2
f (x) = 1 x2 - x +
2
(0, 1)
(2, –1)
Since (0, 1) is a point on the parabola: f (0) = a(0 – 2)2 – 1
1 = 4a –1 and
a = 1
2
24. 24
Example: Find an equation for the parabola with vertex (0, 0)
passing through the point (3, 4).
f (x) = a(x – h)2 + k standard form
f (x) = a(x –0)2 + (0) vertex (0, 0) = (h, k)
Since (3, 4) is a point on the parabola:
f (3) = a(3 – 0)2 + 0
4 = 9a and
2
a = 4
f(x) = 4 x
9
9
25. 25
Example: Find an equation for the parabola with vertex (3, 1)
passing through the point (4, 3).
f (x) = a(x – h)2 + k standard form
f (x) = a(x –3)2 + (1) vertex (3, 1) = (h, k)
Since (4, 3) is a point on the parabola:
f (4) = a(4 – 3)2 + 1
3 = a + 1 and
a = 2
f(x) = 2(x - 3)2 +1
27. Identifying Quadratic Functions
Notice that the quadratic function y = x2 does not
have constant first differences. It has constant second
differences. This is true for all quadratic functions.
28. Identifying Quadratic Functions
Tell whether the function is quadratic. Explain.
Since you are given a table of
ordered pairs with a constant
change in x-values, see if the
second differences are
constant.
Find the first differences,
then find the second
differences.
–9
–2
The function is not quadratic. The second differences
are not constant.
x y
–2
–1
0
1
2
–1
0
7
+7
+1
+1
+7
+1
+1
+1
+1
–6
+0
+6
29. Caution!
Be sure there is a constant change in x-values
before you try to find first or second differences.
30. Your Turn
Tell whether the function is quadratic. Explain.
{(–2, 4), (–1, 1), (0, 0), (1, 1), (2, 4)}
List the ordered pairs in a
table of values. Since there
is a constant change in the
x-values, see if the
differences are constant.
Find the first differences,
then find the second
differences.
4
1
The function is quadratic. The second differences are
constant.
x y
–2
–1
0
1
2
0
1
4
–3
–1
+1
+3
+1
+1
+1
+1
+2
+2
+2
31. Deriving Quadratic Functions
Given the Table of Values
31
X 0 1 2 3 4
y 0 1 4 9 16
The quadratic function is f(x) = x2
32. Deriving Quadratic Functions
Given the Table of Values
32
Given table of values, determine the quadratic
function.
X 0 1 2 3 4
y -6 -6 -4 0 6
0 2 4 6
2 2 2
33. Deriving Quadratic Functions
Given the Table of Values
33
f(x) = ax2+ bx + c
X 0 1 2 3 4
y c a+ b + c 4a+2b+c 9a+3b+c 16a+4b+c
a + b 3a + b 5a + b 7a + b
2a 2a 2a
38. Deriving Quadratic Functions
Given the Three Points on the
Graph
38
Given the three points on its graph, find the
equation of a parabola with axis parallel to the y-axis
and passing through
(1, 9), (-2, 9) and (-1, 1).
39. Deriving Quadratic Functions
Given the Three Points on the
Graph
39
Given the three points on its graph, find the
equation of a parabola with axis parallel to the y-axis
and passing through
(-2,-3), (3, 12) and (0, -3).
40. Deriving Quadratic Functions
Given the Three Points on the
Graph
40
Given the three points on its graph, find the
equation of a parabola with axis parallel to the y-axis
and passing through
(2,-1), (0, 1) and (-2, 11).
43. 43
Example: A basketball is thrown from the free throw line from
a height of six feet. What is the maximum height of the ball if
the path of the ball is: y = - 1 x 2 + 2 x +
6.
9
The path is a parabola opening downward.
The maximum height occurs at the vertex.
y = -1 x2 + 2 x + 6 ®a = - 1
b =
, 2
9
9
9.
At the vertex, x b
= - =
2
a
(9) 15
æ - f
a
ö 2
çè
= = ÷ø
f b
So, the vertex is (9, 15).
The maximum height of the ball is 15 feet.
44. 44
Example: A fence is to be built to form a
rectangular corral along the side of a barn
65 feet long. If 120 feet of fencing are
available, what are the dimensions of the
corral of maximum area?
barn
x corral x
120 – 2x
Let x represent the width of the corral and 120 – 2x the length.
Area = A(x) = (120 – 2x) x = –2x2 + 120 x
The graph is a parabola and opens downward.
The maximum occurs at the vertex where x = -b
,
2a
120
x b
® = - = -
a = –2 and b = 120 30.
4
2
=
-
a
120 – 2x = 120 – 2(30) = 60
The maximum area occurs when the width is 30 feet and the
length is 60 feet.
45. 45
Example: A garments store sells about 40 t-shirts per
week at a price of Php 100 each. For each Php 10
decrease in price, the sales lady found out that 5 more
t-shirts per week were sold.
How much is the sales amount if the t-shirts are priced at Php 100
each?
How much would be their sales if they sell each t-shirt at Php 90?
How much would be their sales if they sell each t-shirt at Php 80?
How much would be their sales if they sell each t-shirt at Php 70?
How much should she sell the t-shirt? Justify
46. 46
Example: The ticket of film showing costs Php 20. At
this price, the organizer found out that all the 300
seats are filled. The organizer estimates that if the
price is increased, the number of viewers will fall by
50 for every Php 5 increase.
How much is the sales amount if the tickets are priced at Php 20 each?
How much would be their sales if they sell each ticket at Php 25?
How much would be their sales if they sell each ticket at Php 30?
How much would be their sales if they sell each ticket at Php 35?
How much should she sell the ticket? Justify
