Graphing Quadratic Functions

2

Let a, b, and c be real numbers a ≠ 0. The function

f (x) = ax2 + bx + cis called a quadratic function. The graph of a quadratic function is a parabola.Every parabola 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.

x

y

axis

f (x) = ax2 + bx + cvertex

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.

When the leading coefficient is negative, the parabola opens downward

and the vertex is a maximum.

x

y

f(x) = ax2 + bx + ca > 0 opens upward

vertex minimum

xy

f(x) = -ax2 + bx + c

a < 0 opens

downward

vertex maximum

4

5

y

x5-5

The simplest quadratic functions are of the form f (x) = ax2 (a ≠ 0) These are most easily graphed by comparing them with the graph of y = x2.

Example: Compare the graphs of

, and2xy = 2

2

1)( xxf = 22)( xxg =

2

2

1)( xxf =

22)( xxg =

2xy =

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 ofg (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

g (x) = (x – 3)2y = x 2

- 4x

y

4

4

vertex (3, 2)

6

The standard form for the equation of a quadratic function is: f (x) = a(x – h)2 + k (a ≠ 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).

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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) = – ( x – 3)2 + 16 add 16 to both sides

a < 0 → parabola opens downward.

h = 3, k = 16 → axis x = 3, vertex (3, 16).

f (x) – 16 = – ( x - 3)2 factor out

8

x

y

4

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

(7, 0)(–1, 0)

(0, 7)

(3, 16)

x2 - 6x - 7 = 0

Find the y-intercept let x = 0 and solve for y. f(x)=- (0)2 + 6(0) + 7 = 7.

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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

a > 0 → parabola opens upward like y = 2x2.

h = –1, k = –3 → axis x = –1, vertex (–1, –3).

f (x) = 2( x + 1)2 – 3 add -3 both sides

10

x

y

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).

x = –1

f (x) = 2x2 + 4x – 1

(–1, –3)

262+−=x

262−−=x

x- intercepts are:

y- intercept:y = -1

),( 0262+−

),( 0262−−

),( 10 −

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.

321 2 ++= xxxf )(.

262 2 −−= xxxf )(.

5843 2 −+= xxxf )(.

11234 2 ++−= xxxf )(.

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Vertex of a Parabola

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

The vertex of the graph of f (x) = ax2 + bx + c (a ≠ 0)

is ,2 2

b bf

a a

− −

At the vertex, 5210

1210

2==−−=−=

)()(

abx

So, the vertex is (5, -3).

322)5(105)5(2

2 −=+−==

−

fa

bf

13

Vertex of a Parabola

1. Find the vertex of the graph of f (x) = 7x2 – 12x + 5.

The vertex of the graph of f (x) = ax2 + bx + c (a ≠ 0)

is ,2 2

b bf

a a

− −

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

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.

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Zeros of a Quadratic Function

15

x

y

4

4

(7, 0)(–1, 0)

(0, 7)

(3, 16)

Ans. -1 and 7

Zeros of a Quadratic Function

16

Ans. -2 and 4

(1, 9)

(-2, 0) (4, 0)

Determine the zeros of the given quadratic function1.f(x) = x2 – 2x – 24

2. f(x) = x2 + 8x -20

3. f(x) = 2x2 + 11x -21

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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.

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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.

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Deriving Quadratic Functions from Zeros of the functions

Example. If are zeros of a

quadratic function, find the quadratic function.

20

323±

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

23

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

y

x

Example: Find an equation for the parabola with vertex (2, –1) 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)

→−−= 1)2(2

1)( 2xxf 12

2

1)( 2 +−= xxxf

(0, 1)

(2, –1)

Since (0, 1) is a point on the parabola: f (0) = a(0 – 2)2 – 1

1 = 4a –1 and 2

1=a

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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)

2

94 xxf =)(

Since (3, 4) is a point on the parabola: f (3) = a(3 – 0)2 + 0

4 = 9a and

94=a

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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)

132 2 +−= )()( xxf

Since (4, 3) is a point on the parabola: f (4) = a(4 – 3)2 + 1

3 = a + 1 and

2=a

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.

Identifying Quadratic Functions

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.

The function is not quadratic. The second differences are not constant.

x y

–2

–1

0

1

2

–1

0

–2

–9

7

+7

+1

+1

+7

+1

+1

+1

+1

–6

+0

+6

Be sure there is a constant change in x-values before you try to find first or second differences.

Caution!

Your TurnTell 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.

The function is quadratic. The second differences are constant.

x y

–2

–1

0

1

2

0

1

1

4

4

–3

–1

+1

+3

+1

+1

+1

+1

+2

+2

+2

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

Deriving Quadratic Functions Given the Table of Values

32

X 0 1 2 3 4

y -6 -6 -4 0 6

Given table of values, determine the quadratic function.

0 2 4 6

2 2 2

Deriving Quadratic Functions Given the Table of Values

33

X 0 1 2 3 4

y c a+ b + c 4a+2b+c 9a+3b+c 16a+4b+c

f(x) = ax2+ bx + c

a + b 3a + b 5a + b 7a + b

2a 2a 2a

• Equate 2a with the second difference in the table of values to find a:

2a = 2

a = 1

Equate a + b with the first difference in the table of values and use the value of a

a + b = 0

1 + b = 0

b = -1

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 34

To find c, locate the ordered pair where x equals zero.

c= -6

The quadratic function is

f(x) = x2- x – 6

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 35

Deriving Quadratic Functions Given the Table of Values

36

x 0 1 2 3 4y 40 28 18 10 4

Deriving Quadratic Functions Given the Table of Values

37

x -2 -1 0 1 2y -12 -9 0 15 36

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).

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).

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).

Deriving Quadratic Function

41

Ans. -2 and 4

(1, 9)

(-2, 0) (4, 0)

Applications on Quadratic Function

42

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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: 21

2 6.9

y x x= − + +

The path is a parabola opening downward. The maximum height occurs at the vertex.

2 ,9

162

9

1 2 =−=→++−= baxxy

.92

=−=a

bxAt the vertex,

( ) 1592

==

−

fa

bf

So, the vertex is (9, 15). The maximum height of the ball is 15 feet.

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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

corralx x

120 – 2xLet 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 ,

2a

bx

−=

a = –2 and b = 120 .304

120

2=

−−=−=→

a

bx

120 – 2x = 120 – 2(30) = 60The maximum area occurs when the width is 30 feet and the length is 60 feet.

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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

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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