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Graphing Quadratic Functions A “Shortcut” and A “Summary”
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Graphing Quadratic Functions A “Shortcut” and A “Summary”

Dec 13, 2015

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Frank Gilmore
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Page 1: Graphing Quadratic Functions A “Shortcut” and A “Summary”

Graphing Quadratic Functions

A “Shortcut” and A “Summary”

Page 2: Graphing Quadratic Functions A “Shortcut” and A “Summary”

All the slides in this presentation are timed.

Trying to advance the slides before you are asked to do so will result in skipping part of the presentation on that slide.

When each slide is finished a box will appear to let you know there is nothing left on that slide.

DONE

Page 3: Graphing Quadratic Functions A “Shortcut” and A “Summary”

Before we begin …

Things you should know/understand before we begin.

1. What is a parabola?

2. What is the vertex of a parabola?

3. How do you find the vertex of a parabola?

a) In Standard Form … y = ax2 + bx + c

b) In Vertex Form … y = a (x – h)2 + k

c) In Intercept Form … y = a (x – p)(x – q)

4. What is the line of symmetry of a parabola?

5. The symbol, , means “change”. Thus x, means “the change in x”.

DONE

Page 4: Graphing Quadratic Functions A “Shortcut” and A “Summary”

A Little Exploration

Fill out the following table. Click the screen when you are done.

x x y = x2 y

0

1

2

3

4

5

6

7

8

0

1

4

9

16

25

36

49

64

1 1

1 3

1 5

1

7

1

9

1

11

1

13

1 15

DONE

Page 5: Graphing Quadratic Functions A “Shortcut” and A “Summary”

A Little Exploration

Fill out the following table. Click the screen when you are done.

x x y = 2x2 y

0

1

2

3

4

5

6

7

8

0

2

8

18

32

50

72

98

128

1 2 = 2(1)

1 6 = 2(3)

1 10 = 2(5)

1

14 = 2(7)

1

18 = 2(9)

1

22 = 2(11)

1

26 = 2(13)

1 30 = 2(15)

DONE

Page 6: Graphing Quadratic Functions A “Shortcut” and A “Summary”

A Little Exploration

Fill out the following table. Click the screen when you are done.

x x y = 3x2 y

0

1

2

3

4

5

6

7

8

0

3

12

27

48

75

108

147

192

1 3 = 3(1)

1 9 = 3(3)

1 15 = 3(5)

1

21 = 3(7)

1

27 = 3(9)

1

33 = 3(11)

1

39 = 3(13)

1 45 = 3(15)

DONE

Page 7: Graphing Quadratic Functions A “Shortcut” and A “Summary”

A Little Exploration

Fill out the following table. Click the screen when you are done.

x x y = 0.5 x2 y

0

1

2

3

4

5

6

7

8

0

0.5

2

4.5

8

12.5

18

24.5

32

1 0.5 = 0.5(1)

1 1.5 = 0.5(3)

1 2.5 = 0.5(5)

1

3.5 = 0.5(7)

1

4.5 = 0.5(9)

1

5.5 = 0.5(11)

1

6.5 = 0.5(13)

1 7.5 = 0.5(15)

DONE

Page 8: Graphing Quadratic Functions A “Shortcut” and A “Summary”

A Little Exploration

Fill out the following table. Click the screen when you are done.

x x y = ax2 y

0

1

2

3

4

5

6

7

8

0

a

4a

9a

16a

25a

36a

49a

64a

1 1a

1 3a

1 5a

1

7a

1

9a

1

11a

1

13a

1 15a

DONE

Page 9: Graphing Quadratic Functions A “Shortcut” and A “Summary”

So What?First notice that in every example we just did the vertex was at (0, 0).

Second, as we moved right on the x-axis, how far did we go every time?

Third, as we moved up the y-axis, we followed a pattern. Did you see that pattern?

To put it together, if we want to graph a parabola by hand …

FIRST: Find the vertex

SECOND: To find other points on the parabola without making a table of values, start at the vertex and move right 1, up a … then move right 1, up 3a … then move right 1 , up 5a … etc.

THIRD: To finish graphing the parabola, reflect the points you graphed in the last step over the line of symmetry.

Up a, 3a, 5a, 7a, 9a, etc.

1

The vertex of a parabola is not always at (0, 0), but the patterns we are going to employ will still work IF you start at the vertex.

Notice that in every example so far, the a value was positive. That is why we moved UP. If a is negative, then the pattern of odd multiples of a still holds, but you would move DOWN.

DONE

Page 10: Graphing Quadratic Functions A “Shortcut” and A “Summary”

Let’s Try a Few

Graph each parabola below. Click on either the answer to see just the graph or the “HELP” button to see a step by step explanation.

#1: y = 2x2 – 4x + 1

ANSWER HELP

ANSWER HELP

ANSWER HELP

#2: y = –½(x – 5)2 + 4

#3: y = –3(x + 2)(x + 4)

Page 11: Graphing Quadratic Functions A “Shortcut” and A “Summary”

x

y

#1: y = 2x2 – 4x + 1 STANDARD FORM

I’VE GOT IT! TRY ANOTHER ONE.

HOW DO YOU DO THIS PROBLEM?

Click one

Page 12: Graphing Quadratic Functions A “Shortcut” and A “Summary”

x

y

#1: y = 2x2 – 4x + 1 First, find the line of symmetry:

41

2 2(2)

bx

a

- += = =

Second, find the vertex by plugging in the line of symmetry for the x-value.

( ) ( )22 1 4 1 1 1y = - + =-

So, the vertex is (1, –1).

Since a = 2 …

Move Right 1, Up 2 … (1a)

Move Right 1, Up 6 … (3a)

Reflect the points across the line of symmetry and draw the parabola.

STANDARD FORM

DONEClick here to go back to problems

Page 13: Graphing Quadratic Functions A “Shortcut” and A “Summary”

x

y

#2: y = –½(x – 5)2 + 4 VERTEX FORM

I’VE GOT IT! TRY ANOTHER ONE.

HOW DO YOU DO THIS PROBLEM?

Click one

Page 14: Graphing Quadratic Functions A “Shortcut” and A “Summary”

x

y

First, plot the vertex at (5, 4).

Since a = – ½ …

Move Right 1, Down ½ … (1a)

32Move Right 1, Down … (3a)

Reflect the points across the line of symmetry and draw the parabola.

#2: y = –½(x – 5)2 + 4

Move Right 1, Down … (5a)52

The line of symmetry of a parabola is a vertical line through the vertex.

VERTEX FORM

DONEClick here to go back to problems

Page 15: Graphing Quadratic Functions A “Shortcut” and A “Summary”

x

y

INTERCEPT FORM#3: y = –3(x + 2)(x + 4)

I’VE GOT IT! TRY ANOTHER ONE.

HOW DO YOU DO THIS PROBLEM?

Click one

Page 16: Graphing Quadratic Functions A “Shortcut” and A “Summary”

x

y

First, the x-intercepts are the values of x such that each factor equals 0. So, the x-intercepts are x = –2 and x = –4.

Since a = –3…

Move Right 1, Down 3 … (1a)

Move Right 1, Down 9 … (3a)

Reflect the points across the line of symmetry and draw the parabola.

The line of symmetry is halfway between these two values. So, the line of symmetry is x = –3.

INTERCEPT FORM#3: y = –3(x + 2)(x + 4)

Plug this value of x into the original equation to find the y-value of the vertex.

( )( )3 3 2 3 4 3y =- - + - + =

So the vertex is (–3, 3)

DONEClick here to go back to problems