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

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

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

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



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



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



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



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

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

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)

x

y

#1: y = 2x2 – 4x + 1 STANDARD FORM

I’VE GOT IT! TRY ANOTHER ONE.

HOW DO YOU DO THIS PROBLEM?

Click one

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

x

y

#2: y = –½(x – 5)2 + 4 VERTEX FORM



Click one

x

y

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

Since a = – ½ …

Move Right 1, Down ½ … (1a)

32Move Right 1, Down … (3a)


#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


x

y

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



Click one

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)


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)


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

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