Graphing%20Quadratic%20Functions - PowerPoint PPT Presentation

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Graphing%20Quadratic%20Functions

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The vertex of a parabola is not always at (0, 0), but the patterns we are going ... Graph each parabola below. ... the line of symmetry and draw the parabola. ... – PowerPoint PPT presentation

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Title: Graphing%20Quadratic%20Functions


1
Graphing Quadratic Functions
  • A Shortcut and A Summary

2
  • 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
3
Before we begin
Things you should know/understand before we begin.
  • What is a parabola?
  • What is the vertex of a parabola?
  • How do you find the vertex of a parabola?
  • In Standard Form y ax2 bx c
  • In Vertex Form y a (x h)2 k
  • c) In Intercept Form y a (x p)(x q)
  • What is the line of symmetry of a parabola?
  • The symbol, ?, means change. Thus ?x, means
    the change in x.

DONE
4
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
1
1
4
1
3
9
1
5
16
1
7
25
1
9
36
1
11
49
1
13
64
1
15
DONE
5
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
1
2 2(1)
8
1
6 2(3)
18
1
10 2(5)
32
1
14 2(7)
50
1
18 2(9)
72
1
22 2(11)
98
1
26 2(13)
128
1
30 2(15)
DONE
6
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
1
3 3(1)
12
1
9 3(3)
27
1
15 3(5)
48
1
21 3(7)
75
1
27 3(9)
108
1
33 3(11)
147
1
39 3(13)
192
1
45 3(15)
DONE
7
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
1
0.5 0.5(1)
2
1
1.5 0.5(3)
4.5
1
2.5 0.5(5)
8
1
3.5 0.5(7)
12.5
1
4.5 0.5(9)
18
1
5.5 0.5(11)
24.5
1
6.5 0.5(13)
32
1
7.5 0.5(15)
DONE
8
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
1
1a
4a
1
3a
9a
1
5a
16a
1
7a
25a
1
9a
36a
1
11a
49a
1
13a
64a
1
15a
DONE
9
So What?
First notice that in every example we just did
the vertex was at (0, 0).
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.
Second, as we moved right on the x-axis, how far
did we go every time?
1
Third, as we moved up the y-axis, we followed a
pattern. Did you see that pattern?
Up a, 3a, 5a, 7a, 9a, etc.
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.
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.
DONE
10
Lets 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
2 y ½(x 5)2 4
ANSWER
HELP
3 y 3(x 2)(x 4)
ANSWER
HELP
11
1 y 2x2 4x 1
STANDARD FORM
IVE GOT IT! TRY ANOTHER ONE.
HOW DO YOU DO THIS PROBLEM?
12
1 y 2x2 4x 1
First, find the line of symmetry
STANDARD FORM
Second, find the vertex by plugging in the line
of symmetry for the x-value.
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.
Click here to go back to problems
DONE
13
2 y ½(x 5)2 4
VERTEX FORM
IVE GOT IT! TRY ANOTHER ONE.
HOW DO YOU DO THIS PROBLEM?
14
First, plot the vertex at (5, 4).
2 y ½(x 5)2 4
VERTEX FORM
Since a ½
Move Right 1, Down ½ (1a)
The line of symmetry of a parabola is a vertical
line through the vertex.
Reflect the points across the line of symmetry
and draw the parabola.
Click here to go back to problems
DONE
15
3 y 3(x 2)(x 4)
INTERCEPT FORM
IVE GOT IT! TRY ANOTHER ONE.
HOW DO YOU DO THIS PROBLEM?
16
3 y 3(x 2)(x 4)
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.
INTERCEPT FORM
The line of symmetry is halfway between these two
values. So, the line of symmetry is x 3.
Plug this value of x into the original equation
to find the y-value of the vertex.
So the vertex is (3, 3)
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.
Click here to go back to problems
DONE
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