The document discusses quadratic functions and their graphs. It explores expressing quadratic functions in different forms, and transforming quadratic functions and their graphs through translations, stretches, and other changes. Key aspects covered include the vertex and line of symmetry of parabolas, and how changing the coefficients affects the graph shape and position. Examples of specific quadratic functions are analyzed and graphed.

1. To explore the properties of quadratic functions
and their graphs.
To investigate the different forms in which
quadratic functions can be expressed.
To explore the transformations of quadratic
functions and their graphs.
http://www.youtube.com/watch?v=VSUKNxVXE4E&feature=player_embedded#
http://evmaths.jimdo.com/year­11/functions/?logout=1

2. If a, b and c are real numbers and a≠0, then the
function f (x) = a x2
+b x +c is a quadratic
function.
the expression a x2
+b x +c is a polynomial of
degree 2
a, b and c are coefficients

3. The graph of a quadratic function is a parabola.
What is the graph of a quadratic function?
Use your GDC to plot the graph of
f (x) = 2 x2
­ 8 x +1

4. Use your GDC to plot the graph of
f (x) = 2 x2
­ 8 x +1
1) select menu Graph
2) imput the formula
use this button
for variable ''x''

5. Use your GDC to plot the graph of
f (x) = 2 x2
­ 8 x +1
3) press EXE to enter the
formula
4) Press F6 to DRAW
5) Adjust the window to
view the whole graph.
(F3)

6. vertex
A parabola has a
single turning point
that is called its
vertex and a line of
symmetry that passes
through the vertex
line of symmetry

7. You can use your
calculator to find the
vertex of a parabola
For this parabola the
vertex is its minimum
value
Vertex is the point
(2, ­7)

8. You are going to investigate the effect that
the coefficients have on the graph of a
parabola.
Use your calculator to sketch the graph of the
following functions:
y = x2
y = x2
+ 3 y = x2
­ 2

9. Now you are going to work with dynamic
graphs by using sliders in GeoGebra.
Verify your conclusion using a slider k and
y = x2
+ k.
Vertical translation of parabola.ggb

10. Sketch the graphs of and
y=x2
y=x2
­2
y=x2
+3
vertex:
line of symmetry:
vertex:
line of symmetry:

11. Conclusions:
parabola moves upwards
parabola moves downwards
vertex:
line of symmetry:
( 0 , k )
x = 0
Translation k units along
y­axis
k > 0
k < 0

12. Use GeoGebra to plot the following graphs:
Verify your conclusion using a slider h and
y = (x ­ h) 2
.
y = (x ­ 2 )2
y = (x +3 )2
y = x 2

13. Consider the graphs of
vertex:
line of symmetry:
vertex:
line of symmetry:
y = ( x + 3)2
y = ( x ­ 2 )2
y = x2
and
What can you say about vertex and symmetry line?
Horizontal translation of parabola.ggb

14. Conclusions:
parabola moves to the right
parabola moves to the left
vertex:
line of symmetry:
vertex:
line of symmetry:
( h , 0 )
x = h
( ­ h , 0 )
x = ­ h
Translation along x­axis
( h > 0 )

15. Where would you expect the vertex of
y = (x ­ 4) 2
+ 5 to be?
Describe the shape and the position of the
graph of y = (x ­ h)2
+ k.
Translations in parabola.ggb

16. Conclusions:
vertex
(h , k)
(­h , k)

17. Consider , and
y = x2 y = 2 x2
vertex:
line of symmetry:

18. What do you expect if ?
y = x2
y = ­ x2
vertex:
line of symmetry:
Vertical stretch of parabola.ggb

19. Conclusions:
y = a x 2
The graph of is a parabola with
vertex: (0,0) line of symmetry : x = 0
a > 0 a < 0
|a | > 1 as "a" increases the parabola gets "thinner"
0 < |a| <1 the parabola looks "fatter"
" a " produces a stretch along the y­axis

20. y = ( x ­ 1 ) 2
+ 3
vertex:
line of symmetry:
(1,3)
x=1

21. y = 2 ( x ­ 3 ) 2
vertex:
line of symmetry:
(3,0)
x=3

22. y = ­ 3 x2
+ 4
vertex:
line of symmetry:
(0,4)
x=0

23. y = 3 ( x + 1 ) 2
­ 2
vertex:
line of symmetry:
(-1,-2)
x=-1
http://members.shaw.ca/ron.blond/QFA.CSF.APPLET/index.html
Transformaciones Función Cuadrática.ggb

24. For
Parabolas of the form
What is the y­intercept ?
Find the roots of f.
Concavity?
factorising (if possible)
by formula
(y = 0)
=0

25. y­intercept = ­8
roots : ­ 4 and 2
line of symmetry?
vertex?

26. Line of symmetry is in the middle between the roots :
The vertex will be on the line of symmetry:
We can find the line of symmetry by doing :
y - intercept: ( 0 , c )
a < 0a >0
Cambios cuadratica.ggb

27. For find:
y- intercept:
line of symmetry:
vertex:
roots:
Now draw a sketch of the function.

28.
y- intercept: line of symmetry:
vertex: roots:
Now draw a sketch of the function.
Express f(x) in the form

30. y = a (x ­ x1) ( x ­ x2)Parabolas of the form :
y = ( x ­ 3 ) ( x + 1 )
Roots:
Line of symmetry:
Vertex:
In general:
x1 and x2

31. axis of symmetry
vertex
root
root
y­ intercept
(0 , c )

32. http://www.univie.ac.at/future.media/moe/tests/fun1/erkennen.html
Solve worksheet :
Quadratic functions
At the end of the lesson:QUADRATIC FUNCTIONS I 2010.doc

33. y = (x­2)2
y = ­ x 2
+ 1
y = x2
­ 2
y = x2
+ 3
y = (x ­ 3 )2
+5
y= ­ 2 x 2
+ 1

34. Slide to print:

36. Attachments
Parabola canonica.ggb
Cambios cuadratica.ggb
QUADRATIC FUNCTIONS I 2010.doc
Transformaciones Función Cuadrática.ggb
Vertical translation of parabola.ggb
Vertical stretch of parabola.ggb
Horizontal translation of parabola.ggb
Translations in parabola.ggb