Quadratic   Functions

1­E Precalculus

Galileo  Galilei  and  the  Leaning  Tower  of  PisaGalileo  Galilei  and  the  Leaning  Tower  of  Pisa

1­1 Precalculus

Galileo  Galilei  (1564­1642)

According the law of falling bodies, all objects are falling in vacuo with the same speed – whether a feather or an iron ball, the speed is the same.

The formula

relates the drop height d, g to the time t, the time taken for an object tofall through distance d. Here g is the gravitational acceleration.

1­2

d t =12g t 2

Galileo  Galilei  and  the  Leaning  Tower  of  PisaGalileo  Galilei  and  the  Leaning  Tower  of  Pisa

Precalculus

The famous italien scientist Galileo Galilei was the first to formulatethe laws of falling bodies. In 1590 he performed drop tests in Pisa.His student and biographer Vincenzo Viviani told, that he droppedbodies from the leaning tower to ground, but it is not really known whether he actually used the tower for his experiments.

Examples  of  quadratic  functionsExamples  of  quadratic  functions

The area of a circle with radius r is a quadraticfunction of the radius

The kinetic energy E of a body with mass m is aquadratic function of the velocity v

r

1­3

E kin v =12m v2

S r = r2

Precalculus

Fig. 1­1:   Body  thrown  upward

A body is thrown at time t = 0  with an angle α above the horizontal with initial velocity . The path can be parametrised as follows

1­4

x = v0 ⋅ cos t

y = h0 v0 ⋅ sin t −12g t 2

Examples  of  quadratic  functionsExamples  of  quadratic  functions

v0

Precalculus

1­5

Fig. 1­2:   The  curve  describing  the  flight  of  baron  Münchhausen  can  be  approximated  by  a                  quadratic  function.

Examples  of  quadratic  functionsExamples  of  quadratic  functions

Precalculus

Quadratic   functionsQuadratic   functions

Definition:

Functions satisfying the equation

are called quadratic functions.

– quadratic term

– linear term

– constant term

2­1

f x = a x2 b x c

x , a , b , c ∈ ℝ , a ≠ 0

a x2

b x

c

Precalculus

Quadratc   Functions:   Quadratc   Functions:   the  simplest   parabolathe  simplest   parabola

Fig. 2­1:   simplest  parabola  y = x²

y = x² : parabola opens upward

The vertex S (0, 0) is the lowest point (minimum)

2­2

y = x2 : domain D = ℝ , range R = [ 0, ∞ )

Precalculus

Simplest  Parabola:  Simplest  Parabola:  SymmetriesSymmetries

2­3

Fig. 2­2:  the  parabola  y = x²  is  symmetric  with  respect  to  the  y­axis

f x = x2 : f −x1 = f x1 , f −x2 = f x2

Precalculus

2­4

f x = x2 , x1 = −2, x2 = −1, x1 x2 , f x1 f x2

Fig. 2­3:  Parabola  y = x²  monotonically  decreasing  in  the  negative  domain

Simplest  Parabola:  Simplest  Parabola:  MonotonicityMonotonicity

Precalculus

2­5

Fig. 2­4:  Parabola  y = x²  monotonically  increasing  in  the  positive  domain

f x = x2 , x1 = 1, x2 = 2, x1 x2 , f x1 f x2

Simplest  Parabola:  Simplest  Parabola:  MonotonicityMonotonicity

Precalculus

2­6

A function y = f (x) is monotonically increasing in an interval I of the domain, if for all

Definition:

Definition:

MonotonicityMonotonicity

x1 , x2 ∈ I ⊂ D , x1 x2

f x1 f x2

x1 , x2 ∈ I ⊂ D , x1 x2

f x1 f x2

The function f (x) = x²  is monotonically decreasing for x ≤ 0 andmonotonically increasing for x ≥ 0.

A function y = f (x) is monotonically decreasing in an interval I of the domain, if for all

Precalculus

Quadratic   functions  y = a xQuadratic   functions  y = a x²²:   :   Exercise  1, 2Exercise  1, 2

Exercise 1:

Plot the quadratic functions

Explain their properties by means of the graph.

Exercise 2:

Determine by means of the graph of the functiony = x²   the two values

3­A

f x = x2 , g x = 3 x2 , h x = 0.2 x2

p x = − x2 , q x = −2 x2 , r x = −0.2 x2

2 and 3 .

Precalculus

Simplest  parabolaSimplest  parabola

3­1a

Fig. L1­1:  Parabola   f (x) = x²  on  some  background.  In  the  following  figures  the  background                    is  changing  corresponding  to  the  parameter  a  in  the equation  y = a x² 

http://www.fotocommunity.de/search?q=blatt&index=fotos&options=YToyOntzOjU6InN0YXJ0IjtpOjA7czo3OiJkaXNwbGF5IjtzOjg6IjIyNjA2ODY5Ijt9/pos/116

Precalculus

Abb. L1­2:  Parabolas  f (x) = x²,   g (x) = 3 x².  The  background  is  dilated   corresponding                   to  parameter  a  of  equation  g (x) = a x²

a = 1:           f (x) = x² : simplest parabola

a = 3:           g (x) = 3 x² : dilation by scale factor 33­1b

Dilation  of  simplest  parabola:  Dilation  of  simplest  parabola:  Solution  1Solution  1

Precalculus

a = 1:           f (x) = x² : simplest parabola

a = 0.2:        h (x) = 0.2 x² : dilation by scale factor 0.2

Fig. L1­3:  Parabolas  f (x) = x²,   h (x) = 0.2 x².  The  background  is  dilated   corresponding                   to  parameter  a  of  equation  h (x) = a x²

3­1c

Dilation  of  simplest  parabola:  Dilation  of  simplest  parabola:  Solution  1Solution  1

Precalculus

Fig. L1­4:  Parabolas  f (x) = x²,   p (x) = ­ x²

a = 1:           f (x) = x² : simplest parabola

a = ­1:          p (x) = ­ x² : reflection 3­1d

Reflection  of  simplest  parabola:  Reflection  of  simplest  parabola:  Solution  1Solution  1

Precalculus

Fig. L1­5:  Parabolas  f (x) = x²,    f' (x) = ­ x²,  q (x) = ­2 x².  The  background  is  dilated                  corresponding  to  parameter  a  of  equation  q (x) = a x²

a = ­1:           f ' (x) = ­ x² : reflected parabola

a =  3:          q (x) = ­2 x² : Dilation of parabola f '  by scale factor  23­1e

Dilation  of  simplest  parabola:  Dilation  of  simplest  parabola:  Solution  1Solution  1

Precalculus

fig. L1­6:  Parabolas  f (x) = x²,   f ' (x) = ­ x²,   r (x) =­  0.2 x².  The  background  is  dilated                  corresponding  to  parameter  a  of  equation  r (x) = a x²  

a = ­1:           f ' (x) = ­ x² : reflected parabola

a = 3:           r (x) = ­0.2 x² : Dilation of parabola f '  by scale factor  0.23­1f

Dilation  of  parabola:  Dilation  of  parabola:  Solution  1Solution  1

Precalculus

Quadratic  functions  y = a xQuadratic  functions  y = a x²²:   :   Solution  2Solution  2

3­2

Fig. L2:  graphical  solution

Precalculus

Quadratic   functions:   Quadratic   functions:   y = xy = x² + n² + n

The graph of the function y = x²  + n is obtained by shifting thegraph of the function y = x² in positive y  direction by n units.

4­1

Fig. 3­1:   parabola  f (x) = x²  is moved  by  n = 3  units  in  positive  y  direction

f x = x 2 , g x = x 2 n

http://www.youtube.com/watch?v=5z5sgNl6cDg

Precalculus

Quadratic  functions:   Quadratic  functions:   y = xy = x² + n² + n

For n > 0 the shift is in positive y  direction, for n < 0 in negative y  direction.

4­2

Fig. 3­2:   The  parabola   f (x) = x²  is  shifted  by  1  unit  in  positive  y  direction                  and  by 2  units  in  negative  y  direction

f x = x 2 , g1 x = x 2 1, g 2 x = x 2 − 2Precalculus

5­A

Quadratic   functions:   Quadratic   functions:   y = a xy = a x² + n² + n

Exercise 3:

Draw the given functions and explain the characteristics:

a ) g1 x = 2 x2 2, g2 x = 0.2 x2 − 1

b ) g1 x = −3 x2 − 1, g2 x = −x2

4

32

Exercise 4:

Draw the given functions using GeoGebra

http://www.geogebra.org/cms/

f x = −0.17 x2 1.5, g x = −0.35 x2 0.3

h x = −1.2 x2 − 0.8

Compare the representation with that of page 5-4.

Precalculus

5­1

Fig. L3­1:   Graphical  representation  of  the  functions

f x = x2 , g1 x = 2 x2 2, g2 x = 0.2 x2 − 1

Quadratic   functions:   Quadratic   functions:   Solution  3aSolution  3a

Precalculus

5­2

f x = − x2 , g1 x = −3 x2 − 1, g2 x = −x2

4

32

Quadratic   functions:   Quadratic   functions:   Solution  3bSolution  3b

Fig. L3­2:   Graphical  representation  of  the  functions

Precalculus

Dilation: a = 2, shift by n = 2  units in positive y direction

a ) g1 x = 2 x2 2 :

g2 x = 0.2 x2− 1 :

b ) g1 x = −3 x2 − 1

g2 x = −x2

4

32

Reflection over the x-axis, dilation: a = 3, shift by n = 1  unitsin negative y  direction

5­3

Quadratic   functions:   Quadratic   functions:   Solution  3Solution  3

Dilation: a = 0.2, shift by n = 1  units in negative y direction

Reflection over the x-axis, dilation: a = 0.25, shift by n = 3/2 units in positive y  direction

Precalculus

http://www.fotocommunity.de/pc/pc/cat/575/display/4323393

f x = −0.17 x2 1.5, g x = −0.35 x2 0.3, h x = −1.2 x2 − 0.85­4

Fig. L3­3:  Graphical  representation,  exercise 4

Quadratic   functions:   Quadratic   functions:   Solution  4Solution  4

Precalculus

Quadratic  functions:  Quadratic  functions:  y = (x y = (x – d)²– d)²

Fig. 4­1:   The  parabola  f (x) = x²  is  shifted  by  d  units  in  positive  x  direction

5­5

f x = x 2 , g x = x − d 2

http://www.youtube.com/watch?v=5z5sgNl6cDg

The graph of the function y = g (x) is obtained by shifting thegraph of the function y = f (x) in positive x  direction by d units.

Precalculus

For d > 0  the shift is in direction of positive x,for d < 0 in direction of negative x.

f x = x 2 , g1 x = x − 4 2 d = 4 , g 2 x = x 2 2 d = −2

5­6

Quadratic  functions:  Quadratic  functions:  y = (x y = (x – d)²– d)²

Fig. 4­2:   The  parabola  f (x) = x²  is  shifted  by  4  units  in  positive  x  direction  and  by  2  units  in  negative  x  direction

Precalculus

Quadratic  functions:   Quadratic  functions:   y = (x y = (x – d)² + n– d)² + n

6­1

g1 x = x2 g2 x = x − 32 f x = x − 32 − 2

Fig.  5:  The  representation  of  the  function  y = f (x)  is  obtained  by  two  subsequent  shifts

Precalculus