Quadratic Functions 1 E. Precalculus

Transcription

1 Quadratic Functions 1 E

2 Galileo Galilei and the Leaning Tower of Pisa Galileo Galilei ( ) 1 1

3 Galileo Galilei and the Leaning Tower of Pisa The famous italien scientist Galileo Galilei was the first to formulate the laws of falling bodies. In 1590 he performed drop tests in Pisa. His student and biographer Vincenzo Viviani told, that he dropped bodies from the leaning tower to ground, but it is not really known whether he actually used the tower for his experiments. 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 d t = 1 g t2 2 relates the drop height d, g to the time t, the time taken for an object to fall through distance d. Here g is the gravitational acceleration. 1 2

4 Examples of quadratic functions The kinetic energy E of a body with mass m is a quadratic function of the velocity v E kin v = 1 m v2 2 The area of a circle with radius r is a quadratic function of the radius S r = r 2 r 1 3

5 Examples of quadratic functions Fig. 1 1: Body thrown upward A body is thrown at time t = 0 with an angle α above the horizontal with initial velocity v0. The path can be parametrised as follows x = v 0 cos t 1 4 y = h 0 v 0 sin t 1 g t2 2

6 Examples of quadratic functions Fig. 1 2: The curve describing the flight of baron Münchhausen can be approximated by a quadratic function. 1 5

7 Quadratic functions Definition: Functions satisfying the equation f x = a x 2 b x c x, a, b, c ℝ, a 0 are called quadratic functions. 2 1 a x2 quadratic term bx linear term c constant term

8 Quadratc Functions: the simplest parabola Fig. 2 1: simplest parabola y = x² y = x 2 : domain D = ℝ, range R = [ 0, ) y = x² : parabola opens upward The vertex S (0, 0) is the lowest point (minimum) 2 2

9 Simplest Parabola: Symmetries Fig. 2 2: the parabola y = x² is symmetric with respect to the y axis f x = x 2 : 2 3 f x 1 = f x 1, f x 2 = f x 2

10 Simplest Parabola: Monotonicity Fig. 2 3: Parabola y = x² monotonically decreasing in the negative domain f x = x2, 2 4 x 1 = 2, x 2 = 1, x1 x 2, f x1 f x 2

11 Simplest Parabola: Monotonicity Fig. 2 4: Parabola y = x² monotonically increasing in the positive domain f x = x 2, 2 5 x 1 = 1, x 2 = 2, x1 x 2, f x1 f x 2

12 Monotonicity Definition: A function y = f (x) is monotonically increasing in an interval I of the domain, if for all x1, x 2 I D, x1 x 2 f x1 f x2 Definition: A function y = f (x) is monotonically decreasing in an interval I of the domain, if for all x1, x 2 I D, x1 x 2 f x1 f x2 The function f (x) = x² is monotonically decreasing for x 0 and monotonically increasing for x

13 Quadratic functions y = a x²: Exercise 1, 2 Exercise 1: Plot the quadratic functions f x = x 2, g x = 3 x 2, h x = 0.2 x 2 p x = x 2, q x = 2 x2, r x = 0.2 x 2 Explain their properties by means of the graph. Exercise 2: Determine by means of the graph of the function y = x² the two values 2 3 A and 3.

14 Simplest parabola Fig. L1 1: 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 3 1a

15 Dilation of simplest parabola: Solution 1 Abb. L1 2: Parabolas f (x) = x², g (x) = 3 x². The background is dilated corresponding to parameter a of equation g (x) = a x² 3 1b a = 1: a = 3: f (x) = x² : g (x) = 3 x² : simplest parabola dilation by scale factor 3

16 Dilation of simplest parabola: Solution 1 Fig. L1 3: Parabolas f (x) = x², h (x) = 0.2 x². The background is dilated corresponding to parameter a of equation h (x) = a x² 3 1c a = 1: f (x) = x² : simplest parabola a = 0.2: h (x) = 0.2 x² : dilation by scale factor 0.2

17 Reflection of simplest parabola: Solution 1 Fig. L1 4: Parabolas f (x) = x², p (x) = x² 3 1d a = 1: f (x) = x² : simplest parabola a = 1: p (x) = x² : reflection

18 Dilation of simplest parabola: Solution 1 Fig. L1 5: 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 3 1e a = 1: a = 3: f ' (x) = x² : reflected parabola q (x) = 2 x² : Dilation of parabola f ' by scale factor 2

19 Dilation of parabola: Solution 1 fig. L1 6: 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 3 1f a = 1: f ' (x) = x² : a = 3: r (x) = 0.2 x² : reflected parabola Dilation of parabola f ' by scale factor 0.2

20 Quadratic functions y = a x²: Solution 2 Fig. L2: graphical solution 3 2

21 Quadratic functions: y = x² + n Fig. 3 1: parabola f (x) = x² is moved by n = 3 units in positive y direction The graph of the function y = x² + n is obtained by shifting the graph of the function y = x² in positive y direction by n units. 4 1 f x = x 2, g x = x 2 n

22 Quadratic functions: y = x² + n Fig. 3 2: The parabola f (x) = x² is shifted by 1 unit in positive y direction and by 2 units in negative y direction For n > 0 the shift is in positive y direction, for n < 0 in negative y direction. 4 2 f x = x 2, g 1 x = x 2 1, g2 x = x 2 2

23 Quadratic functions: y = a x² + n Exercise 3: Draw the given functions and explain the characteristics: a ) g 1 x = 2 x 2 2, 2 b ) g 1 x = 3 x 1, g 2 x = 0.2 x 2 1 x2 3 g2 x = 4 2 Exercise 4: Draw the given functions using GeoGebra f x = 0.17 x 2 1.5, g x = 0.35 x h x = 1.2 x Compare the representation with that of page A

24 Quadratic functions: Solution 3a Fig. L3 1: Graphical representation of the functions f x = x 2, 5 1 g 1 x = 2 x 2 2, g 2 x = 0.2 x 2 1

25 Quadratic functions: Solution 3b Fig. L3 2: Graphical representation of the functions 2 f x = x, g 1 x = 3 x 1, x2 3 g2 x = 4 2

26 Quadratic functions: Solution 3 a ) g 1 x = 2 x 2 2 : Dilation: a = 2, shift by n = 2 units in positive y direction g 2 x = 0.2 x 2 1 : Dilation: a = 0.2, shift by n = 1 units in negative y direction b ) g 1 x = 3 x2 1 Reflection over the x-axis, dilation: a = 3, shift by n = 1 units in negative y direction x2 3 g2 x = 4 2 Reflection over the x-axis, dilation: a = 0.25, shift by n = 3/2 units in positive y direction 5 3

27 Quadratic functions: Solution 4 Fig. L3 3: Graphical representation, exercise 4 f x = 0.17 x 2 1.5, 5 4 g x = 0.35 x 2 0.3, h x = 1.2 x 2 0.8

28 Quadratic functions: y = (x d)² Fig. 4 1: The parabola f (x) = x² is shifted by d units in positive x direction The graph of the function y = g (x) is obtained by shifting the graph of the function y = f (x) in positive x direction by d units. f x = x 2, 5 5 g x = x d 2

29 Quadratic functions: y = (x d)² Fig. 4 2: The parabola f (x) = x² is shifted by 4 units in positive x direction and by 2 units in negative x direction For d > 0 the shift is in direction of positive x, for d < 0 in direction of negative x. f x = x 2, 5 6 g 1 x = x 4 2 d = 4, g 2 x = x 2 2 d = 2

30 Quadratic functions: y = (x d)² + n Fig. 5: The representation of the function y = f (x) is obtained by two subsequent shifts g 1 x = x g 2 x = x 3 2 f x = x 3 2 2