IB Mathematics A&I SL Functions #2 Test [36 marks] 5 /34 Name Sage Gimse Hour_ Teacher_Johnson 1. A quadratic function f is given by f(x) = ax2 + bx + c. The points (0, 5) and (-4, 5) lie on the graph of y = f(x). 1a. [1 mark] Write down the value of c. 1b. [2 marks] Write down the equation of the axis of symmetry. 1c. [1 mark] The y-coordinate of the minimum of the graph is 3. Write down the coordinates of the vertex of the graph of y = f (x). 1d. [3 marks] Find the values of a and b. 5 - 5 1b) * = - b x = - 5 20 X = 3 25 X=0) - 2 2. 0 1 c ) 1 2 ( 5 - 2 = 3 , 1 + 2 = 3 ) - 1 1d) a= 0 , 1b= 5 - 35 143 + 5- 2749 30 ) from -2+2 - 2 3. The function f is defined by / (x) = 2x+ 5. 3d. [2 marks) Write down the domain and range 3b. (2 marks) Determine the value of f '(7). 3a. [2 marks) Write down the range of / 21) f-'(1)= 2( 7)45 3c. (2 marks) Solve f (x) = -1 2. Maegan designs a decorative glass face for a new Fine Arts Centre. The glass face is made up of small triangular panes. The first three levels of the glass face are illustrated in the following diagram. diagram not to scale The 1st level, at the bottom of the glass face, has 5 triangular panes. The 2nd level has 7 triangular paces, and the 3rd level has 9 triangular panes, Each additional level has 2 more triangular panes than the level below it 21 Za. [2 marks) Find the number of triangular panes in the 12th level 2b. (2 marks) Show that the total number of triangular panes, S,, in the first n levels is given by: 2c (1 marks) Hence, find the total number of panes in a glass face with 18 levels. 24 (3 marks) Maegan has 1000 triangular panes to build the decorative glass face and does not want it to have any incomplete levels, Find the maximum number of complete levels that Maegan can build. 20) 3/9 , 4/ 11, 5/ 13, 6/ 15, 7 /12, 8/19, 9/21, 10123, 11/ 25 12 / 21 2 3 27 panes 2b ) on = n2 +4 25 82= 32 + 4(3)2 83= 9+12-21 =$1 2 in the fiat 3 levels, there are 21 triangular panes v 2 20 ) 81 = 18 4 4 (18 ) 2 3 519 : 2 24 + 72 27 396 theageer paved 21 ) 1000 = 12 +4 solve with graph table ?, ov down Quadratic Formula - 2 1941 4 1/31 Levels3. The function f is defined by f (x) = 2x3 + 5, -2 5 x 52. 3a. [2 marks] Write down the range of f. 3b. [2 marks] Determine the value of f-1(7). 3c. [2 marks] Solve f-1(x) = -1. 3d. [2 marks] Write down the domain and range of f-1. 30 ) from - 2- 2 - 2 30 f- () = 2(7)$5 , -# 7:2 4 143 + 5 - 279415= 274927 2749 - 2 30) -1= 2x3+5 - 6- 273 13- x3 70 - 2 2 2 - 24. (4 marks) The gravitational force (in Newtons) exerted on an object by the Earth, referred to as the object's weight, varies inversely with the square of the object's distance from the centre of the Earth. The radius of the Earth may be estimated to be 6370 km, and an astronaut with a mass of 100 kg weighs approximately 980 N on the Earth's surface. Find the approximate weight of the astronaut when he is 11 km above the Earth's surface.5. A player throws a basketball. The height of the basketball is modelled by h(t) = -4.75t2 + 8.75t + 1.5, t 2 0, where h is the height of the basketball above the ground, in metres, and t is the time in seconds. 5a. [2 marks] Find how long it takes for the basketball to reach its maximum height. 5b. [2 marks] Assuming that no player catches the basketball, find how long it would take for the basketball to hit the ground. 5c. [2 marks] Another player catches the basketball when it is at a height of 1.2 metres. Find the value of t when this player catches the basketball. 5d. [2 marks] Write down two limitations of using h(t) to model the height of the basketball