Math Functions and Graphs

1. Let g(x) = 9 + kx Given that g(− 2) = 3, find the value of k. • g (x) = 9 + kx g(− 2) = 9 + k(− 2) = 3 9 – 2k = 3 – 2k = 3 – 9 – 2k = – 6 k = 3

2. The graph shown represents the flight of a golf ball. The height of the ball, after tseconds, is given by the function h(t) = at − t2, where t ∈ ℝ. (i) • After how many seconds is the ball at a height of 8 m? Go to 8 on the y-axis. Draw a horizontal line across until you touch the graph. Draw vertical lines downwards until you touch the x-axis. The ball is at a height of 8 m after 2 seconds and 4 seconds.

2. The graph shown represents the flight of a golf ball. The height of the ball, after tseconds, is given by the function h(t) = at − t2, where t ∈ ℝ. (ii) • Find the two times when the height of the ball is 5 m. Go to 5 on the y-axis. Draw a horizontal line across until you touch the graph. Draw vertical lines downwards until you touch the x-axis. After 1 second and 5 seconds.

2. The graph shown represents the flight of a golf ball. The height of the ball, after tseconds, is given by the function h(t) = at − t2, where t ∈ ℝ. (iii) • Find the value of a. h(t) = at – t2 h(1) = 5 • h(t) = a(1) – (1)2 = 5 a – 1 = 5 a = 6

3. The diagram shows part of the graph of the function f : x → x2 + 2x − 3, where x ∈ ℝ. • The graph intersects the x-axis at the points P and Q and the y-axis at R. Find the coordinates of the points P, Q and R. (i) Cuts the x-axis when y = 0 f (x) =x2 + 2x − 3 x2 + 2x – 3 = 0 (x + 3)(x – 1) = 0 x + 3 = 0 x – 1 = 0 x = – 3 x = 1 P(− 3, 0) Q(1, 0)

3. The diagram shows part of the graph of the function f : x → x2 + 2x − 3, where x ∈ ℝ. • The graph intersects the x-axis at the points P and Q and the y-axis at R. Find the coordinates of the points P, Q and R. (i) Cuts the y-axis when x = 0 y = x2 + 2x – 3 y = (0)2 + 2(0) – 3 = 0 + 0 – 3 = – 3 R(0, – 3)

3. The diagram shows part of the graph of the function f : x → x2 + 2x − 3, where x ∈ ℝ. • Hence, write down the range of values of x for which x2 + 2x − 3 ≤ 0 (ii) x2 + 2x – 3  0 is where the curve is below the x-axis –3 x 1

4. Which of the following show the graph of a function? Give a reason for your answer. (a) A vertical line will, at most, cross this graph only once. This is a function as there is only one output value corresponding to any input value.

4. Which of the following show the graph of a function? Give a reason for your answer. (c) (b) (b) and (c) fail the vertical line test as a vertical line will pass through the graph more than once, in some places.

5. f : x → 3x2 − 1 and g : x → 2x − 3. Find: • f (2) (i) f (x) = 3x2 – 1 f (2) = 3(2)2 – 1 = 12 – 1 = 11

5. f : x → 3x2 − 1 and g : x → 2x − 3. Find: • f g (0) (ii) g(x) = 2x – 3 f g (0) = f (− 3) g(0) = 2(0) – 3 = 3(− 3)2 – 1 = 0 – 3 = 3(9) – 1 = – 3 = 27 – 1 = 26

5. f : x → 3x2 − 1 and g : x → 2x − 3. Find: • g f (− 2) (iii) f (− 2) = 3(− 2)2 – 1 g f (− 2) = g(11) = 3(4) – 1 = 2(11) – 3 = 12 – 1 = 22 – 3 = 11 = 19

5. f : x → 3x2 − 1 and g : x → 2x − 3. Find: (f ◦ g)(x) (iv) (f ◦ g)(x) = f (2x – 3) = 3(2x – 3)2 – 1 = 3(2x – 3)(2x – 3) – 1 = 3(4x2 – 12x + 9) – 1 = 12x2 – 36x + 27 – 1 = 12x2 – 36x + 26

6. g(x) = x2 + x + k is a function defined on ℝ, where k ∈ ℤ. • If g(1) = − 4, find the value of k. (i) g(x) = x2 + x + k g(1) = (1)2 + (1) + k = – 4 1 + 1 + k = – 4 2 + k = – 4 k = – 4 – 2 k = – 6

6. g(x) = x2 + x + k is a function defined on ℝ, where k ℤ. • Hence, solve the equation g (x + 5) = 0 (ii) g(x + 5) = (x + 5)2 + (x + 5) + k = 0 x2 + 10x + 25 + x + 5 – 6 = 0 x2 + 11x + 24 = 0 (x + 8)(x + 3) = 0 x + 8 = 0 x + 3 = 0 x = – 8 x = – 3

7. (i) Graph the function g (x) = 2x3 − 5x2 − 2x + 5 in the domain −1 ≤ x ≤ 3.

7. (ii) Use your graph to estimate: • the roots of the equation g(x) = 0 (a) Roots are the points where the graph crosses the x-axis. When x = – 1, 1, 2·5

7. (ii) Use your graph to estimate: • the values of x for which g(x) = −1 (b) Go to −1 on the y-axis. Draw a horizontal line across until you touch the graph. Draw vertical lines upwards until you touch the x-axis. When g(x) = −1, x = 1·2 and 2·4.

7. (ii) Use your graph to estimate: • g(0·5) (c) Go to 0·5 on the x-axis. Draw a vertical line upwards until you touch the graph. Draw a horizontal line across until you touch the y-axis. g(0·5) = 3

7. (ii) Use your graph to estimate: • g (− 0·5) (d) Go to − 0·5 on the x-axis. Draw a vertical line upwards until you touch the graph. Draw a horizontal line across until you touch the y-axis. g(− 0·5) = 4·5

7. Check the accuracy of your estimated roots in part (ii) (a), by checking if they satisfy the equation g(x) = 0. • Substitute each root into g(x) = 2x3 − 5x2 − 2x + 5 x = – 1: 2(−1)3 – 5(−1)2 – 2(−1) + 5 2(−1) – 5(1) + 2 + 5 – 2 – 5 + 2 + 5 = 0 x = 1: 2(1)3 – 5(1)2 – 2(1) + 5 All three roots satisfy g(x) = 0 2(1) – 5(1) – 2 + 5 2 – 5 – 2 + 5 = 0 x = 2·5: 2(2·5)3 – 5(2·5)2 – 2(2·5) + 5 2(15·625) – 5(6·25) – 5 + 5 31·25 – 31·25 – 5 + 5 = 0

8. A puddle contains 500 litres of water. The water is evaporating at a rate of 6% per hour. The volume of the water remaining in the puddle, after t hours, is given by: v(t) = 500(0∙94)t • Is the value of v(t) increasing or decreasing? (i) Give a reason for your answer. v(t) = 500(0·94)t Decreasing since 0·94 < 1.

8. A puddle contains 500 litres of water. The water is evaporating at a rate of 6% per hour. The volume of the water remaining in the puddle, after t hours, is given by: v(t) = 500(0∙94)t • If t = 0 represents 8 am, find the volume of water remaining at 3 pm. Give your answer to two decimal places. (ii) t = 0 = 8 am t = 7 = 3 pm v(7) = 500(0·94)7 = 324·24 litres

8. A puddle contains 500 litres of water. The water is evaporating at a rate of 6% per hour. The volume of the water remaining in the puddle, after t hours, is given by: v(t) = 500(0∙94)t • Copy and complete the following table and hence graph the function v(t) using the same axes and scale. (iii)

8. A puddle contains 500 litres of water. The water is evaporating at a rate of 6% per hour. The volume of the water remaining in the puddle, after t hours, is given by: v(t) = 500(0∙94)t • Use your graph to: Estimate, to the nearest hour, when the volume of water is at 350 litres. (iv) From the graph, when volume = 350, t = 5·75 hours  6 hours or 2 pm.

8. A puddle contains 500 litres of water. The water is evaporating at a rate of 6% per hour. The volume of the water remaining in the puddle, after t hours, is given by: v(t) = 500(0∙94)t • Use your graph to: Estimate the volume of water remaining in the puddle at 5 pm. (v) t = 0 = 8 am t = 9 = 5 pm When t = 9, volume = 286·5 litres

8. A puddle contains 500 litres of water. The water is evaporating at a rate of 6% per hour. The volume of the water remaining in the puddle, after t hours, is given by: v(t) = 500(0∙94)t • Use your graph to: Estimate, to the nearest hour, when the volume of the water has been reduced by 40%. (vi) 40 % of 500 = 200 litres 500 – 200 = 300 litres From the graph, when volume = 350, t = 8·25 hours  8 hours or 4 pm.

9. The function g : x → 2x3 − 4x2 − 5x + 4 is defined for x ∈ ℝ. Find the coordinates of the y-intercept of the graph of g (x). That is, the point where the graph of g cuts the y-axis. (i) Cuts y-axis when x = 0 g(x) = 2x3 – 4x2 – 5x + 4 g(x) = 2(0)3 – 4(0)2 – 5(0) + 4 = 4 Coordinates are (0, 4)

9. The function g : x → 2x3 − 4x2 − 5x + 4 is defined for x ∈ ℝ. • Verify, using algebra, that the point (− 1, 3) is on the graph of g. (ii) g(x) = 2x3 – 4x2 – 5x + 4 3 = 2(−1)3 – 4(−1)2 – 5(−1) + 4 = 2(−1) – 4(1) + 5 + 4 3 = –2 – 4 + 5 + 4 3 = 3 (−1, 3) is on the graph of g.

10. The graph of f (x) = − x2 + 2x + 3, where x ∈ℝ, is shown in the diagram. Copy the graph into your copybook. Using the same axes and scale, graph each of the following functions: • g(x) = f (x) − 2 (i) 2 is subtracted from the f(x) function, so the graph drops by 2 units. • h(x) = f (x − 2) + 1 (ii) 2 is subtracted from the x part and 1 is added to the f(x) function, so the graph moves 2 units to the right, then up 1 unit.

11. Clay pigeon shooting is a sport which involves a clay target, being released from a trap to simulate the flight of a bird. The objective of the sport is to shoot the clay. The path of the clay is modelled by the function h(t) = − 5t2 + 32t + 2, where hrepresents the height of the clay, in metres, t seconds after the clay is released. Donald shoots his gun. The path of Donald’s bullet is modelled by the function b(t) = 31·5t + 1, where brepresents the height of the bullet, in metres, tseconds after the clay is released.

the time when the bullet hits the clay. (i) 11. Clay pigeon shooting is a sport which involves a clay target, being released from a trap to simulate the flight of a bird. The objective of the sport is to shoot the clay. The path of the clay is modelled by the function h(t) = − 5t2 + 32t + 2, where hrepresents the height of the clay, in metres, t seconds after the clay is released. Donald shoots his gun. The path of Donald’s bullet is modelled by the function b(t) = 31·5t + 1, where brepresents the height of the bullet, in metres, tseconds after the clay is released. Height of bullet = height of clay 31·5t + 1 = – 5t2 + 32t + 2 5t2 + 31·5t – 32t + 1 – 2 = 0 5t2 – t – 1 = 0 10t2 – t – 2 = 0 (5t + 2)(2t – 1) = 0 5t + 2 = 0 2t – 1 = 0 5t = –2 2t = 1 (Reject, as t cannot be negative)

the height of the clay at this time. (ii) 11. Clay pigeon shooting is a sport which involves a clay target, being released from a trap to simulate the flight of a bird. The objective of the sport is to shoot the clay. The path of the clay is modelled by the function h(t) = − 5t2 + 32t + 2, where hrepresents the height of the clay, in metres, t seconds after the clay is released. Donald shoots his gun. The path of Donald’s bullet is modelled by the function b(t) = 31·5t + 1, where brepresents the height of the bullet, in metres, tseconds after the clay is released. Height of clay = height of bullet b(t) = 31·5t + 1 = 15·75 + 1 Height = 16·75 metres

11. The device releasing the clay is adjusted, and the clay now follows a path modelled by the function h(t) = −5t2 + 44t + 2. If the path of the bullet remains the same, find: • after how many seconds the bullet will hit the clay. (iii) Height of bullet = height of the clay 31·5t + 1 = – 5t2 + 44t + 2 5t2 + 31·5t – 44t + 1 – 2 = 0 5t2 – 12·5t – 1 = 0 a = 10 10t2 – 25t – 2 = 0 b = – 25 c = – 2

11. The device releasing the clay is adjusted, and the clay now follows a path modelled by the function h(t) = −5t2 + 44t + 2. If the path of the bullet remains the same, find: • after how many seconds the bullet will hit the clay. (iii) t = 2·577 t = – 0·0775 t = 2·6 sec (Reject)

11. The device releasing the clay is adjusted, and the clay now follows a path modelled by the function h(t) = −5t2 + 44t + 2. If the path of the bullet remains the same, find: • the height of the clay, at this time. (iv) Height of clay = height of bullet Let t = 2·6 sec b(t) = 31·5t + 1 b(2·6) = 31·5(2·6) + 1 = 82·9 metres

Math Functions and Graphs