Name: Class: Date: ID: A Midterm Review Short Answer 1. For each graph, write the equation of a radical function of the form y = a b(x h) + k. a) b) c) 2. Determine the domain and range of each function. a) f(x) = x + 1 2 b) f(x) = 2 2 (x 2) + 1 3 c) y + 2 = 2 2(x 1) 1
Name: ID: A 3. a) Sketch the graph of each function f(x), and then sketch the graph of g(x) = f(x). i) f(x) = x ii) f(x) = x 2 4 iii) f(x) = 0.5x + 2 b) State the domain and range of each f(x) and g(x). 2
Name: ID: A 4. Using each graph of y = f(x), sketch the graph of y = f(x). a) b) c) 5. Sketch the graph of f(x) = 2x + 3 6 and use it to sketch the graph of y = f(x). 3
Name: ID: A 6. Solve the equation 3x 6 = 12 graphically. 7. Factor fully. a) x 2 (x 2)(x + 2) + 3x + 6 b) 16x 4 (x + 1) 2 8. Factor 2x 3 + 5x 2 14x 8 fully 9. Solve by factoring. a) x 3 2x 2 5x + 6 = 0 b) x 3 3x 2 4x + 12 = 0 c) x 3 = 4x 2 10. Sketch the graph of the function f(x) = 5 using transformations and identify x + 3 a) the vertical asymptotes (if any) b) the horizontal asymptotes (if any) c) the domain and range d) the behaviour near any non-permissible values e) the end behaviour 4
Name: ID: A 11. Consider the function f(x) = 3x + 8 x 2. a) Determine the key features of the function: i) domain and range ii) intercepts iii) equations of any asymptotes b) Sketch the graph of the function. x + 3 12. Consider the function f(x) = x 2 x 12. a) Determine the key features of the function: i) domain and range ii) intercepts iii) equations of any asymptotes b) Sketch the graph of the function. Problem 1. Consider the function f(x) = 2 x + 1 4. a) State the domain and range of the function. b) Use this information to determine the domain and range of the inverse of the function. c) Determine the inverse of the function. d) Graph the function and its inverse on the same set of axes (include the line y = x to verify the inverse). 5
Name: ID: A 2. Determine an equation in expanded form for the polynomial function represented by the graph. 3. Determine an equation in factored form for the polynomial function represented by the graph 4. Determine an equation in factored form for the polynomial function represented by the graph 6
Midterm Review Answer Section SHORT ANSWER 1. ANS: a) y = x + 4 3 b) y = 2 x 1 + 2 c) y = 1 2(x + 2) + 1 2 PTS: 1 DIF: Average OBJ: Section 2.1 NAT: RF13 TOP: Radical Functions and Transformations KEY: graph transformations 2. ANS: a) {x x 0, x R},{y y 1 2, y R} b) {x x 2, x R},{y y 1, y R} c){x x 1, x R},{y y 2, y R} PTS: 1 DIF: Average OBJ: Section 2.1 NAT: RF13 TOP: Radical Functions and Transformations KEY: domain range 1
3. ANS: a) i) The graph of f(x) = x is shown in blue and the graph of g(x) = f(x) is shown in red. ii) The graph of f(x) = x 2 4 is shown in blue and the graph of g(x) = f(x) is shown in red. iii) The graph of f(x) = 0.5x + 2 is shown in blue and the graph of g(x) = f(x) is shown in red. b) i) f(x): {x x R},{y y R} g(x): {x x 0, x R},{y y 0, y R} ii) f(x): {x x R},{y y 4, y R} g(x): {x x 2, x 2, x R},{y y 0, y R} iii)f(x): {x x R},{y y R} g(x): {x x 4, x R},{y y 0, y R} PTS: 1 DIF: Average OBJ: Section 2.2 NAT: RF13 TOP: Square Root of a Function KEY: graph domain range square root of a function 2
4. ANS: The graph of y = f(x) is shown in black, and the graph of y = a) f(x) is shown in blue. b) c) PTS: 1 DIF: Average OBJ: Section 2.2 NAT: RF13 TOP: Square Root of a Function KEY: graph square root of a function 3
5. ANS: The graph of y = f(x) is shown in black, and the graph of y = f(x) is shown in blue. PTS: 1 DIF: Difficult OBJ: Section 2.2 NAT: RF13 TOP: Square Root of a Function KEY: graph square root of a function 6. ANS: PTS: 1 DIF: Average OBJ: Section 2.3 NAT: RF13 TOP: Solving Radical Equations Graphically KEY: graphical solution 4
7. ANS: a) x 2 (x 2)(x + 2) + 3x + 6 = x 2 (x 2)(x + 2) + 3(x + 2) = (x + 2)[x 2 (x 2) + 3] = (x + 2)(x 3 2x 2 + 3) Use the factor theorem on the second factor. Try x = 1. P(x) = x 3 2x 2 + 3 P( 1) = ( 1) 3 2( 1) 2 + 3 P( 1) = 0 Divide. = 1 2 + 3 1 1 2 0 3 1 3 3 1 3 3 0 The quotient is not factorable. Thus, x 2 (x 2)(x + 2) + 3x + 6 = (x + 2)(x + 1)(x 2 3x + 3) b) 16x 4 (x + 1) 2 = (4x 2 ) 2 (x + 1) 2 = [4x 2 (x + 1)][4x 2 + (x + 1)] = (4x 2 x 1)(4x 2 + x + 1) PTS: 1 DIF: Average OBJ: Section 3.3 NAT: RF11 TOP: The Factor Theorem KEY: factor theorem factor integral zero theorem grouping rational zero theorem NOT: A variety of factoring techniques is required. 5
8. ANS: Possible values of x in the factor theorem are ±1, ± 1, ±2, ±4, and ±8. 2 Try x = 2. P(x) = 2x 3 + 5x 2 14x 8 P(2) = 2(2) 3 + 5(2) 2 14(2) 8 = 16 + 20 28 8 P(2) = 0 Thus, x 2 is a factor of P(x). Divide. 2 2 5 14 8 Thus, 4 18 8 2 9 4 0 2x 3 + 5x 2 14x 8 = (x 2)(2x 2 + 9x + 4) = (x 2)(2x 2 + x + 8x + 4) = (x 2)[x(2x + 1) + 4(2x + 1)] = (x 2)(2x + 1)(x + 4) PTS: 1 DIF: Difficult + OBJ: Section 3.3 NAT: RF11 TOP: The Factor Theorem KEY: factor theorem factor integral zero theorem grouping rational zero theorem NOT: A variety of factoring techniques is required. 6
9. ANS: a) The possible values of x for the factor theorem are ±1, ±2, ±3, and ±6. Try x = 1. P(x) = x 3 2x 2 5x + 6 P(1) = 1 3 2(1) 2 5(1) + 6 = 1 2 5 + 6 P(1) = 0 Thus, x 1 is a factor. Find another factor using synthetic division. 1 1 2 5 6 Thus, 1 1 6 1 1 6 0 x 3 2x 2 5x + 6 = 0 (x 1)(x 2 x 6) = 0 (x 1)(x + 2)(x 3) = 0 x = 1, 2, 3 b) x 3 3x 2 4x + 12 = 0 x 2 (x 3) 4(x 3) = 0 (x 3)(x 2 4) = 0 (x 3)(x 2)(x + 2) = 0 c) x 3 = 4x 2 x = 3, 2, 2 x 3 4x 2 = 0 x 2 (x 4) = 0 x = 0, 4 PTS: 1 DIF: Average OBJ: Section 3.3 Section 3.4 NAT: RF11 TOP: The Factor Theorem Equations and Graphs of Polynomial Functions KEY: factor theorem factor integral zero theorem roots 7
10. ANS: a) vertical asymptote with equation x = 3 b) horizontal asymptote with equation y = 0 c) domain {x x 3, x R}; range {y y 0, y R} d) As x approaches 3, y becomes very large. e) As x becomes very large, y approaches 0. PTS: 1 DIF: Easy OBJ: Section 9.1 NAT: RF14 TOP: Exploring Rational Functions Using Transformations KEY: graph characteristics transformations 8
11. ANS: a) i) {x x 2, x R}, {y y 3, y R} ii) x-intercept: 8, y-intercept: 4 3 iii) x = 2, y = 3 b) PTS: 1 DIF: Average OBJ: Section 9.2 Section 9.3 NAT: RF14 TOP: Analysing Rational Functions Connecting Graphs and Rational Equations KEY: linear expressions in numerator and denominator key features graph 9
12. ANS: a) i) {x x 3,x 4, x R}, {y y 0, y R} ii) x-intercept: none, y-intercept: 1 4 iii) x = 4, y = 0 Ê b) Note the hole at the point 3, 1 ˆ ËÁ 7. PTS: 1 DIF: Difficult OBJ: Section 9.2 Section 9.3 NAT: RF14 TOP: Analysing Rational Functions Connecting Graphs and Rational Equations KEY: rational function key features graph 10
PROBLEM 1. ANS: a) domain {x 1, x R}, range {y 4, y R} b) domain {x 4, x R}, range {y 1, y R} c) y = 2 x + 1 4 ( x + 4) 2 Ê Ê x + 4 ˆ ˆ ËÁ 2 ËÁ x = 2 y + 1 4 x + 4 = 2 y + 1 2 = y + 1 = y + 1 Ê y = x + 4 ˆ ËÁ 2 2 1 f 1 (x) = 1 4 ( x + 4) 2 1 d) The graph of f(x) is shown in blue and the graph of f 1 (x) is shown in red. PTS: 1 DIF: Difficult OBJ: Section 1.4 NAT: RF6 TOP: Inverse of a Relation KEY: graph inverse of a function function notation domain range 11
2. ANS: The graph has a single zero at x = 0 and a double zero at x = 2. The graph also passes through the point (3, 6). Thus, the graph is of the form y = ax(x 2) 2. Substitute the point (3, 6) to find a. y = ax(x 2) 2 6 = a(3)(3 2) 2 6 = 3a a = 2 Thus, y = 2x(x 2) 2 = 2x(x 2 4x + 4) y = 2x 3 8x 2 + 8x PTS: 1 DIF: Average OBJ: Section 3.4 NAT: RF12 TOP: Equations and Graphs of Polynomial Functions KEY: polynomial equation graph zeros 3. ANS: The graph has triple zeros at x = 2 and x = 1, and it passes through the point (0, 2). Thus, the equation of the graph is of the form y = a(x + 2) 3 (x 1) 3. Substitute the point (0, 2) to find a. y = a(x + 2) 3 (x 1) 3 2 = a(0 + 2) 3 (0 1) 3 2 = 8a a = 1 4 Thus, y = 1 4 (x + 2) 3 (x 1) 3 PTS: 1 DIF: Average OBJ: Section 3.4 NAT: RF11 RF12 TOP: Equations and Graphs of Polynomial Functions KEY: graph polynomial equation zeros 12
4. ANS: There are single zeros at x = 1 and x = 1, and a double zero at x = 2. The graph passes through the point (0, 4). Thus, the equation is of the form y = a(x + 1)(x 1)(x 2) 2. Substitute the point (0, 4) to find a. y = a(x + 1)(x 1)(x 2) 2 4 = a(0 + 1)(0 1)(0 2) 2 4 = 4a a = 1 Thus, y = (x + 1)(x 1)(x 2) 2. PTS: 1 DIF: Average OBJ: Section 3.4 NAT: RF12 TOP: Equations and Graphs of Polynomial Functions KEY: graph polynomial equation zeros y-intercept 13