Chp 1 Midterm Review Solutions (1) PDF

Transformations Practice Exam - ANSWERS Use this sheet to record your answers 1. A 11. C 20. B NR 1. 5 12. A...

Transformations Practice Exam - ANSWERS Use this sheet to record your answers 1. A 11. C 20. B NR 1. 5 12. A NR 6. 57 2. C NR 3. 231 21. C 3. B 13. D 22. D NR 2. 4 14. B 23. D 4. B 15. A 24. B 5. A 16. C 25. B 6. B 17. A 26. A 7. D NR 4. 3 27. D 8. C NR 5. 8212 28. B 9. A 18. D 29. C 10. C 19. D Copyright © Barry Mabillard, 2006 www.math12.com Principles of Math 12 - Transformations Practice Exam - ANSWERS 1 www.math12.com  Transformations Practice Exam - Answers 1. If f ( x ) = x 2 − 1 , then it is a parabola as shown below. Domain: x ∈ R Range: y ≥ −1 Look at the graph in each response: A. B. Domain: x ∈ R Domain: x ∈ R Range: y ≥ −1 Range: y ≥ −2 C. D. Domain: x ≥ −1 Domain: x ∈ R, x ≠ ±1 Range: y ∈ R Range: y ∈ R, y ≠ 0 The only graph with the same domain and range as the original is A. Principles of Math 12 - Transformations Practice Exam - ANSWERS 2 www.math12.com  1 NR1) Draw the graph of the transformation y = f ( −2 x ). Multiply the x-values by 2 and reflect across the y-axis. Original Transformed y = f ( −2 x ) As can be seen in the transformed graph, the largest x-intercept is 5. 2. You can rewrite this as: y = f ( x − 2 ) + 4. Move the original right 2 units, and up 4 units. Transformed Original y = f ( x − 2) + 4 The range of the transformed graph is y ≥ −4 The answer is C. Principles of Math 12 - Transformations Practice Exam - ANSWERS 3 www.math12.com 3. The invariant points on the graph of a reciprocal occur when the original has a y-value equal to 1 or -1. This happens on the graph four times, at the points indicated. (The relative maximum in the middle is too low and does not reach a y-value of -1.) The answer is B. NR2. The y-intercept is three units away from the line y = -5. When those three units are stretched by a factor of 3, the new point is nine units away from the line, at the point (0, 4). The answer is 4. 4. The inverse graph is reflected about the line y = x. The inverse graph does not pass the vertical line test, therefore it is not a function. The answer is B. Principles of Math 12 - Transformations Practice Exam - ANSWERS 4 www.math12.com 5. The graph of the inverse is reflected in the line y = x. In the case of a parabola with the vertex at the origin, there must be overlap to properly draw in the graph. The answer is A. ⎛1 ⎞ 6. The graph of y = f ⎜ x ⎟ is drawn by applying a horizontal stretch by a factor of 2 ⎝2 ⎠ to the original graph. ⎛1 ⎞ y = f ⎜ x⎟ ⎝2 ⎠ The answer is B. 7. The graph of y = −2 f ( x + 5 ) has the following transformations: a vertical stretch by a factor of 2, a reflection in the x-axis, and a shift of five units left. The answer is D. Principles of Math 12 - Transformations Practice Exam - ANSWERS 5 www.math12.com 8. The graph of g ( x ) = f ( − x ) is a reflection in the y-axis. g ( x) = f ( −x) The graph of the transformation has a range of y ≥ 0 The answer is C. 9. Apply the transformation y = f ( 2 x + 4 ) to the point (8, -5) Rewrite as y = f ⎡⎣2 ( x + 2 ) ⎤⎦ 1 Horizontal stretch by a factor of gives (4, -5) Æ (multiply x – values by ½) 2 Shift two units left: (2, -5) Æ (subtract 2 from the x – coordinate) The answer is A. 10. In a reciprocal graph, invariant points occur when the y-value is ±1. The graph has two invariant points. The answer is C Principles of Math 12 - Transformations Practice Exam - ANSWERS 6 www.math12.com 11. The transformation f ( x + 1) − 2 tells you to replace x with x + 1. Original: f ( x ) = x 2 − 2 Transformed: f ( x + 1) = ( x + 1) − 2 2 Replace x with x+1 The result is ( x + 1) − 2 2 The answer is C. 12. Rewrite as y = f ( x + 1) − 2 and apply the transformation y = f ( x + 1) − 2 From the graph, it can be seen that the domain, but not the range, is the same. The answer is A. NR3. Quadrant II is a reflection in the y – axis, and that is equation 2. Quadrant III is a reflection in both axis, and that is equation 3. Quadrant IV is a reflection in the x – axis, and that is equation 1. The answer is 231. Principles of Math 12 - Transformations Practice Exam - ANSWERS 7 www.math12.com  1 13. The graph of f ( x ) is stretched by a factor of about the line y = 2, then is 2 reflected across the y-axis. The answer is D. 14. The graph of f ( x ) is moved one unit right, then 8 units down. The answer is B. 15. Piece the transformation together starting with y = f ( x ) 1 ⎛1 ⎞ A horizontal stretch of 3 means we use with x: y = f ⎜ x⎟ 3 ⎝3 ⎠ ⎛1 ⎞ A reflection in the x-axis means we put a negative outside: y = − f ⎜ x ⎟ ⎝3 ⎠ ⎛1 ⎞ Translated 4 right and 2 up gives y = − f ⎜ ( x − 4 ) ⎟ + 2 ⎝3 ⎠ The answer is A. 16. If the graph is reflected in the y-axis, then points on the y-axis don’t change. The only point on the y-axis is III. The answer is C. 17. An inverse reflects the graph along the line y = x. Inverse graph The graph that gives the new shape is A. Principles of Math 12 - Transformations Practice Exam - ANSWERS 8 www.math12.com  ( ) NR4. Write the function as y = b 2 x3 − 4 x 2 + 3x − 5 , then plug in the point given. −129 = b ⎡ 2 ( −2 ) − 4 ( −2 ) + 3 ( −2 ) − 5⎤ 3 2 ⎣ ⎦ −129 = b ⎡⎣ 2 ( −8 ) − 4 ( 4 ) + 3 ( −2 ) − 5⎤⎦ −129 = b [ −8 − 16 − 6 − 5] −129 = −43b b=3 The answer is 3. NR5. Rewrite as: y = 2 f ( x ) + 4. Apply the transformation by multiplying the y-values by two, then adding four. (-2, 2) Æ (-2, 8) (1, -1) Æ (1, 2) (5, 4) Æ (5, 12) The answer is 8212 18. The point (2, 2) lies on the line y = x, so it is invariant in an inverse graph. The answer is D. 19. Vertical asymptotes occur at the x-intercepts of the original graph. Therefore, x = 8 will be an asymptote. The answer is D. 1 20. Do exactly as the question says: Replace y with y 3 y = f ( x) 1 y = f ( x) 3 Then cross multiply to get y = 3 f ( x ) This is a vertical stretch by a factor of 3. The answer is B. Principles of Math 12 - Transformations Practice Exam - ANSWERS 9 www.math12.com NR6. Think of the point (7, 8) as being on the graph of g ( x ) = f ( x − 7 ) + 8 If the graph is further transformed by moving it two units left and one unit down, then the point (7, 8) becomes (5, 7). The new equation of the transformation is g ( x ) = f ( x − 5 ) + 7 The answer is 57 21. Point C is three units above y = -3, so when it is stretched vertically by a factor of 2 about that line, the new coordinate is (4, 3) The answer is C. 1 22. The vertical asymptotes in the graph of are found at the x-intercepts, f ( x) There are four x-intercepts, and therefore there are four vertical asymptotes. The answer is D. 1 23. The invariant points in the graph of occur when y = ±1 f ( x) There are four points where y = ±1, so there are four invariant points. The answer is D. 24. Rewrite g ( x ) = f ( 2 x − 4 ) as g ( x ) = f ⎡⎣ 2 ( x − 2 ) ⎤⎦ 1 There is a horizontal stretch by a factor of , and a shift of 2 units right. 2 The point (-2, 1) becomes (1, 1) The answer is B. Principles of Math 12 - Transformations Practice Exam - ANSWERS 10 www.math12.com 25. If the domain is x ≤ 3 , and the graph is shifted 10 units left, the new domain is x ≤ −7 The answer is B. 26. Rewrite y = f ( 3 x − 6 ) − 1 as y = f ⎡⎣3 ( x − 2 ) ⎤⎦ − 1 The point (-3, 4) is transformed as follows: 1 Horizontal stretch by Æ (-1, 4) 3 Two units right and one unit down Æ (1, 3) The answer is A. 27. Write the function as y = bf ( x ) , then plug in the information y = b ( x2 − 5x + 6) 15 = b ( 82 − 5 ( 8 ) + 6 ) 15 = 30b 1 b= 2 The answer is D. The inverse graph of y = ( x + 1) is a horizontal parabola, and does not pass the 2 28. vertical line test. The inverse is not a function. The answer is B. 1 29. The vertex of (0, 0) is 2 units away from the line x = 2, so when the stretch of 2 is applied, it will only be one unit away from that line. The new vertex is at (1, 0) The answer is C. Principles of Math 12 - Transformations Practice Exam - ANSWERS 11 www.math12.com  Written Response 1: 2 Marks Use p = 0, and q = 14 for the Raising the height of the coordinates of the vertex. arch is the same as a vertical stretch, so it should result in To determine the value of b in the the b value getting larger. equation y = b x − p + q , you can pick There is no shift left or another point on the triangular arch, right, so the p value stays such as (3, 10), and use this for x & y. the same. Plugging in all the values into the The vertex gets higher, so equation gives the q value should get bigger. 10 = b 3 − 0 + 14 10 = 3b + 14 2 Marks −4 = 3b 4 b=− 3 4 The equation is y = − x + 14 3 2 Marks Principles of Math 12 - Transformations Practice Exam - ANSWERS 12 www.math12.com  Written Response 2: Written Response 3: Vertical stretch by a factor of 2 y = − f ( x ) Æ B, E (about the x-axis) y = f (−x) Æ D y = f −1 ( x ) Æ D 1 y= Æ A, C, F. G f ( x) 1 V.S. about y = ÆD 2 y = − f ( − x ) Æ None 1 Mark Each Reflection in the x - axis Reciprocal transformation 2 Marks Each. Principles of Math 12 - Transformations Practice Exam - ANSWERS 13 www.math12.com