Algebra - Absolute values - graphs.
Test Yourself 2.
Transformations | 1. (i) Graph y = |x| (ii) Write the equation transformed from y = |x| where there is a horizontal shift of 2 to the right and a shift of 1 down. (iii) Graph the transformed equation. |
2. (i) Write the equation transformed from y = |x| where there is a horizontal dilation of 2 and a horizontal shift of -1. (ii) Graph the transformed equation. |
Graph the two components of each of the following equations on the same set of axes. So for Q3, graph y = |3x + 2| and y = 8. On the basis of your graph, solve the given equation for x value(s). Check you answers with the corresponding question number in Absolute Value equations TYS 1. | ||
3. |3x + 2| = 8
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4. |3 - 2x| = 5 | |
5. |-2x - 3| = 7 | 6. | |
7. | 8. | |
Solving equations graphically. | 9. |x2 - 3| = 6 | 10. |x2 + 3| = 8 |
11. |2x + 3| = 3x NOTE: The lines only cross once - so x = -0.6 cannot be a solution. |
12. |3x + 1| = 2x + 4 |
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13. |4 - 2x| = x - 2 |
14. |2x + 5| = 3x + 9 |
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15. |x - 2| - x = 1 |
16. |2x + 6| = |x + 10| |
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17. 2|x + 8| = 3 |x + 5| |
18. |6x - 7| = 2|4 - 2x| |
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19. 6|x + 3| - 2|x + 1| = 0 Two graphs are drawn for comparison: (i) Here we are looking for the solution where the graph crosses the x axis as the equation was given as the LHS = 0. Solutions at x = -4 and |
(ii) Graphing the two components separately and then equating them at the points of intersection: Here we are looking for the two points of interesction because we have equated the components. Again the solutions shown by these points of intersection are at x = -4 and at x = -2.5. |
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20. (i)
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(ii) Finding all values for k for which the equation The gradient of the lines are +2 and -2 respectively. |
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