Dr. J's Maths.com
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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

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
 

13. |4 - 2x| = x - 2

14. |2x + 5| = 3x + 9
 

15. |x - 2| - x = 1

16. |2x + 6| = |x + 10|
 

17. 2|x + 8| = 3 |x + 5|

18. |6x - 7| = 2|4 - 2x|

 

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
at x = -2.5

(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.
 

20. (i)

 

(ii) Finding all values for k for which the equation
|2x - 4| = kx + 1 has only one solution.

The gradient of the lines are +2 and -2 respectively.
 
  • Our objective is to draw a line which will cross the basic shape only once given that the y intercept is 1.
  • If the line starts at (0,1) it cannot be less that the gradient of the RH line. Hence it cannot have a gradient of less than 2 because then it will cut the RH line.
  • If the line starts at (0,1) it cannot have a gradient of greater than -2 because then it will cut the LH line.
  • If a line starts at (0, 1) and passes through
    the point (2, 0) - with a gradient of 1รท2 =-0.5 - it will cross the x axis at (2, 0) and so there will be one solution.
  • The values which can therefore be taken by the gradient k are therefore:
  • k = -0.5;
  • k < -2 and k < 2.