Square OABC is drawn on a centimetre grid. O is (0,0) A is (2,0) B is (2,2) C is (0,2) Write down how many invariant points ther
e are on the perimeter of the square when OABC is rotated 90 degrees clockwise, centre (2,0).
1 answer:
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Answer:
- vertical scaling by a factor of 1/3 (compression)
- reflection over the y-axis
- horizontal scaling by a factor of 3 (expansion)
- translation left 1 unit
- translation up 3 units
Step-by-step explanation:
These are the transformations of interest:
g(x) = k·f(x) . . . . . vertical scaling (expansion) by a factor of k
g(x) = f(x) +k . . . . vertical translation by k units (upward)
g(x) = f(x/k) . . . . . horizontal expansion by a factor of k. When k < 0, the function is also reflected over the y-axis
g(x) = f(x-k) . . . . . horizontal translation to the right by k units
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Here, we have ...
g(x) = 1/3f(-1/3(x+1)) +3
The vertical and horizontal transformations can be applied in either order, since neither affects the other. If we work left-to-right through the expression for g(x), we can see these transformations have been applied:
- vertical scaling by a factor of 1/3 (compression) . . . 1/3f(x)
- reflection over the y-axis . . . 1/3f(-x)
- horizontal scaling by a factor of 3 (expansion) . . . 1/3f(-1/3x)
- translation left 1 unit . . . 1/3f(-1/3(x+1))
- translation up 3 units . . . 1/3f(-1/3(x+1)) +3
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<em>Additional comment</em>
The "working" is a matter of matching the form of g(x) to the forms of the different transformations. It is a pattern-matching problem.
The horizontal transformations could also be described as ...
- translation right 1/3 unit . . . f(x -1/3)
- reflection over y and expansion by a factor of 3 . . . f(-1/3x -1/3)
The initial translation in this scenario would be reflected to a translation left 1/3 unit, then the horizontal expansion would turn that into a translation left 1 unit, as described above. Order matters.
