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LESSON 4: REFLECTIONS
1. Vertical Reflections (reflections in the x-axis):
Example 1:
The graphs of y = x2 and y = -x2 are given below.
Note that the graphs are congruent. The graph of y = -x2 (red) is a reflection (mirror image) in the x-axis of the graph of y = x2 (blue).
The transformation may be depicted in mapping form. Each point (x, y) on the base curve y = x2 has been transformed as follows:
(x, y) -------------------------------à (x,
-y) [reflection in the x-axis]
For example, the point (1, 1) on the curve y = x2
reflects into the point (1, -1) on the image curve y = - x2
Note that the x-coordinate stays the same and the y-coordinate changes sign.
Other points on the image curve may be obtained using this mapping as follows.
(x, y) -------------------------------à (x,
-y)
(-2, 4) ---------------------à (-2 , -4)
(-1, 1) ---------------------à (-1 , -1)
(0,0) ---------------------à (0, 0)
(1, 1) ---------------------à (1, -1)
(2, 4) ---------------------à (2, -4) yielding the graph as shown above in (red)
(x, y) -------------------------------à (x,
-y)
(x, y) -------------------------------à(x, - y)
(0, 0) --------------------------à(0, 0)
(1, 1) --------------------------à(1, – 1)
(4, 2) --------------------------à(4, -2)
(9, 3) --------------------------à(9, - 3)
(16, 4) -------------------------à(16, -4) yielding the graph at left (red)
In general, the graph of y = -f(x) is a reflection in the x-axis of the graph of y = f(x).
Example 3: Given the graph of y = f(x) as shown, draw the graph of y = -f(x).
The required graph will be a
reflection in the x-axis of the graph of y = f(x) and may be represented in
mapping form:
(x, y)
---------------------------(x, -y)
Now take key points on the graph
of y = f(x) -- {(-5, 1), (-3, 1), (-1,
3),(0,1.5), (1, 0), (3, 2)} and change the sign of the y-coordinates using the
mapping.
(x, y)
---------------------------------à(x, -y)
(-5, 1)
---------------------------à(-5, -1)
(-3, 1)
---------------------------à(-3, -1)
(-1, 3)
---------------------------à(-1, -3)
(1, 0)
----------------------------à(1, 0)
(3, 2)
----------------------------à(3, -2)
2. Horizontal
Reflections (reflections in the
y-axis):
Example 4: Given the function f(x) = 3x – 1, find the equation of f(-x) and draw its graph.
To find f(-x), simply substitute (-x) for x in the equation
f(-x) = 3(-x) – 1 = -3x – 1 or y = -3x - 1
Draw the graph of both functions by completing the tables of values below.
For y = 3x – 1
If x = -2, y = 3(-2) –1 = -7
If x = -1, y = 3(-1) – 1= -4, etc. giving
x |
-2 |
-1 |
0 |
1 |
2 |
y |
-7 |
-4 |
-1 |
2 |
5 |
For y = -3x – 1
If x = -2, y = -3(-2) –1 = 5
If x = -1, y = -3(-1) – 1= 2, etc. giving
x |
-2 |
-1 |
0 |
1 |
2 |
y |
5 |
2 |
-1 |
-4 |
-7 |
Note that the graph of y = f(-x) = -3x – 1 (red) is a reflection (mirror image) of the graph of y = f(x) = 3x – 1 (blue) in the y – axis.
For example, the point (2, 5) on y = 3x – 1 is reflected into its image point (-2, 5) on the graph of y = -3x –1.
Notice the x-values change sign and the y-values stay the same. This reflection could be put in mapping form as follows;
(x, y) --------------------------------à
(-x, y) [reflection in the
y-axis]
(x, y) --------------------------à(-x, y)
(0, 0)
--------------------------à(0, 0)
(1, 1)
--------------------------à(-1, 1 )
(4, 2)
--------------------------à(-4, 2 )
(9, 3)
--------------------------à(-9, 3)
(16, 4)
-------------------------à(-16, 4)
.
(x, y) -------------------------------à(-x - 3, y)
(0, 0)
--------------------------à(-3, 0)
(1, 1)
--------------------------à(-4, 1 )
(4, 2)
--------------------------à(-7, 2 )
(9, 3)
--------------------------à(-12, 3)
(16, 4)
-------------------------à(-19, 4)
Example 7: Given f(x) = x2 + 4x, find the equation for f(-x) and draw both graphs.
Solution: By completing the square, we can rewrite f(x) as follows:
f(x) = x2 + 4x + 4 – 4 ** Recall – divide the
coefficient of x by 2 and square it [
(4/2)2] = 4
= (x2 + 4x + 4) – 4
= (x + 2)2 - 4 ** trinomial x2
+ 4x + 4 gets factored as (x+2)(x+2) =
(x + 2)2
The graph of f(x) (blue) will be a translation (shift) left 2 and down 4 relative to the basic function y = x2 . In mapping form we have
(x, y) ---------------------à (x - 2, y - 4) and using the key points for the function y = x2, we obtain the points of the transformed function.
(x, y) ---------------------à (x - 2, y - 4)
(-2, 4) --------------------à (-2-2 , 4-4) = (-4, 0)
(-1, 1) --------------------à (-1-2 , 1-4) = (-3, -3)
(0,0) ----------------------à (-2, -4)
(1, 1) ---------------------à (-1, -3)
(2, 4) ---------------------à (0, 0) yielding the
graph as shown with vertex at (-2, -4)
Now determine f(-x) = (-x)2
+ 4(-x)
= x2 – 4x
The graph of f(-x) (red) will be a reflection in the y-axis of
the above parabola (blue) with vertex at (-2, -4),
Take the points on this parabola
and use the reflection mapping: (x, y)
-------------------------------à (-x, y).
(x,
y) ---------------------------à (-x, y) [reflection in
the y-axis]
(-4, 0) --------------------------à (4, 0)
(-3, -3) -------------------------à(3, -3)
(-2, -4) -------------------------à(2, -4)
(-1, -3) -------------------------à (1, -3)
(0, 0) --------------------------à (0, 0)