### UNIT 2  : FUNCTIONS & TRANSFORMATIONS

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)
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