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 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) ------------------------------- (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) ------------------------------- (x, -y)

(-2, 4) --------------------- (-2 , -4)

(-1, 1) --------------------- (-1 , -1)

(0,0) --------------------- (0, 0)

(1, 1) --------------------- (1, -1)

(2, 4) --------------------- (2, -4) yielding the graph as shown above in (red)

 

 

(x, y) ------------------------------- (x, -y)

 

 

(x, y) -------------------------------(x, - y)

(0, 0) --------------------------(0, 0)

(1, 1) --------------------------(1, 1)

(4, 2) --------------------------(4, -2)

(9, 3) --------------------------(9, - 3)

(16, 4) -------------------------(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) ---------------------------------(x, -y)

(-5, 1) ---------------------------(-5, -1)

(-3, 1) ---------------------------(-3, -1)

(-1, 3) ---------------------------(-1, -3)

(1, 0) ----------------------------(1, 0)

(3, 2) ----------------------------(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) -------------------------------- (-x, y) [reflection in the y-axis]

 

 
 

 

 

 


 

(x, y) --------------------------(-x, y)

(0, 0) --------------------------(0, 0)

(1, 1) --------------------------(-1, 1 )

(4, 2) --------------------------(-4, 2 )

(9, 3) --------------------------(-9, 3)

(16, 4) -------------------------(-16, 4)

 

 

 

 

.

 

 

 

(x, y) -------------------------------(-x - 3, y)

(0, 0) --------------------------(-3, 0)

(1, 1) --------------------------(-4, 1 )

(4, 2) --------------------------(-7, 2 )

(9, 3) --------------------------(-12, 3)

(16, 4) -------------------------(-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) --------------------- (x - 2, y - 4) and using the key points for the function y = x2, we obtain the points of the transformed function.

 

(x, y) --------------------- (x - 2, y - 4)

(-2, 4) -------------------- (-2-2 , 4-4) = (-4, 0)

(-1, 1) -------------------- (-1-2 , 1-4) = (-3, -3)

(0,0) ---------------------- (-2, -4)

(1, 1) --------------------- (-1, -3)

(2, 4) --------------------- (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) ------------------------------- (-x, y).

 

 

(x, y) --------------------------- (-x, y) [reflection in the y-axis]

(-4, 0) -------------------------- (4, 0)

(-3, -3) -------------------------(3, -3)

(-2, -4) -------------------------(2, -4)

(-1, -3) ------------------------- (1, -3)

(0, 0) -------------------------- (0, 0)

 

 

 

 

 

 

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