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15

REFLECTIONS


CONSIDER THE FIRST QUADRANT point (a, b), and let us reflect it about the y-axis.  It is reflected to the second quadrant point (−a, b).

If we reflect (a, b) about the x-axis, then it is reflected to the fourth quadrant point (a, −b).

Finally, if we reflect (a, b) through the origin, then it is reflected to the third quadrant point (−a, −b).  The distance from the origin to (a, b) is equal to the distance from the origin to (−a, −b).

Example 1.

Fig. 1 is the graph of the parabola

f(x) = x² − 2x − 3 = (x + 1)(x − 3).

The roots −1, 3 are the x-intercepts.

Fig. 2 is its reflection about the x-axis.  Every point that was above the x-axis gets reflected to below the x-axis.  And every point below the x-axis gets reflected above the x-axis.  Only the roots, −1 and 3, are invariant.

Again, Fig. 1 is  y = f(x).  Its reflection about the x-axis is  y = −f(x).  Because every y-value is the negative of the original f(x).

Fig. 3 is the reflection of  Fig. 1 about the y-axis.  Every point that was to the right of the origin gets reflected to the left.  And every point that was on the left gets reflected to the right.  Every  x  becomes  −x. Only the y-intercept is invariant.  The equation of the reflection of f(x) about the y-axis is  y = f(−x).

If y = f(x), then

y = f(−x) is its reflection about the y-axis,

y = −f(x) is its reflection about the x-axis.

Problem 1.   Let f(x) = x² + x − 2.

a)  Sketch the graph of f(x).

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x² + x − 2 = (x + 2)(x − 1).  The x-intercepts are at −2 and 1.

b)  Write the function −f(x), and sketch its graph.

f(x) = −(x² + x − 2) = −x² − x + 2.  Its graph is the reflection of f(x) about the x-axis.

   

c)  Write the function f(−x), and sketch its graph.

Replace each x with −x.  f(−x) = x² − x − 2 = (x − 2)(x + 1).  Its graph is the reflection of f(x) about the y-axis.

        

Problem 2.   Let  f(x) = (x + 3)(x + 1)(x − 2).

Sketch the graph of f(x), then sketch the graphs of f(−x) and −f(x).

        

The graph on the left is f(x).  The roots -- the x-intercepts -- are −3, −1, 2.

The middle graph is f(−x), which is its reflection about the y-axis.

The graph on the right is −f(x), which is its reflection about the x-axis.

Problem 3.   Let f(x) = x² − 4.

Sketch the graph of f(x), then sketch the graph of f(−x).

x² − 4 = (x + 2)(x − 2).

Here, the graph of f(−x) -- its reflection about the y-axis -- is equal to the graph of f(x).

Problem 4.   Let f(x) = x³.

Sketch the graph of f(x), then sketch the graphs of  f(−x)  and  −f(x).

      

The graph on the left is f(x).

The graph on the right is f(−x), which is its reflection about the y-axis .  But (−x )³ = −x³, so that
f(−x) is equal to −f(x) -- which is its reflection about the x-axis!

Example 2.   Sketch the graph of  y = −x² + x + 6.

 Solution.   It is best to consider a graph when the leading coefficient is positive.  Therefore, let us call the given function −f(x):

f(x)  =  x² + x + 6
 
   =  −(x² − x − 6)
 
   =  −(x + 2)(x − 3)

f(x), then, is (x + 2)(x − 3).  Its x-intercepts are at −2 and 3.  The graph we want is −f(x), which is the reflection of f(x) about the x-axis:

Problem 5.   Sketch the graph of y = −x² − 2x + 8.

f(x)  =  x² − 2x + 8
 
   =  −(x² + 2x − 8)
 
   =  −(x + 4)(x − 2)

Here is the graph:

It is the reflection about the x-axis of

f(x) = x² + 2x − 8

Problem 6.   Sketch the graph of  y = −x3 − 2x² + x + 2.

[Hint:  Call the function −f(x), then factor f(x) by grouping.]

f(x)  =  x³ − 2x² + x + 2
   =  −(x³ + 2x² − x − 2)
 
   =  −[x²(x + 2) − (x + 2)]
 
   =  −(x² − 1)(x + 2)
 
   =  −(x + 1)(x − 1)(x + 2)

The graph is the reflection about the x-axis of

f(x) = (x + 2)(x + 1)(x − 1)

Problem 7.   Sketch the graph of  y = −.

Problem 8.      Sketch the graph of  y = −|x|.


Next Topic:  Symmetry


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