Chapter 1.6, Problem 55E

Solution

Show that the combine result of reflection and translation differ when applied in opposite order and explain the reason.

The combine result of translation and reflection is different in opposite order.

Given:

Take an equation $y=f\left(x\right)$

for example whose graph is as follows:

1

Concept Used:

The graph of $y=f\left(x\right)$ is translated or shifted horizontally left or right according as:

1) If $y=f\left(x-c\right)$ , then graph shifted by c units right from $y=f\left(x\right)$ .

2) If $y=f\left(x+c\right)$ , then graph shifted by c units left from $y=f\left(x\right)$ .

And, the graph of $y=f\left(x\right)$ is translated or shifted vertically up or down according as:

1) If $y=f\left(x\right)+c$ , then graph shifted by c units up from $y=f\left(x\right)$ .

2) If $y=f\left(x\right)-c$ , then graph shifted by c units down from $y=f\left(x\right)$ .

The transformation of horizontal stretches or shrinks from $y=f\left(x\right)$ is as follows:

The graph of $y=f\left(\frac{x}{c}\right)$ transform to horizontal stretches by factor c , when $c>1$ and horizontal shrinks by factor c when $c<1$ .

The transformation of vertical stretches or shrinks from $y=f\left(x\right)$ is as follows:

The graph of $y=c\cdot f\left(x\right)$ transform to vertical stretches by factor c , when $c>1$ and shrinks by factor c when $c<1$ .

The graph of $y=-f\left(x\right)$ is the reflection of graph of $y=f\left(x\right)$ across x -axis.

Calculation:

Transform the graph of $y=f\left(x\right)$ sequence of transformation, a reflection across x -axis, then a shift of one unit left and finally a shift of one unit up, ,as follows

1) The reflection of $y=f\left(x\right)$ across x -axis.

  

2) A shift of one unit left.

  

3) The shift of $1$ units up, transform.

  

Now, apply the same transformation in opposite order, that is, a shift of one unit up then a shift of one unit left and finally reflection across x -axis.

1) Translate by a shift of one unit up.

  

2) Apply transformation of translation by a shift of one left,

  

3) Finally apply the reflection across x -axis.

  

Clearly, the final graph obtained in the two sequences of transformation is different.

The transformation in first sequence, a reflection across x- axis transforms equation $y=f\left(x\right)$ to $y=-f\left(x\right)$ .

Then, a shift of one unit left transform equation $y=-f\left(x\right)$ to $y=-f\left(x+1\right)$ .

And finally, the shift of one unit up, transform $y=-f\left(x+1\right)$ to $y=-f\left(x+1\right)+1$

Hence, the final transformed equation is $y=-f\left(x+1\right)+1$ .

Now, consider the second sequence, a shift of one unit up transform equation $y=f\left(x\right)$ to $y=f\left(x\right)+1$ , then a shift of one unit left transform $y=f\left(x\right)+1$ to $y=f\left(x+1\right)+1$ and finally a reflection across x -axis transform $y=f\left(x+1\right)+1$ to $y=-\left[f\left(x+1\right)+1\right]=-f\left(x+1\right)-1$ .

Hence the final transformed equation is $y=-f\left(x+1\right)-1$ .

The final equation in the two orders of transformations is not same because in reflection across x -axis the whole function is multiplied by $-1$ which will not be same if applied in different order. Hence combine result of reflection and translation depends on order and the correct transformation is obtain when reflection to be applied first.

Conclusion The combine result of translation and reflection is different in opposite order.