Chapter 1.7, Problem 7E

(a)

To determine

To show shifts in the graph with c = -2, 0 and 2.

Explanation of Solution

Given information: For each function sketch a graph on the same set of coordinate axes.

Graph:

  

Interpretation: We have modulus graph:- $y=\left[x\right]$ . Now putting values of c in function $y=\left[x\right]+c$ we get :-

Values of c	Function
-2	$y=\left[x\right]-2$
0	$y=\left[x\right]+0$
2	$y=\left[x\right]+2$

CaseI. $y=\left[x\right]-2$

The Vertical Shift in the graph, $y=f(x)$ is represented as :-

  $h(x)=f(x)-c$ which implies vertical shift c units down.

On comparing the function $y=\left[x\right]-2$ with $h(x)=f(x)-c$ where c =2

So,we can say that the graph is drawn downward by three unit.

Case II. $y=\left[x\right]+0$

The Vertical Shift in the graph, $y=f(x)$ is represented as :-

  $h(x)=f(x)+c$ which implies vertical shift c units down.

On comparing the function $y=\left[x\right]+0$ with $h(x)=f(x)+c$ where c =0

So, the greatest integer graph with is drawn without shifts.

Case III. $y=\left[x\right]+2$

The Vertical Shift in the graph, $y=f(x)$ is represented as :-

  $h(x)=f(x)+c$ which implies vertical shift c units up.

On comparing the function $y=\left[x\right]+2$ with $h(x)=f(x)+c$ where c =2

So,we can say that the graph is drawn downward by three unit.

(b)

To determine

To sketch a graph of each function for c = -2 , 0 and 2.

Explanation of Solution

Given information: For each function sketch a graph on the same set of coordinate axes.

Graph:

  

Interpretation: We have modulus graph:- $y=\left[x\right]$ |. Now putting values of c in function

  $y=\left[x+c\right]$ we get :-

Values of c	Function
-2	$y=\left[x-2\right]$
0	$y=\left[x+0\right]$
2	$y=\left[x+2\right]$

CaseI. $y=\left[x-2\right]$

The Horizontal Shift in the graph, $y=f(x)$ is represented as :-

  $h(x)=f(x-c)$ implies horizontal shift c units to the right .

On comparing $y=\left[x-2\right]$ with $h(x)=f(x-c)$ where c =2 we have drawn the graph towards right hand side by two unit.

CaseII. $y=\left[x+0\right]$

The Horizontal Shift in the graph, $y=f(x)$ is represented as :-

  $h(x)=f(x-c)$ implies horizontal shift c units to the right .

On comparing $y=\left[x+0\right]$ with $h(x)=f(x-c)$ where c =0

This represent the greatest integer graph with no shifts.

Case III. $y=\left[x+2\right]$

Horizontal Shift in the graph, $y=f(x)$ is represented as :-

  $h(x)=f(x+c)$ implies horizontal shift c units to the left.

On comparing $y=\left[x+2\right]$ with $h(x)=f(x+c)$ where c =2 we have drawn the graph towards left hand side by two unit.