Chapter 1.4, Problem 4E

To determine

To fill in the blanks according to the given question

Answer to Problem 4E

A non-rigid transformation of $y=f(x)$ represented by $cf(x)$ is vertical stretch when $c>1$ and a vertical shrink when $0<c<1$

Explanation of Solution

The best way to understand this problem is by taking an example and solving it accordingly.

Let, $f(x)=x^2$

For vertical stretch$(c>1)$:

  $\begin{array}{l}h(x)=3f(x)\\ \Rightarrow h(x)=3x^2\end{array}$

Calculation for graph:

Consider $f(x)=x^2$

Values of x	Values of f (x )
0	0
1	1
-1	1
2	4
3	9

By taking different values of x , the graph can be plotted.

Calculation for graph:

Consider $h(x)=3x^2$

Values of x	Values of h (x)
0	0
1	3
-1	3
2	12
3	27

By taking different values of x , the graph can be plotted.

Graph for both $f(x)$ and $h(x)$ on same plane:

  

Interpretation:

From the above graph, it is clear that $h(x)$ is the vertically stretched form of $f(x)$ .

For vertical shrink$(0<c<1)$:

  $\begin{array}{l}h(x)=\frac{1}{2}f(x)\\ \Rightarrow h(x)=\frac{1}{2}{x}^2\end{array}$

Calculation for graph:

Consider $f(x)=x^2$

Values of x	Values of f (x )
0	0
1	1
-1	1
2	4
3	9

By taking different values of x , the graph can be plotted.

Calculation for graph:

Consider $h(x)=\frac{x^2}{2}$

Values of x	Values of h (x)
0	0
1	0.5
-1	0.5
2	2
3	4.5

By taking different values of x , the graph can be plotted.

Graph for both $f(x)$ and $h(x)$ on same plane:

  

Interpretation:

From the above graph, it is clear that $h(x)$ is the vertically shrinked form of $f(x)$ .

Conclusion:

Hence,

A non-rigid transformation of $y=f(x)$ represented by $cf(x)$ is vertical stretch when $c>1$ and a vertical shrink when $0<c<1$ .