Chapter 4, Problem 5CA

(a)

To determine

To Find:The model for each transformation of the graph $f(x)=x$ .

Answer to Problem 5CA

  $g(x)=1(x-(-3))+2$

Explanation of Solution

Given:The graph is a translation $2units$ up and $3units$ left.

  $g(x)=\mathrm{....}(x-\mathrm{....})+\mathrm{....}$ .

The translation $2units$ up involves adding $2units$ to the function and translation $3units$ left involves subtracting $3units$ .

Thus

  $\begin{array}{l}f(x)=x\\ g(x)=f(x)+2\\ g(x)=x+2\\ g(x)=(x-(-3))+2\\ thus\\ g(x)=1(x-(-3))+2\end{array}$

Hence, $g(x)=1(x-(-3))+2$ .

(b)

To determine

To Find: The model for each transformation of the graph $f(x)=x$ .

Answer to Problem 5CA

  $g(x)=1(x-2)+(-3)$

Explanation of Solution

Given: The graph is a translation $2units$ right and $3units$ down.

  $g(x)=\mathrm{....}(x-\mathrm{....})+\mathrm{....}$ .

The translation $2units$ right involves subtracting $2units$ to the variable and translation $3units$ down involves subtracting $3units$ to the function value.

Thus

  $\begin{array}{l}g(x)=f(x-2)\\ g(x)=x-2\\ g(x)=(x-2)-3\\ thus\\ g(x)=1(x-2)+(-3)\end{array}$

Hence, $g(x)=1(x-2)+(-3)$ .

(c)

To determine

To Find: The model for each transformation of the graph $f(x)=x$ .

Answer to Problem 5CA

  $g(x)=2(x-0)+2$

Explanation of Solution

Given: The graph is vertical stretch by a factor of $2units$ followed by the translation $2units$ up.

  $g(x)=\mathrm{....}(x-\mathrm{....})+\mathrm{....}$ .

Vertical stretch by two means mean multiply the function by two and translation of $2units$ up means adding $2units$ to the function.

  $\begin{array}{l}g(x)=2f(x)\\ g(x)=2x\\ g(x)=2x+2\\ thus\\ g(x)=2(x-0)+2\end{array}$

Hence, $g(x)=2(x-0)+2$ .

(d)

To determine

To Find: The model for each transformation of the graph $f(x)=x$ .

Answer to Problem 5CA

  $g(x)=\frac{1}{2}(x-3)+(-4)$

Explanation of Solution

Given: The graph is translation $3units$ right and vertical shrink by a factor of $\frac{1}{2}$ followed by the translation $4units$ down.

  $g(x)=\mathrm{....}(x-\mathrm{....})+\mathrm{....}$ .

  $\begin{array}{l}g(x)=f(x-3)\\ g(x)=x-3\\ g(x)=\frac{1}{2}(x-3)-4\\ thus\\ g(x)=\frac{1}{2}(x-3)+(-4)\end{array}$

Hence, $g(x)=\frac{1}{2}(x-3)+(-4)$ .