4.3E: Exercises - Understanding Transformations of Functions
- Page ID
- 147267
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Exercise \(\PageIndex{1}\)
Match the graph to the function definition.






- \(f(x) = \sqrt{x + 4}\)
- \(f(x) = |x − 2| − 2\)
- \(f(x) = \sqrt{x + 1} -1\)
- \(f(x) = |x − 2| + 1\)
- \(f(x) = \sqrt{x + 4} + 1\)
- \(f(x) = |x + 2| − 2\)
- Answer
-
1. e
3. d
5. f
Exercise \(\PageIndex{2}\)
Graph the given function. Identify the basic function and translations used to sketch the graph. Then state the domain and range.
- \(g(x) = −4\)
- \(g(x) = 2\)
- \(f(x) = x + 3\)
- \(f(x) = x − 2\)
- \(g(x) = x^{2} + 1\)
- \(g(x) = x^{2} − 4\)
- \(g(x) = (x − 5)^{2} + 2\)
- \(g(x) = (x + 2)^{2} − 5\)
- \(h(x) = |x − 1| − 3\)
- \(h(x) = |x + 2| − 5\)
- \(g(x) = \sqrt{x − 2} + 1\)
- \(g(x) = \sqrt{x + 2} + 3\)
- \(h(x) = (x − 1)^{3} − 4\)
- \(h(x) = (x + 1)^{3} + 3\)
- \(f(x) = \frac{1}{x+1} − 2\)
- \(f(x) = \frac{1}{x−3} + 3\)
- \(f ( x ) = \sqrt [ 3 ] { x - 2 } + 6\)
- \(f ( x ) = \sqrt [ 3 ] { x + 8 } - 4\)
- Answer
-
1. Basic graph \(y = −4\); domain: \( (- \infty , \infty) \); range: \(\{−4\}\)
Figure 2.5.37 3. \(y = x\); Shift up \(3\) units; domain: \((- \infty , \infty)\); range: \((- \infty , \infty)\)
Figure 2.5.24 5. \(y = x^{2}\); Shift up \(1\) unit; domain: \((- \infty , \infty)\); range: \([1, ∞)\)
Figure 2.5.25 7. \(y = x^{2}\); Shift right \(5\) units and up \(2\) units; domain: \( (- \infty , \infty) \); range: \([2, ∞)\)
Figure 2.5.27 9. \(y = |x|\); Shift right \(1\) unit and down \(3\) units; domain: \( (- \infty , \infty) \); range: \([−3, ∞)\)
Figure 2.5.29 11. \(y = \sqrt{x}\); Shift right \(2\) units and up \(1\) unit; domain: \([2, ∞)\); range: \([1, ∞)\)
Figure 2.5.31 13. \(y = x^{3}\); Shift right \(1\) unit and down \(4\) units; domain: \( (- \infty , \infty) \); range: \( (- \infty , \infty) \)
Figure 2.5.33 15. \(y = \frac{1}{x}\); Shift left \(1\) unit and down \(2\) units; domain: \((−∞, −1) ∪ (−1, ∞)\); range: \((−∞, −2) ∪ (−2, ∞)\)
Figure 2.5.36 17. \(y = \sqrt [ 3 ] { x }\); Shift up \(6\) units and right \(2\) units; domain: \( (- \infty , \infty) \); range: \( (- \infty , \infty) \)
Figure 2.5.38
Exercise \(\PageIndex{3}\)
Graph the piecewise functions.
- \(h ( x ) = \left\{ \begin{array} { l l } { x ^ { 2 } + 2 } & { \text { if } x < 0 } \\ { x + 2 } & { \text { if } x \geq 0 } \end{array} \right.\)
- \(h ( x ) = \left\{ \begin{array} { l l } { x ^ { 2 } - 3 \text { if } x < 0 } \\ { \sqrt { x } - 3 \text { if } x \geq 0 } \end{array} \right.\)
- \(h ( x ) = \left\{ \begin{array} { l l } { x ^ { 3 } - 1 } & { \text { if } x < 0 } \\ { | x - 3 | - 4 } & { \text { if } x \geq 0 } \end{array} \right.\)
- \(h ( x ) = \left\{ \begin{array} { c c } { x ^ { 3 } } & { \text { if } x < 0 } \\ { ( x - 1 ) ^ { 2 } - 1 } & { \text { if } x \geq 0 } \end{array} \right.\)
- Answer
-
1.
Figure 2.5.39 3.
Figure 2.5.40
Exercise \(\PageIndex{4}\)
Write an equation that represents the function whose graph is given.
1.

2.

3.

4.

5.

6.

7.

8.

- Answer
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1. \(f ( x ) = \sqrt { x - 5 }\)
3. \(f ( x ) = ( x - 15 ) ^ { 2 } - 10\)
5. \(f ( x ) = \frac { 1 } { x + 8 } + 4\)
7. \(f ( x ) = \sqrt { x + 16 } - 4\)
Exercise \(\PageIndex{5}\)
Match the graph to the given function definition.






- \(f ( x ) = - 3 | x |\)
- \(f ( x ) = - ( x + 3 ) ^ { 2 } - 1\)
- \(f ( x ) = - | x + 1 | + 2\)
- \(f ( x ) = - x ^ { 2 } + 1\)
- \(f ( x ) = - \frac { 1 } { 3 } | x |\)
- \(f ( x ) = - ( x - 2 ) ^ { 2 } + 2\)
- Answer
-
1. b
3. d
5. f
Exercise \(\PageIndex{6}\)
Use the transformations to graph the following functions.
- \(f ( x ) = - x + 5\)
- \(f ( x ) = - | x | - 3\)
- \(g ( x ) = - | x - 1 |\)
- \(f ( x ) = - ( x + 2 ) ^ { 2 }\)
- \(h ( x ) = \sqrt { - x } + 2\)
- \(h ( x ) = - \sqrt { x - 2 } + 1\)
- \(g ( x ) = - x ^ { 3 } + 4\)
- \(f ( x ) = - x ^ { 2 } + 6\)
- Answer
-
1.
Figure 2.5.57 3.
Figure 2.5.58 5.
Figure 2.5.59 7.
Figure 2.5.61