4.3E: Exercises - Understanding Transformations of Functions

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Jun 28, 2024
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- 4.3: Understanding Transformations of Functions
- 4.4: Quadratic Functions and Applications

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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: $\{−4\}$

3. $y = x$ ; Shift up $3$ units; domain: $(- \infty , \infty)$ ; range: $(- \infty , \infty)$

5. $y = x^{2}$ ; Shift up $1$ unit; domain: $(- \infty , \infty)$ ; range: $[1, ∞)$

7. $y = x^{2}$ ; Shift right $5$ units and up $2$ units; domain: $(- \infty , \infty)$ ; range: $[2, ∞)$

9. $y = |x|$ ; Shift right $1$ unit and down $3$ units; domain: $(- \infty , \infty)$ ; range: $[−3, ∞)$

11. $y = \sqrt{x}$ ; Shift right $2$ units and up $1$ unit; domain: $[2, ∞)$ ; range: $[1, ∞)$

13. $y = x^{3}$ ; Shift right $1$ unit and down $4$ units; domain: $(- \infty , \infty)$ ; range: $(- \infty , \infty)$

15. $y = \frac{1}{x}$ ; Shift left $1$ unit and down $2$ units; domain: $(−∞, −1) ∪ (−1, ∞)$ ; range: $(−∞, −2) ∪ (−2, ∞)$

17. $y = \sqrt [ 3 ] { x }$ ; Shift up $6$ units and right $2$ units; domain: $(- \infty , \infty)$ ; range: $(- \infty , \infty)$

Exercise $\PageIndex{3}$

Graph the piecewise functions.

$h ( x ) = \left\{ \begin{array} { l l } { x ^ { 2 } + 2 } & { \text { if } x < 0 } \\ { x + 2 } & { \text { if } x \geq 0 } \end{array} \right.$
$h ( x ) = \left\{ \begin{array} { l l } { x ^ { 2 } - 3 \text { if } x < 0 } \\ { \sqrt { x } - 3 \text { if } x \geq 0 } \end{array} \right.$
$h ( x ) = \left\{ \begin{array} { l l } { x ^ { 3 } - 1 } & { \text { if } x < 0 } \\ { | x - 3 | - 4 } & { \text { if } x \geq 0 } \end{array} \right.$
$h ( x ) = \left\{ \begin{array} { c c } { x ^ { 3 } } & { \text { if } x < 0 } \\ { ( x - 1 ) ^ { 2 } - 1 } & { \text { if } x \geq 0 } \end{array} \right.$