Graph the following functions, give coordinates of three points, and give the domains in interval notation.\[ 20. \quad f(x) = \left| \frac{1}{3}x - 2 \right| - 1 \]---To graph the function, $ f(x) = \left| \frac{1}{3}x - 2 \right| - 1 $, we start by identifying key components and plotting essential points.1. **Vertex**: The expression inside the absolute value, $ \frac{1}{3}x - 2 = 0 $, gives the x-coordinate of the vertex. Solving for $ x $, we find: \[ \frac{1}{3}x = 2 \] \[ x = 6 \] Substituting $ x = 6 $ into the function gives: \[ f(6) = \left| \frac{1}{3}(6) - 2 \right| - 1 = |2 - 2| - 1 = -1 \] So, the vertex is at $ (6, -1) $.2. **Additional Points**: Choose points on either side of the vertex: - For $ x = 3 $: \[ f(3) = \left| \frac{1}{3}(3) - 2 \right| - 1 = |1 - 2| - 1 = 0 \] Coordinates: $ (3, 0) $ - For $ x = 9 $: \[ f(9) = \left| \frac{1}{3}(9) - 2 \right| - 1 = |3 - 2| - 1 = 0 \] Coordinates: $ (9, 0) $3. **Domain**: This function is defined for all real numbers. Thus, the domain in interval notation is: \[ (-\infty, \infty) \]Graphing the function will involve plotting these points and drawing a V-shaped graph with the vertex at $ (6, -1) $ and lines extending through $ (3, 0) $ and $ (9, 0) $.

Answered: Graph the following functions, give…

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

See similar textbooks

Related questions

Concept explainers

Question

Graph the following functions, give coordinates of three points, and give the domains in interval notation.

\[ 20. \quad f(x) = \left| \frac{1}{3}x - 2 \right| - 1 \]

---

To graph the function, $ f(x) = \left| \frac{1}{3}x - 2 \right| - 1 $, we start by identifying key components and plotting essential points.

1. **Vertex**: The expression inside the absolute value, $ \frac{1}{3}x - 2 = 0 $, gives the x-coordinate of the vertex. Solving for $ x $, we find:

\[ \frac{1}{3}x = 2 \]
\[ x = 6 \]

Substituting $ x = 6 $ into the function gives:

\[ f(6) = \left| \frac{1}{3}(6) - 2 \right| - 1 = |2 - 2| - 1 = -1 \]

So, the vertex is at $ (6, -1) $.

2. **Additional Points**: Choose points on either side of the vertex:

- For $ x = 3 $:
\[ f(3) = \left| \frac{1}{3}(3) - 2 \right| - 1 = |1 - 2| - 1 = 0 \]
Coordinates: $ (3, 0) $

- For $ x = 9 $:
\[ f(9) = \left| \frac{1}{3}(9) - 2 \right| - 1 = |3 - 2| - 1 = 0 \]
Coordinates: $ (9, 0) $

3. **Domain**: This function is defined for all real numbers. Thus, the domain in interval notation is:

\[ (-\infty, \infty) \]

Graphing the function will involve plotting these points and drawing a V-shaped graph with the vertex at $ (6, -1) $ and lines extending through $ (3, 0) $ and $ (9, 0) $.

Expert Solution

Step 1

Given

$f (x) = |\frac{1}{3} x - 2| - 1$

Step 2

Let

$\begin{array}{rcl} \frac{1}{3} x - 2 & = & 0 \\ \frac{1}{3} x & = & 2 \\ x & = & 6 \end{array}$

Thus

$f (x) = \{\begin{cases} - (\frac{1}{3} x - 2) - 1, x < 6 \\ (\frac{1}{3} x - 2) - 1, x \geq 6 \end{cases} f (x) = \{\begin{cases} - \frac{1}{3} x + 1, x < 6 \\ \frac{1}{3} x - 3, x \geq 6 \end{cases}$

Step 3

Thus graph of $f (x)$ is

Step by step

Solved in 5 steps with 1 images

Knowledge Booster

$Limits and Continuity$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, algebra and related others by exploring similar questions and additional content below.

Similar questions