Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 2 - 2.2 - Functions - 2.2 Exercises - Page 183: 32


$1; \dfrac{1}{2},0,\dfrac{1}{2},1$

Work Step by Step

We are given the function: $h(t)=\dfrac{1}{2}|t+3|$ Compute $h(t)$ for $t=-5,-4,-3,-2,-1$: $h(-5)=\dfrac{1}{2}|-5+3|=\dfrac{1}{2}|-2|=\dfrac{1}{2}(2)=1$ $h(-4)=\dfrac{1}{2}|-4+3|=\dfrac{1}{2}|-1|=\dfrac{1}{2}(1)=\dfrac{1}{2}$ $h(-3)=\dfrac{1}{2}|-3+3|=\dfrac{1}{2}|0|=\dfrac{1}{2}(0)=0$ $h(-2)=\dfrac{1}{2}|-2+3|=\dfrac{1}{2}|1|=\dfrac{1}{2}(1)=\dfrac{1}{2}$ $h(-1)=\dfrac{1}{2}|-1+3|=\dfrac{1}{2}|2|=\dfrac{1}{2}(2)=1$ Place the values in a table:
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.