At Westonci.ca, we make it easy to get the answers you need from a community of informed and experienced contributors. Find reliable answers to your questions from a wide community of knowledgeable experts on our user-friendly Q&A platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
To determine the rule for [tex]\( g(x) \)[/tex] based on the given transformations of [tex]\( f(x) \)[/tex], let’s go through each transformation step-by-step applied to the function [tex]\( f(x) = -\frac{1}{2} x + 1 \)[/tex]:
1. Vertical compression by a factor of [tex]\(\frac{1}{5}\)[/tex]:
[tex]\[ g_1(x) = \frac{1}{5} \left(-\frac{1}{2} x + 1\right) \][/tex]
Simplifying inside the parentheses:
[tex]\[ g_1(x) = \frac{1}{5} \left(-\frac{1}{2} x + 1\right) = \frac{1}{5}\left(-\frac{1}{2} x \right) + \frac{1}{5}(1) = -\frac{1}{10} x + \frac{1}{5} \][/tex]
2. Horizontal translation right by [tex]\(1\)[/tex] unit (note: moving 'right' means we adjust the input [tex]\( (x - 1) \)[/tex]):
[tex]\[ g_2(x) = -\frac{1}{10}(x - 1) + \frac{1}{5} \][/tex]
Distribute [tex]\(-\frac{1}{10}\)[/tex]:
[tex]\[ g_2(x) = -\frac{1}{10}x + \frac{1}{10} + \frac{1}{5} = -\frac{1}{10}x + \frac{1}{10} + \frac{2}{10} = -\frac{1}{10}x + \frac{3}{10} \][/tex]
3. Horizontal translation left by [tex]\(-\frac{1}{2}\)[/tex] unit (multiply [tex]\( x \)[/tex] by the factor [tex]\(\frac{1}{2}\)[/tex] inside the function, which the problem seems to incorrectly quote):
[tex]\[ g_3(x) = -\frac{1}{10}\left(x + \frac{1}{2}\right) + \frac{3}{10} \][/tex]
Distribute [tex]\(-\frac{1}{10}\)[/tex] inside:
[tex]\[ g_3(x) = -\frac{1}{10} x - \frac{1}{20} + \frac{3}{10} = -\frac{1}{10} x + \frac{3}{10} - \frac{1}{20} \][/tex]
Combine the fractions:
[tex]\[ g_3(x) = -\frac{1}{10} x + \frac{3}{10} - \frac{1}{20} = -\frac{1}{10} x + \frac{6}{20} - \frac{1}{20} \][/tex]
Simplify the constant terms:
[tex]\[ g_3(x) = -\frac{1}{10} x + \frac{5}{20} = -\frac{1}{10} x + \frac{1}{4} \][/tex]
Thus, after parsing through potential confusion from problem statement (`g(x) = -10x + 1/5`), this shows progressive term handling:
Final Function:
[tex]\[ g(x) = -\frac{1}{10}x + \frac{1}{4} \][/tex]
1. Vertical compression by a factor of [tex]\(\frac{1}{5}\)[/tex]:
[tex]\[ g_1(x) = \frac{1}{5} \left(-\frac{1}{2} x + 1\right) \][/tex]
Simplifying inside the parentheses:
[tex]\[ g_1(x) = \frac{1}{5} \left(-\frac{1}{2} x + 1\right) = \frac{1}{5}\left(-\frac{1}{2} x \right) + \frac{1}{5}(1) = -\frac{1}{10} x + \frac{1}{5} \][/tex]
2. Horizontal translation right by [tex]\(1\)[/tex] unit (note: moving 'right' means we adjust the input [tex]\( (x - 1) \)[/tex]):
[tex]\[ g_2(x) = -\frac{1}{10}(x - 1) + \frac{1}{5} \][/tex]
Distribute [tex]\(-\frac{1}{10}\)[/tex]:
[tex]\[ g_2(x) = -\frac{1}{10}x + \frac{1}{10} + \frac{1}{5} = -\frac{1}{10}x + \frac{1}{10} + \frac{2}{10} = -\frac{1}{10}x + \frac{3}{10} \][/tex]
3. Horizontal translation left by [tex]\(-\frac{1}{2}\)[/tex] unit (multiply [tex]\( x \)[/tex] by the factor [tex]\(\frac{1}{2}\)[/tex] inside the function, which the problem seems to incorrectly quote):
[tex]\[ g_3(x) = -\frac{1}{10}\left(x + \frac{1}{2}\right) + \frac{3}{10} \][/tex]
Distribute [tex]\(-\frac{1}{10}\)[/tex] inside:
[tex]\[ g_3(x) = -\frac{1}{10} x - \frac{1}{20} + \frac{3}{10} = -\frac{1}{10} x + \frac{3}{10} - \frac{1}{20} \][/tex]
Combine the fractions:
[tex]\[ g_3(x) = -\frac{1}{10} x + \frac{3}{10} - \frac{1}{20} = -\frac{1}{10} x + \frac{6}{20} - \frac{1}{20} \][/tex]
Simplify the constant terms:
[tex]\[ g_3(x) = -\frac{1}{10} x + \frac{5}{20} = -\frac{1}{10} x + \frac{1}{4} \][/tex]
Thus, after parsing through potential confusion from problem statement (`g(x) = -10x + 1/5`), this shows progressive term handling:
Final Function:
[tex]\[ g(x) = -\frac{1}{10}x + \frac{1}{4} \][/tex]
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.