Shrink the graph of f(x)=x+5 horizontally by a factor of 1\/2<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To shrink the graph of ( f(x) = x + 5 ) horizontally by a factor of ( \\frac{1}{2} ), we need to modify the function’s equation in a way that compresses it along the x-axis. A horizontal shrink is achieved by altering the argument of the function, specifically by multiplying ( x ) by a constant.<\/p>\n\n\n\n The graph of ( f(x) = x + 5 ), after shrinking horizontally by a factor of ( \\frac{1}{2} ), becomes ( f(x) = 2x + 5 ). This transformation causes the graph to compress, and the x-values need to change more to achieve the same output values for ( y ).<\/p>\n","protected":false},"excerpt":{"rendered":" Shrink the graph of f(x)=x+5 horizontally by a factor of 1\/2 The Correct Answer and Explanation is : To shrink the graph of ( f(x) = x + 5 ) horizontally by a factor of ( \\frac{1}{2} ), we need to modify the function’s equation in a way that compresses it along the x-axis. A […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-162531","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/162531","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=162531"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/162531\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=162531"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=162531"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=162531"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step-by-step Explanation:<\/h3>\n\n\n\n
\n
A horizontal shrink by a factor of ( \\frac{1}{2} ) means that the function will be compressed so that the x-values will need to increase more rapidly in order to achieve the same output. For any general function ( f(x) ), if we apply a horizontal shrink by a factor of ( k ), the new function will be ( f(kx) ), where ( k > 1 ) results in a shrink. In this case, the factor is ( \\frac{1}{2} ), so the transformation will be applied by replacing ( x ) with ( 2x ). This is because a horizontal shrink by ( \\frac{1}{2} ) is the same as scaling ( x ) by ( 2 ), causing the graph to compress towards the y-axis.<\/li>\n\n\n\n
The original function is:
[
f(x) = x + 5
]
After applying the horizontal shrink by a factor of ( \\frac{1}{2} ), we replace ( x ) with ( 2x ). Thus, the transformed function is:
[
f(2x) = 2x + 5
]<\/li>\n\n\n\n\n
Conclusion:<\/h3>\n\n\n\n