{"id":23436,"date":"2025-06-17T13:16:05","date_gmt":"2025-06-17T13:16:05","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=23436"},"modified":"2025-06-17T13:16:08","modified_gmt":"2025-06-17T13:16:08","slug":"which-equation-is-the-inverse-of-y-16x2-1","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/which-equation-is-the-inverse-of-y-16x2-1\/","title":{"rendered":"Which equation is the inverse of y = 16×2 + 1"},"content":{"rendered":"\n

Which equation is the inverse of y = 16×2 + 1? D. y= +or- sqrt x-1\/4 is the correct answer<\/p>\n\n\n\n

The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n

To find the inverse<\/strong> of the equation: y=16×2+1y = 16x^2 + 1<\/p>\n\n\n\n

we must follow a step-by-step algebraic process<\/strong> to switch the input and output variables and solve for the new dependent variable.<\/p>\n\n\n\n

\n\n\n\n

Step 1: Replace yy with f(x)f(x)<\/h3>\n\n\n\n

f(x)=16×2+1f(x) = 16x^2 + 1<\/p>\n\n\n\n

Step 2: Swap xx and yy<\/h3>\n\n\n\n

x=16y2+1x = 16y^2 + 1<\/p>\n\n\n\n

This step is essential because inverses reverse the roles<\/strong> of inputs and outputs.<\/p>\n\n\n\n

\n\n\n\n

Step 3: Solve for yy<\/h3>\n\n\n\n

Start with: x=16y2+1x = 16y^2 + 1<\/p>\n\n\n\n

Subtract 1 from both sides: x\u22121=16y2x – 1 = 16y^2<\/p>\n\n\n\n

Divide both sides by 16: x\u2212116=y2\\frac{x – 1}{16} = y^2<\/p>\n\n\n\n

Now take the square root of both sides: y=\u00b1x\u2212116y = \\pm \\sqrt{\\frac{x – 1}{16}}<\/p>\n\n\n\n

This simplifies to: y=\u00b1x\u221214y = \\pm \\frac{\\sqrt{x – 1}}{4}<\/p>\n\n\n\n

\n\n\n\n

Final Answer:<\/h3>\n\n\n\n

The inverse equation is: y=\u00b1x\u221214y = \\pm \\frac{\\sqrt{x – 1}}{4}<\/p>\n\n\n\n

Which matches choice D<\/strong> (when written as): y=\u00b1x\u2212116ory=\u00b1x\u221214y = \\pm \\sqrt{\\frac{x – 1}{16}} \\quad \\text{or} \\quad y = \\pm \\frac{\\sqrt{x – 1}}{4}<\/p>\n\n\n\n

\n\n\n\n

Interpretation:<\/h3>\n\n\n\n
    \n
  • The \u00b1 (plus or minus)<\/strong> symbol appears because the original function y=16×2+1y = 16x^2 + 1 is not one-to-one<\/strong> (it is a parabola), and its inverse will not pass the vertical line test<\/strong> unless we restrict the domain.<\/li>\n\n\n\n
  • This means for each xx in the inverse, there are two possible values<\/strong> of yy, unless a restriction is applied (e.g., x\u22650x \\geq 0).<\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    Conclusion:<\/h3>\n\n\n\n

    \u2705 Correct answer: D.<\/strong> y=\u00b1x\u2212116ory=\u00b1x\u221214y = \\pm \\sqrt{\\frac{x – 1}{16}} \\quad \\text{or} \\quad y = \\pm \\frac{\\sqrt{x – 1}}{4}<\/p>\n\n\n\n

    This is the inverse of y=16×2+1y = 16x^2 + 1.<\/p>\n","protected":false},"excerpt":{"rendered":"

    Which equation is the inverse of y = 16×2 + 1? D. y= +or- sqrt x-1\/4 is the correct answer The correct answer and explanation is: To find the inverse of the equation: y=16×2+1y = 16x^2 + 1 we must follow a step-by-step algebraic process to switch the input and output variables and solve for […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-23436","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/23436","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=23436"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/23436\/revisions"}],"predecessor-version":[{"id":23438,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/23436\/revisions\/23438"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=23436"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=23436"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=23436"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}