/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 54 Write an equation for the functi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Write an equation for the function described by the given characteristics. The shape of \(f(x)=\sqrt{x},\) but shifted nine units down and then reflected in both the \(x\) -axis and the \(y\) -axis

Short Answer

Expert verified
The function is \(h(x) = 9 - \sqrt{-x}\) for \(x \le 0\).

Step by step solution

01

Shift Nine Units Down

Shifting a function nine units down effectively translates to subtracting 9 from each y-coordinate of the function. For the function \(f(x)=\sqrt{x}\), let's denote \(g(x)\) as \(g(x) = \sqrt{x} - 9\).
02

Reflect Along the X-Axis

Reflecting a function along the x-axis means changing the sign of the y-coordinate. Therefore, after the reflection, say \(\varphi(x)\), we get \(\varphi(x) = -g(x) = -(\sqrt{x} - 9)\). Simplifying this gives us \(\varphi(x) = 9 - \sqrt{x}\).
03

Reflect Along the Y-Axis

Reflecting a function along the y-axis means changing the sign of the x-coordinate. Therefore, after the reflection, let's denote the final function as \(h(x)\), we get \(h(x) = \varphi(-x) = 9 - \sqrt{-x}\). We should swiftly note that the square root of a negative number is not a real number. Therefore, this function is not continuous and it has its domain limited to \(x \le 0\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.