Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Determine whether \(y\) is a function of \(x\). $$ y^{2}=x^{2}-1 $$

Short Answer

Expert verified
No, \(y\) is not a function of \(x\) in the equation \(y^{2}=x^{2}-1\).

Step by step solution

01

Understand the Definition of Function

The first step is to understand what makes an equation a function. In simple terms, each input \(x\) in a function must correspond to exactly one output \(y\). This is generally known as the vertical line test. It means if the vertical line crosses the graph at more than one point, then the graph does not represent a function.
02

Rearrange the Equation to Solve for y

In order to determine if \(y\) is a function of \(x\) in the equation \(y^{2}=x^{2}-1\), we need to solve the equation for \(y\). Taking the square root of both sides to isolate \(y\) we get \(y=\pm\sqrt{x^2-1}\). In this case, for each calculated \(x\) value, there will be two possible \(y\) values (a positive root and a negative root), unless \(x = 1\) or \(x = -1\), in which case there will be one real \(y\) value (0).
03

Determine Whether the Equation is a Function

Since there is more than one possible \(y\) value for most \(x\) values, the equation \(y=\pm\sqrt{x^2-1}\) does not meet the criterion of a function requiring each input \(x\) to correspond to exactly one output \(y\). Therefore, \(y\) is not a function of \(x\) in the given equation.

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 Vaia!

One App. One Place for Learning.

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

Get started for free

Study anywhere. Anytime. Across all devices.

Sign-up for free