Call Now to Set Up Tutoring:

(888) 888-0446

Precalculus : Determine if a Relation is a Function

Study concepts, example questions & explanations for Precalculus

Create An Account

All Precalculus Resources

12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Determine If A Relation Is A Function

Which of the following expressions is not a function?

Possible Answers:

y=5x+99 $\displaystyle y=5x+99$

$\displaystyle y=2^x+343$

$y=\pm\sqrt{x^2-7x+5}$ $\displaystyle y=\pm\sqrt{x^2-7x+5}$

$\displaystyle y=6x^2-54x+76$

y=sinx $\displaystyle y=sinx$

Correct answer:

$y=\pm\sqrt{x^2-7x+5}$ $\displaystyle y=\pm\sqrt{x^2-7x+5}$

Explanation:

Recall that an expression is only a function if it passes the vertical line test. Test this by graphing each function and looking for one which fails the vertical line test. (The vertical line test consists of drawing a vertical line through the graph of an expression. If the vertical line crosses the graph of the expression more than once, the expression is not a function.)

Functions can only have one y value for every x value. The only choice that reflects this is:

$y=\pm\sqrt{x^2-7x+5}$ $\displaystyle y=\pm\sqrt{x^2-7x+5}$

Report an Error

Example Question #1 : Determine If A Relation Is A Function

Suppose we have the relation $\small (a,b)$ $\displaystyle \small (a,b)$ on the set of real numbers $\mathbb{R}$ $\displaystyle \mathbb{R}$ whenever $\small a<b$ $\displaystyle \small a< b$ . Which of the following is true.

Possible Answers:

The relation is a function because for every $\small a$ $\displaystyle \small a$ , there is only one $\small b$ $\displaystyle \small b$ such that $\small (a,b)$ $\displaystyle \small (a,b)$ holds.

The relation is a function because every relation is a function, since that's how relations are defined.

The relation is not a function because $\small (2,3)$ $\displaystyle \small (2,3)$ and $\small (2,4)$ $\displaystyle \small (2,4)$ both hold.

The relation is a function because $\small (2,4)$ $\displaystyle \small (2,4)$ holds and $\small (3,4)$ $\displaystyle \small (3,4)$ also holds.

The relation is not a function because $\small (0,1)$ $\displaystyle \small (0,1)$ holds but $\small (1,0)$ $\displaystyle \small (1,0)$ does not.

Correct answer:

The relation is not a function because $\small (2,3)$ $\displaystyle \small (2,3)$ and $\small (2,4)$ $\displaystyle \small (2,4)$ both hold.

Explanation:

The relation is not a function because $\small (2,3)$ $\displaystyle \small (2,3)$ and $\small (2,4)$ $\displaystyle \small (2,4)$ hold. If it were a function, $\small (2,b)$ $\displaystyle \small (2,b)$ would hold only for one $\small b$ $\displaystyle \small b$ . But we know it holds for $\small b=3,4$ $\displaystyle \small b=3,4$ because $\small 2<3$ $\displaystyle \small 2< 3$ and $\small 2<4$ $\displaystyle \small 2< 4$ . Thus, the relation $\small (a,b)$ $\displaystyle \small (a,b)$ on the set of real numbers $\small \mathbb{R}$ $\displaystyle \small \mathbb{R}$ is not a function.

Report an Error

Example Question #3 : Determine If A Relation Is A Function

Consider a family consisting of a two parents, Juan and Oksana, and their daughters Adriana and Laksmi. A relation $\small (a,b)$ $\displaystyle \small (a,b)$ is true whenever $\small b$ $\displaystyle \small b$ is the child of $\small a$ $\displaystyle \small a$ . Which of the following is not true?

Possible Answers:

Even if the two parents had only one daughter, the relation would not be a function.

If the two parents had only one daughter, the relation would be a function.

The relation is not a function because (Laksmi,Juan) and (Laksmi,Oksana) both hold.

(Adriana,Laksmi) does not hold because Laksmi is not Adriana's child and is a boy.

(Laksmi,Adriana) does not hold because Adriana is not Laksmi's child.

Correct answer:

Even if the two parents had only one daughter, the relation would not be a function.

Explanation:

The statement

"Even if the two parents had only one daughter, the relation would not be a function."

is not true because if they had only one daughter, say Adriana, then the only relations that would exist would be (Juan, Adriana) and (Oksana,Adriana), which defines a function.

Report an Error

Example Question #3 : Determine If A Relation Is A Function

Which of the following relations is not a function?

Possible Answers:

y=2 $\displaystyle y=2$

x = y^2 $\displaystyle x = y^2$

$y = x^{\frac{3}{2}}$ $\displaystyle y = x^{\frac{3}{2}}$

$y = \sqrt{x}$ $\displaystyle y = \sqrt{x}$

Correct answer:

x = y^2 $\displaystyle x = y^2$

Explanation:

The definition of a function requires that for each input (i.e. each value of $\displaystyle x$ ), there is only one output (i.e. one value of $\displaystyle y$ ). For x = y^2 $\displaystyle x = y^2$ , each value of $\displaystyle x$ corresponds to two values of $\displaystyle y$ (for example, when x = 4 $\displaystyle x = 4$ , both y=2 $\displaystyle y=2$ and y = -2 $\displaystyle y = -2$ are correct solutions). Therefore, this relation cannot be a function.

Report an Error

Example Question #61 : Relations And Functions

Given the set of ordered pairs, determine if the relation is a function

(0,2) $\displaystyle (0,2)$

(1,4) $\displaystyle (1,4)$

(2,6) $\displaystyle (2,6)$

(3,8) $\displaystyle (3,8)$

(2,5) $\displaystyle (2,5)$

Possible Answers:

Yes

Cannot be determined

Correct answer:

Explanation:

A relation is a function if no single x-value corresponds to more than one y-value.

Because the mapping from $\displaystyle 2$ goes to $\displaystyle 5$ and $\displaystyle 6$

the relation is NOT a function.

Report an Error

Example Question #1 : Determine If A Relation Is A Function

What equation is perpendicular to $y = \frac{5}{2}x+12$ $\displaystyle y = \frac{5}{2}x+12$ and passes throgh (2,5) $\displaystyle (2,5)$ ?

Possible Answers:

$y = -\frac{5}{2}x + \frac{22}{5}$ $\displaystyle y = -\frac{5}{2}x + \frac{22}{5}$

$y = -\frac{2}{5}x + \frac{29}{5}$ $\displaystyle y = -\frac{2}{5}x + \frac{29}{5}$

$y = -\frac{4}{5}x + \frac{26}{5}$ $\displaystyle y = -\frac{4}{5}x + \frac{26}{5}$

$y = \frac{2}{5}x + \frac{21}{5}$ $\displaystyle y = \frac{2}{5}x + \frac{21}{5}$

$y = -\frac{2}{5}x + \frac{27}{5}$ $\displaystyle y = -\frac{2}{5}x + \frac{27}{5}$

Correct answer:

$y = -\frac{2}{5}x + \frac{29}{5}$ $\displaystyle y = -\frac{2}{5}x + \frac{29}{5}$

Explanation:

First find the reciprocal of the slope of the given function.
$m_2 = -\frac{1}{m_1}$ $\displaystyle m_2 = -\frac{1}{m_1}$

$m_2 = -\frac{2}{5}$ $\displaystyle m_2 = -\frac{2}{5}$

The perpendicular function is:

$y = -\frac{2}{5}x+c$ $\displaystyle y = -\frac{2}{5}x+c$

Now we must find the constant, $\displaystyle c$ , by using the given point that the perpendicular crosses.

$5 = -\frac{2}{5}(2)+c$ $\displaystyle 5 = -\frac{2}{5}(2)+c$

solve for $\displaystyle c$ :

$c = \frac{29}{5}$ $\displaystyle c = \frac{29}{5}$

$y = -\frac{2}{5}x + \frac{29}{5}$ $\displaystyle y = -\frac{2}{5}x + \frac{29}{5}$

Report an Error

Example Question #72 : Functions

Is the following relation of ordered pairs a function?

$\displaystyle (1,1); (2,2); (3,1); (4,2); (5,1); (6,7)$

Possible Answers:

Cannot be determined

Yes

Correct answer:

Yes

Explanation:

A set of ordered pairs is a function if it passes the vertical line test.

Because there are no more than one corresponding $\displaystyle y$ value for any given $\displaystyle x$ value, the relation of ordered pairs IS a function.

Report an Error

View Pre-Calculus Tutors

Peter
Certified Tutor

Nottingham University UK, Bachelor in Arts, Mathematics.

View Pre-Calculus Tutors

Alejandro
Certified Tutor

Kennesaw State University, Bachelor of Science, Computer Software Engineering.

View Pre-Calculus Tutors

Shane
Certified Tutor

Temple University, Bachelor of Science, Mathematics and Computer Science.

All Precalculus Resources

12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

MCAT Tutors in Dallas Fort Worth, MCAT Tutors in Miami, Algebra Tutors in Washington DC, Spanish Tutors in Atlanta, MCAT Tutors in Atlanta, Computer Science Tutors in New York City, Math Tutors in Washington DC, ACT Tutors in Dallas Fort Worth, ACT Tutors in Boston, Chemistry Tutors in Dallas Fort Worth

Popular Courses & Classes

LSAT Courses & Classes in San Francisco-Bay Area, LSAT Courses & Classes in Chicago, GMAT Courses & Classes in Denver, ISEE Courses & Classes in Miami, Spanish Courses & Classes in Denver, SAT Courses & Classes in Houston, LSAT Courses & Classes in Miami, ISEE Courses & Classes in New York City, ISEE Courses & Classes in San Diego, SAT Courses & Classes in Boston

Popular Test Prep

SSAT Test Prep in Houston, SSAT Test Prep in Boston, ISEE Test Prep in Dallas Fort Worth, GMAT Test Prep in Seattle, GRE Test Prep in Boston, SSAT Test Prep in Miami, MCAT Test Prep in Atlanta, SAT Test Prep in Houston, ISEE Test Prep in New York City, GMAT Test Prep in Denver

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Precalculus : Determine if a Relation is a Function

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Determine If A Relation Is A Function

Example Question #1 : Determine If A Relation Is A Function

Example Question #3 : Determine If A Relation Is A Function

Example Question #3 : Determine If A Relation Is A Function

Example Question #61 : Relations And Functions

Example Question #1 : Determine If A Relation Is A Function

Example Question #72 : Functions

All Precalculus Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: