Call Now to Set Up Tutoring:

(888) 888-0446

Algebra II : Hyperbolic Functions

Study concepts, example questions & explanations for Algebra II

Create An Account

All Algebra II Resources

10 Diagnostic Tests 630 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Hyperbolic Functions

Which of the following equations represents a vertical hyperbola with a center of (-4, 3) and asymptotes at $y=\pm\, \frac{4}{3}\, x$ ?

Possible Answers:

$\frac{(x+4)^{2}}{9}-\frac{(y-3)^{2}}{16}=1$

$\frac{(y-3)^{2}}{9}-\frac{(x+4)^{2}}{16}=1$

$\frac{(y+3)^{2}}{16}-\frac{(x-4)^{2}}{9}=1$

$\frac{(y-3)^{2}}{16}-\frac{(x+4)^{2}}{9}=1$

Correct answer:

$\frac{(y-3)^{2}}{16}-\frac{(x+4)^{2}}{9}=1$

Explanation:

First, we need to become familiar with the standard form of a hyperbolic equation:

$\frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1$

The center is always at (h,k) . This means that for this problem, the numerators of each term will have to contain $(x+4)^{2}$ and $(y-3)^{2}$ .

To determine if a hyperbola opens vertically or horizontally, look at the sign of each variable. A vertical parabola has a positive $y^{2}$ term; a horizontal parabola has a positive $x^{2}$ term. In this case, we need a vertical parabola, so the $y^{2}$ term will have to be positive.

(NOTE: If both terms are the same sign, you have an ellipse, not a parabola.)

The asymptotes of a parabola are always found by the equation $y=\pm\, \frac{b}{a}\, x$ , where is found in the denominator of the $y^{2}$ term and is found in the denominator of the $x^{2}$ term. Since our asymptotes are $y=\pm\, \frac{4}{3}\, x$ , we know that must be 4 and must be 3. That means that the number underneath the $y^{2}$ term has to be 16, and the number underneath the $x^{2}$ term has to be 9.

Report an Error

Example Question #1 : Hyperbolic Functions

What is the shape of the graph depicted by the equation:

$\small \frac{y^2}{9}-\frac{x^2}{4}=1$

Possible Answers:

Hyperbola

Parabola

Circle

Oval

Correct answer:

Hyperbola

Explanation:

The standard equation of a hyperbola is:

$\small \frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

Report an Error

Example Question #3 : Hyperbolic Functions

Express the following hyperbolic function in standard form:

13(y-7)^2-13(x+6)^2=52

Possible Answers:

$\frac{2(y-7)^2}{13}-\frac{2(x+6)^2}{13}=1$

$\frac{(x-7)^2}{4}+\frac{(y+6)^2}{4}=1$

$\frac{(y-7)^2}{4}-\frac{(x+6)^2}{4}=1$

$\frac{(y-7)^2}{13}+\frac{(x+6)^2}{13}=1$

$\frac{(x-7)^2}{52}-\frac{(y+6)^2}{52}=1$

Correct answer:

$\frac{(y-7)^2}{4}-\frac{(x+6)^2}{4}=1$

Explanation:

In order to express the given hyperbolic function in standard form, we must write it in one of the following two ways:

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$

$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1$

From our formulas for the standard form of a hyperbolic equation above, we can see that the term on the right side of the equation is always 1, so we must divide both sides of the given equation by 52, which gives us:

$\frac{13(y-7)^2}{52}-\frac{13(x+6)^2}{52}=1$

Simplifying, we obtain our final answer in standard form:

$\frac{(y-7)^2}{4}-\frac{(x+6)^2}{4}=1$

Report an Error

Example Question #2 : Hyperbolic Functions

Which of the following answers best represent $2\sinh(x)$ ?

Possible Answers:

$\frac{2}{e^x+e^-^x}$

e^x+e^-^x

$\frac{e^x+e^-^x}{2}$

$\frac{e^x-e^-^x}{2}$

e^x-e^-^x

Correct answer:

e^x-e^-^x

Explanation:

The correct definition of hyperbolic sine is:

$\sinh(x)= \frac{e^x-e^-^x}{2}$

Therefore, by multiplying 2 by both sides we get the following answer,

$2\sinh(x)=e^x-e^-^x$

Report an Error

Example Question #5 : Hyperbolic Functions

Which of the following best represents $\cosh(x)-\sinh(x)$ , if the value of is zero?

Possible Answers:

$\frac{1}{2}$

$-\frac{1}{2}$

Correct answer:

Explanation:

Find the values of hyperbolic sine and cosine when x is zero. According to the properties:

$\sinh(0)=\frac{e^x-e^{-x}}{2}=\frac{e^0-e^{-0}}{2}=\frac{1-1}{2}=0$

$\cosh(0)= \frac{e^x+e^{-x}}{2}=\frac{e^0+e^{-0}}{2}=\frac{1+1}{2}=\frac{2}{2}=1$

Therefore:

$\cosh(0)-\sinh(0)=1-0=1$

Report an Error

Example Question #6 : Hyperbolic Functions

What is the value of $-\tanh(0)$ ?

Possible Answers:

$\frac{1}{2}$

$\text{Does not exist}$

Correct answer:

Explanation:

The hyperbolic tangent will need to be rewritten in terms of hyperbolic sine and cosine.

According to the properties:

$\sinh(0)=0$

$\cosh(0)=1$

Therefore:

$-\tanh(0)= -\frac{\sinh(0)}{\cosh(0)}=-\frac{\frac{e^0-e^0}{2}}{\frac{e^0+e^0}{2}}=-\frac{e^0-e^0}{e^0+e^0}=-\frac{0}{1} = 0$

Report an Error

Example Question #7 : Hyperbolic Functions

Simplify: $-\frac{1}{\sinh^2(x)-\cosh^2(x)}$

Possible Answers:

x^2

$\text{Does not exist}$

Correct answer:

Explanation:

The following is a property of hyperbolics that is closely similar to the problem.

$\cosh^2(x) -\sinh^2(x)= 1$

We will need to rewrite this equation by taking a negative one as the common factor, and divide the negative one on both sides.

$(-1)(-\cosh^2(x) +\sinh^2(x))= 1$

$-\cosh^2(x) +\sinh^2(x)= -1$

$\sinh^2(x)-\cosh^2(x) =-1$

Substitute the value into the problem.

$-\frac{1}{sinh^2(x)-cosh^2(x)} = - \frac{1}{-1} =1$

Report an Error

Example Question #8 : Hyperbolic Functions

$\small f(x)=\tfrac{1}{x+4}-3$

Which of the following is the correct expression for a hyperbola that is shifted $\small 5$ $\small units$ units up and $\small 2$ $\small units$ to the right of $\small f(x)$ ?

Possible Answers:

$\small \small \small \small g(x)=\tfrac{1}{x+2}+5$

$\small \small \small g(x)=\tfrac{1}{x+2}+2$

$\small \small \small \small g(x)=\tfrac{1}{x+9}-1$

$\small \small \small \small g(x)=\tfrac{1}{x-2}+5$

$\small \small \small \small g(x)=\tfrac{1}{x+6}-8$

Correct answer:

$\small \small \small g(x)=\tfrac{1}{x+2}+2$

Explanation:

The parent function of a hyperbola is represented by the function $\small \small \small \small g(x)=\tfrac{1}{x-h}+k$ where $\small (h,k)$ is the center of the hyperbola. To shift the original function up by $\small 5$ $\small units$ simply add $\small -3+5$ . To shift it to the right $\small 2$ $\small units$ take away $\small 4-2$ .

Report an Error

Example Question #9 : Hyperbolic Functions

Find the foci of the hyperbola: $-\frac{x^2}{36}+\frac{y^2}{16} =1$

Possible Answers:

$(\pm 2\sqrt{13},0)$

$(0,\pm 2\sqrt{13})$

$(\pm6,0)$

$(\pm 4\sqrt{13},0)$

$(0,\pm4)$

Correct answer:

$(0,\pm 2\sqrt{13})$

Explanation:

Write the standard forms for a hyperbola.

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2} =1$

OR:

$-\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2} =1$

The standard form is given in the second case, which will have different parameters compared to the first form.

Center: (h,k) =(0,0)

Foci: $(h,k\pm c)$ , where c^2 =a^2+b^2

$-\frac{x^2}{36}+\frac{y^2}{16} =1$

Identify the coefficients a,b and substitute to find the value of .

$a^2 = 36, a=\pm6$

$b^2 =16, b=\pm 4$

$c^2 =a^2+b^2\rightarrow c^2 =36+16$

$c^2=52, c=\pm \sqrt{52} =\pm \sqrt{4}\cdot \sqrt{13}= \pm2\sqrt{13}$

$(h,k\pm c) = (0,0 \pm 2\sqrt{13}) = (0,\pm 2\sqrt{13})$

The answer is: $(0,\pm 2\sqrt{13})$

Report an Error

Example Question #10 : Hyperbolic Functions

Given the hyperbola y^2 +4y = 4x^2-24x+41 , what is the value of the center?

Possible Answers:

Correct answer:

Explanation:

In order to determine the center, we will first need to rewrite this equation in standard form.

$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1$

Isolate 41 on the right side. Subtract 4x^2 and add 24x on both sides.

y^2 +4y -(4x^2)+[24x]= 4x^2-24x+41-(4x^2)+[24x]

The equation becomes:

y^2+4y-4x^2+24x = 41

Group the x and y terms. Be careful of the negative signs.

(y^2+4y)-(4x^2-24x )= 41

Pull out a common factor of 4 on the second parentheses.

(y^2+4y)-4(x^2-6x )= 41

Complete the square twice. Divide the second term of each parentheses by two and square the quantity. Add the terms on both sides.

$(y^2+4y+(\frac{4}{2})^2)-4(x^2-6x+(\frac{-6}{3})^2)= 41+(\frac{4}{2})^2+(-4)(\frac{-6}{3})^2$

This equation becomes:

(y^2+4y+4)-4(x^2-6x+9) = 41+4-36

Factorize the left side and simplify the right.

(y+2^2)-4(x-3)^2 = 9

Divide both sides by nine.

$\frac{(y+2)^2}{9}-\frac{4}{9}(x-3)^2 = \frac{9}{9}$

$\frac{(y+2)^2}{9}-\frac{4(x-3)^2 }{9}=1$

The equation is now in the standard form of a hyperbola.

The center is at: (3,-2)

The answer is:

Report an Error

View Algebra 2 Tutors

Matt
Certified Tutor

Oakland University, Bachelor of Science, Electrical Engineering.

View Algebra 2 Tutors

Joseph
Certified Tutor

Old Dominion University, Bachelor of Science, Electrical Engineering. Western Governors University, Masters in Education, Mat...

View Algebra 2 Tutors

Blossom
Certified Tutor

University of Maryland-Baltimore County, Bachelor of Science, Computer Science.

All Algebra II Resources

10 Diagnostic Tests 630 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Algebra Tutors in Seattle, Algebra Tutors in Phoenix, GMAT Tutors in San Diego, GRE Tutors in San Francisco-Bay Area, Physics Tutors in Miami, MCAT Tutors in Chicago, Reading Tutors in Philadelphia, French Tutors in San Francisco-Bay Area, English Tutors in Boston, ACT Tutors in Phoenix

Popular Courses & Classes

GMAT Courses & Classes in San Diego, GMAT Courses & Classes in Seattle, LSAT Courses & Classes in Washington DC, GRE Courses & Classes in Los Angeles, LSAT Courses & Classes in San Francisco-Bay Area, ISEE Courses & Classes in Los Angeles, MCAT Courses & Classes in Washington DC, SSAT Courses & Classes in Houston, Spanish Courses & Classes in Los Angeles, ACT Courses & Classes in Miami

Popular Test Prep

LSAT Test Prep in Dallas Fort Worth, LSAT Test Prep in Houston, SSAT Test Prep in San Diego, ACT Test Prep in Miami, ACT Test Prep in Chicago, SAT Test Prep in Los Angeles, GRE Test Prep in Los Angeles, ISEE Test Prep in Philadelphia, LSAT Test Prep in Denver, MCAT Test Prep in Houston

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

Algebra II : Hyperbolic Functions

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1 : Hyperbolic Functions

Example Question #1 : Hyperbolic Functions

Example Question #3 : Hyperbolic Functions

Example Question #2 : Hyperbolic Functions

Example Question #5 : Hyperbolic Functions

Example Question #6 : Hyperbolic Functions

Example Question #7 : Hyperbolic Functions

Example Question #8 : Hyperbolic Functions

Example Question #9 : Hyperbolic Functions

Example Question #10 : Hyperbolic Functions

All Algebra II Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: