Problem 20 Graph each hyperbola. Give the d... [FREE SOLUTION]

91影视

Precalculus Student Solutions Manual 5th

Margaret L. Lial, John Hornsby, David I. Schneider

$Math Studyset 91影视 Explanations$ Math

5 Edition

Chapter 10: Problem 20

Graph each hyperbola. Give the domain, range, center, vertices, foci, and equations of the asymptotes for each figure. $$\frac{(y+5)^{2}}{4}-\frac{(x-1)^{2}}{16}=1$$

Short Answer

Expert verified

Center: (1, -5), Vertices: (1, -3) and (1, -7), Foci: (1, -5+2sqrt{5}) and (1, -5-2sqrt{5}), Asymptotes: y = -5 卤 (1/2)(x - 1), Domain: (-鈭�, 鈭�), Range: (-鈭�, -7) 鈭� (-3, 鈭�)

Step by step solution

- Identify the Standard Form

The given equation of the hyperbola is: \[\frac{(y+5)^{2}}{4}-\frac{(x-1)^{2}}{16}=1\]This is in the standard form of \[\frac{(y-k)^{2}}{a^{2}}-\frac{(x-h)^{2}}{b^{2}}=1\]The hyperbola is vertical because the $y$-term is positive and comes first.

- Identify the Center

Compare the given equation to the standard form. Here, $(h, k)$ is the center of the hyperbola.So, we have $h = 1$ and $k = -5$. The center is $(1, -5)$.

- Find the Vertices

For a vertical hyperbola, the vertices are $(h, k+a)$ and $(h, k-a)$.Using $a^{2} = 4$, we get $a = 2$.So, the vertices are $(1, -5+2) = (1, -3)$ and $(1, -5-2) = (1, -7)$.

- Find the Foci

The foci are found using the formula $c^{2} = a^{2} + b^{2}$.Here, $a^{2} = 4$ and $b^{2} = 16$.So, $c^{2} = 4 + 16 = 20$ which means $c = \sqrt{20} = 2\sqrt{5}$.Therefore, the foci are at $(1, -5+2\sqrt{5})$ and $(1, -5-2\sqrt{5})$.

- Find the Equations of the Asymptotes

For a vertical hyperbola, the asymptotes are given by $y - k = \pm \frac{a}{b}(x - h)$.Substituting $a = 2$, $b = 4$, $h = 1$, and $k = -5$, the equations of the asymptotes are:\[y + 5 = \pm \frac{2}{4}(x - 1)\]Simplifying, we get:\[y + 5 = \pm \frac{1}{2}(x - 1)\]Or,\[y = -5 \pm \frac{1}{2}(x - 1)\].

- Determine the Domain

For a hyperbola that opens vertically, the domain is all real numbers, $(\-\infty, \infty)$.

- Determine the Range

The range includes all values of $y$ such that $y < k - a$ or $y > k + a$.Substituting $k = -5$ and $a = 2$, the range is $(\-\infty, -7) \cup (-3, \infty)$.

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 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Domain and Range of a Hyperbola

Understanding the domain and range of a hyperbola is crucial for graphing it correctly.
For a vertical hyperbola, like the one given by the equation $\frac{(y+5)^{2}}{4}-\frac{(x-1)^{2}}{16}=1$, let's break down both terms a bit more.

Domain: The domain of a hyperbola that opens vertically is all real numbers, which is written as $(-fty, +fty)$. This is because, for any real value of $x$, there is a corresponding $y$ value that satisfies the hyperbola equation.

Range: The range of this hyperbola is all $y$ values that make $y < k - a$ or $y > k + a$. For our equation, with $k = -5$ and $a = 2$, we get the intervals $(y < -7) \cup (y > -3)$.

Simply put:

The domain tells you how wide the hyperbola stretches horizontally.
The range tells you the vertical stretch.

Asymptotes of a Hyperbola

Asymptotes are crucial for understanding the hyperbola's behavior, especially as it moves away from the center. They help guide your graph.
For the equation $\frac{(y+5)^{2}}{4}-\frac{(x-1)^{2}}{16}=1$, we'll determine the equations of the asymptotes like this:

Note that it is a vertical hyperbola.
The asymptote equations for vertical hyperbolas are given by:
$y - k = \pm \frac{a}{b}(x - h)$.
Using values $a = 2$, $b = 4$, $h = 1$, and $k = -5$, the asymptote equations become:
$y + 5 = \pm \frac{2}{4}(x - 1)$.
After simplification, we have:
$y + 5 = \pm \frac{1}{2}(x - 1)$ or $y = -5 \pm \frac{1}{2}(x - 1) $.

Remember:

Asymptotes indicate the lines that the hyperbola approaches, but never touches.
They help define the hyperbola's direction and shape.

Vertices of a Hyperbola

The vertices are vital anchor points that provide crucial information about the hyperbola's structure.
For a vertical hyperbola like our given equation $\frac{(y+5)^{2}}{4}-\frac{(x-1)^{2}}{16}=1$, the vertices are found using the formula $h, k \pm \ a$.

Formula: For our hyperbola, where $h = 1$, $k = -5$, and $a = 2$, the vertices are computed as follows:

(1, -5 + 2) = (1, -3)
(1, -5 - 2) = (1, -7)

The vertices tell us where the hyperbola crosses its axis of symmetry.

Key Points:

The distance between the center and each vertex is$ a = 2 $.
The vertices lie directly above and below the center in a vertical hyperbola.

One App. One Place for Learning.

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

Get started for free

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

91影视

Graph each hyperbola. Give the domain, range, center, vertices, foci, and equations of the asymptotes for each figure. $$\frac{(y+5)^{2}}{4}-\frac{(x-1)^{2}}{16}=1$$

Short Answer

Step by step solution

- Identify the Standard Form

- Identify the Center

- Find the Vertices

- Find the Foci

- Find the Equations of the Asymptotes

- Determine the Domain

- Determine the Range

Key Concepts

Domain and Range of a Hyperbola

Asymptotes of a Hyperbola

Vertices of a Hyperbola

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Probability and Statistics

Logic and Functions

Mechanics Maths

Theoretical and Mathematical Physics

Pure Maths

Study anywhere. Anytime. Across all devices.