MHB How can I find the range of a hyperbolic curve using its graph?

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SUMMARY

The discussion focuses on determining the range of the hyperbolic curve defined by the function y = 1/x. The range is established as y ∈ ℝ, y ≠ 0, due to the presence of vertical and horizontal asymptotes at x = 0 and y = 0, respectively. Participants emphasize the importance of understanding these asymptotes for interpreting the graph correctly. Additionally, the discussion highlights the relationship between the hyperbolic curve and its representation through the equation x² - y² = 1, demonstrating how axis rotation can transform the graph.

PREREQUISITES
  • Understanding of hyperbolic functions and their properties
  • Familiarity with asymptotes in graphing functions
  • Basic knowledge of graph transformations, including rotation of axes
  • Ability to interpret mathematical notation and inequalities
NEXT STEPS
  • Study the properties of hyperbolic functions, specifically y = 1/x
  • Learn about vertical and horizontal asymptotes in detail
  • Explore graph transformations, particularly axis rotation techniques
  • Watch educational videos on identifying domain and range from graphs
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Students learning precalculus, mathematics educators, and anyone seeking to deepen their understanding of hyperbolic curves and their graphical representations.

mathdad
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Find the range using the graph of y.

y = 1/x

This function is weird. It has a curve in quadrants 1 and 3 that does not cross the lines y = 0 and x = 0.

How can I determine the range of such a graph?
 
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domain is $x \in \mathbb{R}, \, x \ne 0$

range is $y \in \mathbb{R}, \, y \ne 0$

why so difficult to interpret the graph?[DESMOS=-10.10016694490818,9.89983305509182,-10.45,9.55]y=\frac{1}{x}[/DESMOS]
 
I know what this graph looks like. I've seen it hundreds of times but what does it mean to a novice math learner? There is a curve in quadrants 1 and 3 that does not cross the lines x = 0 and y = 0. The textbook answer is (-infinity, 0) U (0, infinity). What on the graph tells me that this is the correct range?
 
There is a curve in quadrants 1 and 3 that does not cross the lines x = 0 and y = 0.

if a graph of a function does not cross a vertical line like x = 0, then that line is a vertical asymptote ... x = 0 is excluded from the function's domain.

if a graph of a function does not cross a horizontal line like y = 0, then that line is a horizontal asymptote ... y = 0 is excluded from the function's range.

can't put it any plainer than that ... maybe you should select video(s) from the link for alternative, non-textbook explanations

https://www.google.com/search?q=ide..._eHUAhUY6GMKHSHzDhAQ_AUICigB&biw=1366&bih=638
 
I'll seek more video help than textbooks. I can find just about anything on youtube.com. I will use this site when a video clip makes no sense.
 
RTCNTC said:
I know what this graph looks like. I've seen it hundreds of times but what does it mean to a novice math learner? There is a curve in quadrants 1 and 3 that does not cross the lines x = 0 and y = 0. The textbook answer is (-infinity, 0) U (0, infinity). What on the graph tells me that this is the correct range?

It may interest you to know that the curve:

$$y=\frac{1}{x}$$

is actually a hyperbolic curve. Consider the graph of the hyperbolic curve:

$$x^2-y^2=1$$

and its asymptotes, given by:

$$x^2=y^2$$

[DESMOS=-5,5,-1.7346053772766694,1.7346053772766694]x^2-y^2=1;x^2=y^2[/DESMOS]

It can be shown that by rotating our axes by $$\frac{\pi}{4}$$, the graphed hyperbola becomes:

$$y=\frac{1}{x}$$

I'll wait until you get to the section on rotation of axes before we explore that further. :D
 
MarkFL said:
It may interest you to know that the curve:

$$y=\frac{1}{x}$$

is actually a hyperbolic curve. Consider the graph of the hyperbolic curve:

$$x^2-y^2=1$$

and its asymptotes, given by:

$$x^2=y^2$$
It can be shown that by rotating our axes by $$\frac{\pi}{4}$$, the graphed hyperbole becomes:

$$y=\frac{1}{x}$$

I'll wait until you get to the section on rotation of axes before we explore that further. :D

Cool. Continue to answer my questions. I appreciate your guidance through my review of a great course. I will use youtube precalculus clips to review each chapter and post questions here when needed. Check your inbox.
 

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