Radicals and Functions: How to Graph and Interpret Equations with Radicals

[lLovePhysics]

Is: y^{2}=x^{2}-1 a hyperbola or half a hyperbola? My textbook claims that it is a NOTfunction so I guess it is a hyperbola... However, I'm still perplexed by radicals. They simply confuse me because I don't know if there are +/- signs in front of it or not. When graphing equations such as the above, do you take the positive, negative, or both of the radical? The same question applies for sleeping (horizontal) parabolas. I know that the calculator only draws one half, but in reality, are the two sided or just the positive half side?

If the other side is present are these kind of graphs all NOT functions then? Thanks in advance.

EDIT: Sorry I mean to say that my textbook said it was NOT a function...

 

[radou]

A simple view is this one: if there exists a vertical line which has more than one intersection with the graph, then it is not a function.

 

[HallsofIvy]

Is: y^{2}=x^{2}-1 a hyperbola or half a hyperbola? My textbook claims that it is a function so I guess it is a hyperbola... However, I'm still perplexed by radicals. They simply confuse me because I don't know if there are +/- signs in front of it or not. When graphing equations such as the above, do you take the positive, negative, or both of the radical? The same question applies for sleeping (horizontal) parabolas. I know that the calculator only draws one half, but in reality, are the two sided or just the positive half side?

If the other side is present are these kind of graphs all NOT functions then? Thanks in advance.

y^2= x^2- 1 is exactly the same as x^2- y^2= 1, which is, of course, a hyperbola and so not a function. That seems a strange mistake for a textbook to make so I suggest you reread it to see if they aren't saying something a little different.

Of course, if you restrict y to be non-negative, this is the same as y= \sqrt{x^2- 1} which is a function, or if you restrict y to be non-positive,you get y= -\sqrt{x^2- 1} which is a function.

(Your calculator draws only one half of a hyperbola or parabola with horizontal axis because it only, in that mode, graphs functions.)

 

[lLovePhysics]

Sorry, I mean to say that my textbook say it was NOT a function.

I see now. So unless they give you a specific range of y, you have to always assume that it is the whole graph??

For example:y=\sqrt{x} is always a sleeping (sideways) parabola (and thus not a function) unless a specific y range is stated?

y^{2}=x^{2}-1 is *always* a hyperbola (and thus not a function) unless a specifc y range is stated?

So whenever given one of these types of equations without a specific y range stated, you must state that they are NOT a function? Thanks a lot!

 

[lLovePhysics]

y^2= x^2- 1 is exactly the same as x^2- y^2= 1, which is, of course, a hyperbola and so not a function. That seems a strange mistake for a textbook to make so I suggest you reread it to see if they aren't saying something a little different.

Of course, if you restrict y to be non-negative, this is the same as y= \sqrt{x^2- 1} which is a function, or if you restrict y to be non-positive,you get y= -\sqrt{x^2- 1} which is a function.

(Your calculator draws only one half of a hyperbola or parabola with horizontal axis because it only, in that mode, graphs functions.)

In what mode do they graph the whole graph? Is there a specific name for *non-functions* or do you just call them equations or non-functions?

Usually, I know the orientation of the graph for like sleeping parabolas and hyperbolas. However, sometimes the graph is missing a side and I don't know so I want a way to graph the whole thing. I know I can put a negative in front of the equation but is that the only way?

 

[lLovePhysics]

Do hyperbolas always have the equation form: x^{2}-y^{2}=1 or can it take the rational function form too? I've thought (and I think heard from my teachers) that hyperbolas oriented so that the vertex of the graphs are on the y=x axis. Am I getting confused here or what?

 

[HallsofIvy]

Do hyperbolas always have the equation form: x^{2}-y^{2}=1 or can it take the rational function form too? I've thought (and I think heard from my teachers) that hyperbolas oriented so that the vertex of the graphs are on the y=x axis.

?? Hyperbolas are geometric figures. You can have a hyperbola without any coordinate axes at all!

The very simplest form for a hyperbola is
\frac{x^2}{a^2}- \frac{y^3}{b^2}= 1 or
\frac{y^2}{b^2}- \frac{x^3}{a^2}= 1
Both give hyperbolas with center at (0,0) (NOT the "vertex"- a hyperbola has two vertices equally distanced from the center) and with axes along the x and y- axes.

However, hyperbolas with center at (x_0,y_0) rather than (0,0), but still with axes parallel to the coordinate axes, can be written
\frac{(x-x_0)^2}{a^2}- \frac{(y-y_0)^3}{b^2}= 1

Hyperbolas with axes NOT parallel to the coordinate axes are hardest of all. Basically, any equation of the form Ax^2+ Bxy+ Cy^2+ Dx+ Ey+ G= 0] has a graph which is a "conic section": parabola, ellipse, circle, hyperbola, or one of the "degenerate conics" (one point, one line, two parallel lines, two intersecting lines). There is a formula which, using the numbers A, B, C, D, E, G, will tell you which but I can never remember it! I would, instead, rotate the coordinate system, using a substitution like x= x' cos(\theta)- y' sin(\theta), y= x' sin(\theta)+ y' cos(\theta) choosing theta so that the coefficient of x'y' is 0.

 

[lLovePhysics]

?? Hyperbolas are geometric figures. You can have a hyperbola without any coordinate axes at all!

The very simplest form for a hyperbola is
\frac{x^2}{a^2}- \frac{y^3}{b^2}= 1 or
\frac{y^2}{b^2}- \frac{x^3}{a^2}= 1
Both give hyperbolas with center at (0,0) (NOT the "vertex"- a hyperbola has two vertices equally distanced from the center) and with axes along the x and y- axes.

However, hyperbolas with center at (x_0,y_0) rather than (0,0), but still with axes parallel to the coordinate axes, can be written
\frac{(x-x_0)^2}{a^2}- \frac{(y-y_0)^3}{b^2}= 1

Hyperbolas with axes NOT parallel to the coordinate axes are hardest of all. Basically, any equation of the form Ax^2+ Bxy+ Cy^2+ Dx+ Ey+ G= 0] has a graph which is a "conic section": parabola, ellipse, circle, hyperbola, or one of the "degenerate conics" (one point, one line, two parallel lines, two intersecting lines). There is a formula which, using the numbers A, B, C, D, E, G, will tell you which but I can never remember it! I would, instead, rotate the coordinate system, using a substitution like x= x' cos(\theta)- y' sin(\theta), y= x' sin(\theta)+ y' cos(\theta) choosing theta so that the coefficient of x'y' is 0.

<br /> <br /> I see, but how about the equation: y= \frac{1}{x}? That is a &quot;rational function&quot; right? After looking at its graph on my calculator, it seems like a hyperbola that has its transverse axis as y=x. I think I&#039;ve heard those called &quot;hyperbolas&quot; before but I&#039;m not sure if they are or not now, since they obviously don&#039;t have the style of the general equation like you posted.

 

[symbolipoint]

I see, but how about the equation: y= \frac{1}{x}? That is a "rational function" right? After looking at its graph on my calculator, it seems like a hyperbola that has its transverse axis as y=x. I think I've heard those called "hyperbolas" before but I'm not sure if they are or not now, since they obviously don't have the style of the general equation like you posted.

A general quadratic equation has an xy term. When an xy term occurs in a quadratic equation, this represents rotation of the axis of the graph. Most of what you study in Intermediate Algebra for conic section do not include the xy term. You need some trigonometry in PreCalculus or Calculus course to tie these things together.

 

[lLovePhysics]

A general quadratic equation has an xy term. When an xy term occurs in a quadratic equation, this represents rotation of the axis of the graph. Most of what you study in Intermediate Algebra for conic section do not include the xy term. You need some trigonometry in PreCalculus or Calculus course to tie these things together.

I've studied precal/trig and I've learned about hyperbolas already in the form (x/a)^2-(y/b)^2=1 Also, I just found out that the equation y=1/x yields a "rectangular/equilateral" But I just don't see the resemblance in equations.

Here's a chat with a tutor that I had:
You: Also, is this a hyperbola?
Daniel C (Tutor): sounds good
Daniel C (Tutor): that looks like one, yes
You: So doesn't that have the equation y= 1/x?
You: It doesn't seem at all like the equation I have below...
Daniel C (Tutor): okay, let me try this, one moment
You: k, no problem
Daniel C (Tutor): I don't think 1/x is a hyperbola
Daniel C (Tutor): it looks like one
Daniel C (Tutor): but it is not
You: Is it just a rational function then?
You: I'm confused about this because I recall having heard people, or even my teacher, call these kind of graphs hyperbolas
Daniel C (Tutor): undefined at zero though
Daniel C (Tutor): things like 1/x?
You: yes, like the graph I've drawn below
You: but really, I know hyperbolas look like:
You: Oops
You: like two sleeping (sideways) parabolas
Daniel C (Tutor): okay
Daniel C (Tutor): actually
Daniel C (Tutor): I am wrong
Daniel C (Tutor): y = 1/x is a hyperbola
You: how?
You: Does it have resemblance to the "hyperbola" equation below?
You: What classifies a graph as a hyperbola?
You: because one type of hyperbolas is a function while the other kind is not?
Daniel C (Tutor): no, there should be a way
Daniel C (Tutor): to write 1/x in that form
Daniel C (Tutor): if there is not
Daniel C (Tutor): what I do know is that 1/x
Daniel C (Tutor): is called a "rectangular hyprebola"
You: The x is not squared though
You: ooo
You: I see
Daniel C (Tutor): it may have a unique sort of equation
Daniel C (Tutor): but I don't want to say
Daniel C (Tutor): it's impossible to write 1/x like that
Daniel C (Tutor): there may be a way that I'm trying to figure out now
Daniel C (Tutor): A special case of the hyperbola is the equilateral or rectangular hyperbola, in which the asymptotes intersect at right angles. The rectangular hyperbola with the coordinate axes as its asymptotes is given by the equation xy=c, where c is a constant.
Daniel C (Tutor): that is what I found out

 

[HallsofIvy]

If y= 1/x, then xy= 1.

Let x= x&#039; cos(\theta)- y&#039; sin(\theta) and y= x&#039; sin(\theta)+ y&#039; cos(\theta) (those are the equations for a rotation about the origin through angle \theta).

Then xy= ((x&#039; cos(\theta)- y&#039; sin(\theta))(x&#039; sin(\theta)+ y&#039; cos(\theta))
= x&#039;^2 cos(\theta)sin(\theta)- x&#039;y&#039;(cos^2(\theta)- sin^2(\theta)-y&#039;<br /> ^2 sin(\theta)cos(\theta)

We can get rid of the x'y' term by taking cos^2(\theta)- sin^2(\theta)= 0. That is, cos^2(\theta)= sin^2(\theta) for which the simplest solution is \theta= \pi/4. Of course, in that case sin(\theta)= cos(\theta)= \sqrt{2}/2 and sin(\theta)cos(\theta)= 1/2.

In terms of the x'y' coordinate system (rotated 45 degrees from the xy coordinate system) the equation is
\frac{x^2}{2}- \frac{y^2}{2}= 1
a hyperbola.

 

[lLovePhysics]

If y= 1/x, then xy= 1.

Let x= x&#039; cos(\theta)- y&#039; sin(\theta) and y= x&#039; sin(\theta)+ y&#039; cos(\theta) (those are the equations for a rotation about the origin through angle \theta).

Then xy= ((x&#039; cos(\theta)- y&#039; sin(\theta))(x&#039; sin(\theta)+ y&#039; cos(\theta))
= x&#039;^2 cos(\theta)sin(\theta)- x&#039;y&#039;(cos^2(\theta)- sin^2(\theta)-y&#039;<br /> ^2 sin(\theta)cos(\theta)

We can get rid of the x'y' term by taking cos^2(\theta)- sin^2(\theta)= 0. That is, cos^2(\theta)= sin^2(\theta) for which the simplest solution is \theta= \pi/4. Of course, in that case sin(\theta)= cos(\theta)= \sqrt{2}/2 and sin(\theta)cos(\theta)= 1/2.

In terms of the x'y' coordinate system (rotated 45 degrees from the xy coordinate system) the equation is
\frac{x^2}{2}- \frac{y^2}{2}= 1
a hyperbola.

Wow that is crazy, mabye I will try that sometime. Thanks for showing me!

 

[Feldoh]

1/x is a hyperbola that has been rotated around it's center. If you want to look into it most pre-calc texts mention it^^