Hi, this is juriemagic,

I thought I might expand a little on what i have already said.

Think about the most basic hyperbola example

\(y=\frac{1}{x} \)

Now you cannot divide by 0 so x cannot be 0.  x=0 is an asymptote.

Now, the right hand side cannot equal zero because 1 divide by any number cannt be 0 

SO      y cannot be zero either.                         y=0 is also an asymptote.

Lets look at the basic hyperbola formula with horizonal and vertical intercepts.

\(y-k=\frac{a}{x-h}\qquad where \;\;a\ne 0 \)

x-h cannot be 0        SO x cannot be h so      x=h is an asymptote

y-k cannot be 0         so  y cannot be k so      y=k is also an asymptote

Hi Juriemagic,

Around the time you posted this I could see you logged on at the bottom of my page.  

I hope that cookie problem gets fixed soon.  

I am trying to help a grade 11 pupil prepare for his exam, and we have a question here that we know he is going to get. I have tried, but, yes, you guessed it, I'm tooooo stupid!!.  Its a hyperbola related problem. The equation given is k(x)=(a/(x-p))+q where D is the x-intercept (2,0). Now, the graph is given but I cannot plot it here for you. What I have additional is the Horizontal asymptote is Y=1 and the Vertical one is x=3. If someone would Please,..(sigh), walk the xtra mile and plot the graph, which I'm sure is possible to do using the info I have given, and then help me calculate a, p and q. Alland any help will be accepted with the utmost of humility and apreciation!..thank you all kindly.. 

\(k(x)=\frac{a}{(x-p)}+q\\ asymptotes \;\;y=1\;\;and\;\;x=3\\ x\;intercept = 2\)

-----------------------------------------------

The equation of any hyperbola with with a vertical and horizonal axis is 

\(y-k = \frac{c}{x-h}\\ \mbox{The horizontal axis will be y=k and the vertical x=h}\\ \mbox{So we have h=3 and k =1}\\ \mbox{So far the equation of this hyperbola is }\\ y-1 = \frac{c}{x-3}\\ \mbox{Now this equation also passes through the point (2,0) so I will sub this point in to get c}\\ 0-1 = \frac{c}{2-3}\\ -1 = \frac{c}{-1}\\ c=1\\ \mbox{So the equation is } y-1 = \frac{1}{x-3}\\ or\\ y= \frac{1}{x-3}+1\\ \)

sorry, that is 

\(k(x)=\frac{1}{x-3}+1\)

So, a=1,  p=3  and q=1

check.     You shouold get familiar with Desmos graphing calculator.  It is a fabulous tool.

https://www.desmos.com/calculator/polvndqdyn

You're thinking about it back to front, first calculate a,p and q from the information given and then sketch  the graph, (which is easily done on paper, you don't need a fancy graph plotter).

The vertical asymptote happens when x - p is zero, so from that you should know the value of p.

The horizontal one happens when x goes to (plus or minus) infinity, so that gets you the value of q.

Finally x = 2, y = 0 allows you to find the value of a.

If you've got that far, you should be able to sketch the graph.

Guest is right.  You do not need a fancy plotter to graph this.

He and i have both shown you how it is done.

It is good to become familiar with a 'fancy' plotter though because you can use it to check your answers and it can also help you develop an understandng of how graphs work :)

To Melody and Guest,

thank you both for this much informed answer!. Why is it when others do it, it seems so simple??????

Melody?..when can I come over for a 40 Hour lesson?????..

Any time.  I will put the kettle on :))

Very nice, Melody and Guest.....!!!!!