Thread: Write the closest equation of the function that this graph represents--help please

I know how to graph rational functions, but I'm having trouble writing a function given the graph. Can someone help me out with this?

<a href="http://s1015.photobucket.com/albums/af279/tom555_photo/?action=view&amp;current=IMG_1886.jpg" target="_blank"><img src="http://i1015.photobucket.com/albums/af279/tom555_photo/IMG_1886.jpg" border="0" alt="Photobucket"></a>

vertical asymptote at ...

horizontal asymptote at ...

...

...

see what you can do with it now.

Thanks for the reply, but I'm not exactly sure what all that means above (I'm in an honors level algebra 2 course if that helps). I know where the asymptotes and the intercepts are, but how do I use them to figure out the equation? I know that in the denominator there has to be (x-4) and the equation has to look something like 2x/x to satisfy to horizontal asymptote.

- also that to find the y-intercept I plug in zero for x, and to find the x-intercept I set the numerator equal to zero

Originally Posted by mathguy5

Thanks for the reply, but I'm not exactly sure what all that means above (I'm in an honors level algebra 2 course if that helps). I know where the asymptotes and the intercepts are, but how do I use them to figure out the equation?

the equation is the general form for a rational equation where a, b, c, and d are all constants ... I've given you some constraints for the values of the constants based upon information from the graph.

start with c = 1 and see what you get.

Sorry, I'm still a little lost. What do I do besides c=1? I'm not familiar with the "general form," I suppose. I know how to graph a given rational function, but I'm confused as to how to produce the function given a graph.

Is it f(x)= 2x(x-2)/x(x-4)?

Originally Posted by mathguy5

Is it f(x)= 2x(x-2)/x(x-4)?

f(x)= 2(x-2)/(x-4)

the additional x in the numerator and denominator creates a point discontinuity at x = 0 which the graph does not have.

Thanks to all who responded and PM'd me