skg94 said:
Homework Statement
http://tinypic.com/r/14uet5i/6
just in case it didnt work
http://tinypic.com/r/14uet5i/6
how do i find the equation of those two graphs? Also
Homework Equations
The Attempt at a Solution
1. asym is at -2 so bottom must have (x+2) what's the equation if vertical discontinuity at (0,3) just x? so bottom would be (x^2+2x) ? 2. asy at 1 so (x-1) and discontinuity at -3 so, bottom is (x-1)(x+3) = idk/x^2+2x-3
Begin with the simplest rational function: y = 1/x.
You already know that we can shift the vertical asymptote of x=0 by changing the denominator into y = 1/(x-a) for an x-asymptote of x=a.
But what if we want a different horizontal asymptote? Well, what if we shift everything up by 1 by adding 1 to the function? So we have y= 1 + 1/x. This gives us a horizontal asymptote at y=1. So essentially, if we want a horizontal asymptote at y=c, then we make our rational function y= c + 1/x.
But what if want the shape of the function to be a little different? For example, the graph y=1/x has the point (1,1) but what if want the asymptotes to stay the same but instead we want it to have the point (1,2)? Well, we just multiply the function by 2 to get y=2/x because at x=1 we now get y=2 instead of y=1. So in general, if you want y=1/x to have the point (1,k) instead of (1,1) then we multiply by k to get y= k/x.
If our asymptotes change however, then you need to take that into account too. The point (1,k) is 1 unit in the x direction to the right of the vertical asymptote and the point k is k units in the y direction above the horizontal asymptote.
Finally, tying everything together -
If we want the rational function that has a vertical asymptote at x=a, a horizontal asymptote at y=c and the point 1 unit to the right of the vertical asymptote with a value of k units above the horizontal asymptote, then our function will be
[tex]y=c+\frac{k}{x-a}[/tex]
And converting this into the general rational function form [itex]y=f(x)/g(x)[/itex] is easily done by creating a common denominator.
p.s. I just noticed that your functions have holes in them. This is easily added to your function by noticing that if you have a function y=f(x), then the function
[tex]y=f(x)\cdot\frac{x-d}{x-d}[/tex]
is the line y=f(x) with a hole at x=d.