Define rational function

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Sep 16th 2008, 09:12 AM
courteous

Define rational function

From graph of the rational function, you must define $a, b, c, d$ in $f(x)=\frac{ax+b}{cx+d}$ .
http://shrani.si/f/3u/AR/3BerSEhH/scan0004.jpg
(I'm assuming both x-axis and y-axis are crossed at $-\frac{1}{2}$ )
I start with this set-up: $\frac{b}{d}=-\frac{1}{2}$ (because of $f(0)=-\frac{1}{2}$ )
$\frac{a}{c}=2$ (asymptote)
$-\frac{1}{2}a+b=0$ (because at $x=-\frac{1}{2}$ the function cross the x-axis and is equal to zero)
$2c+d=0$ (because pole of the function is at $x=2$ ).

So it comes down to 4 equations with 4 unknown variables, but I've tried many times solving, and I constantly get trapped in loop (you know, where your "new" equation is actually some old ones mixed up?)!!
Where is the flaw in my logic?
Sep 16th 2008, 09:35 AM
Moo

Hello,

Your logic is perfect ! I think I'm getting what's looking wrong.

I get these equations from yours :

$\begin{aligned} d=-2b \\ a=2c \\ a=2b \\ d=-2c \end{aligned}$

From the second and the third (or the first and the fourth), we have $\boxed{b=c}$ .

So it reduces to :
$\boxed{d=-2b}$
$\boxed{a=2b}$

The equation is now $y=\frac{2bx+b}{bx-2b}$

Simplify by b (Tongueout)
Sep 16th 2008, 10:44 AM
courteous

Makes it a lot easier to insert into $f(x)$ (Smirk)... but, shouldn't there be a solution "directly" from the 4 original equations (as there are 4 unknowns)?
How would you solve it without putting variables into equation and without getting trapped in "loop" (is it even possible?)? Is it even possible?
Sep 16th 2008, 11:04 AM
Moo

Quote:

Originally Posted by courteous

Makes it a lot easier to insert into $f(x)$ (Smirk)... but, shouldn't there be a solution "directly" from the 4 original equations (as there are 4 unknowns)?
How would you solve it without putting variables into equation and without getting trapped in "loop" (is it even possible?)? Is it even possible?

It depends.

Here for example, the trap was that there is sort of a division of the coefficients. So it was likely that you had a linear combination of a and b proportional to a linear combination of c and d. No matter what, because it would be simplified while putting it in f(x).

Here, it appeared that all were proportionally related.

You can have exercises where you have to find a,b,c,d and where there won't be any loop. But it won't be a quotient like the one you have here.