Linear Fractional Transformation - find the formula

[anniecvc]

http://math.sfsu.edu/federico/Clase/Math350.S15/linea.JPG 1. Homework Statement
The picture below represents the map from a "green projective line" to a "red projective line." It takes the "green points" 1,3,7,-11 to the "red points" 0,6,10,20, respectively as shown by the ruler. Let f be the corresponding linear fractional, so f(1) = 0, f(3) = 6, f(7) = 10, f(-11) = 20. Find a formula for the "red point" f(x) on the ruler where the "green point" x lands.

Homework Equations

a linear fractional is where f(x) = (ax+b)/(cx+d) where ad-bc is not equal to 0.

The Attempt at a Solution

First a did a system of equations:
(a(1) + b) / ( c(1) + d ) = 0
(a(3) +b) / ( c(3) + d )= 6
(a(7) +b) / ( c(7) + d ) = 10
(a(-11) +b) / (c(-11) + d )= 20

Thinking I could solve for a,b,c,d. However if I get all the letters in terms of let's say d and solve, this becomes nonsensical, since something d over something d is a number (the d's reduce) , and I'm left with a number = another number.

 

[Svein]

a linear fractional is where f(x) = (ax+b)/(cx+d) where ad-bc is not equal to 0.

Since you have a fraction, you can choose either a or c to be 1. So, choose c = 1. The first equation says a⋅1 + b = 0, so b = -a. That's two variables accounted for. Now the rest is easy...

 

[anniecvc]

Choosing a or c to be 1 (or any number) seems to be an invalid assumption since even this is a ratio of a sum not just of a number over a number (choosing c=1 affects d, and choosing a=1 affects b), but please explain if I am incorrect. I chose c to be 1, then since I had everything else worked out in terms of d I plugged it all in. When I did, c=1 did not give f(3) = 6, so I solved for what c would be and got c = 1/6. Then, I tried to test this formula with f(7) = 10. It didn't work.

 

[Svein]

Choosing a or c to be 1 (or any number) seems to be an invalid assumption since even this is a ratio of a sum not just of a number over a number (choosing c=1 affects d, and choosing a=1 affects b), but please explain if I am incorrect. I chose c to be 1, then since I had everything else worked out in terms of d I plugged it all in. When I did, c=1 did not give f(3) = 6, so I solved for what c would be and got c = 1/6. Then, I tried to test this formula with f(7) = 10. It didn't work.

I am sorry, but you have made an error somewhere. One solution is: a = 15, b= -15, c=1, d=2.

About choosing one of the variables: A fundamental fact of fractions is that you can multiply or divide by the same number above or below. Therefore $\frac{a\cdot x + b}{c\cdot x + d}=\frac{a(x+\frac{b}{a})}{a(\frac{c\cdot x}{a}+\frac{d}{a})}=\frac{c(\frac{a}{c}x+\frac{b}{c})}{c(x+\frac{d}{c})}$ (of course, you cannot do that if the one you choose turns out to be 0).

 

[anniecvc]

Thanks very much Svein.

 

[Mentallic]

Another quick simplification is to notice the case where f(x)=0

$$\frac{a+b}{c+d}=0$$
and for a fraction to equal 0, it means that the numerator = 0, hence

$$a+b=0$$

$$a=-b\text{, or } b=-a$$

Therefore you just have an equation of the form

$$f(x)=\frac{ax-a}{cx+d}$$