Describe how F(x)=(ax+b)/(cx+d) is related to f(x) = 1/x

I am struggling with a an extra credit problem, and I was hoping to get some guidance. I've done alot of research and can't come up with anything. Anyone here have a place to look? The problem is this:

"Describe how the graph of f(x) = (ax+b)/(cx+d) can be obtained from the graph of f(x) = 1/x. Make sure to sure to describe the transformation(s) in terms of a, b, c, and/or d."


I'm at a complete loss. As I said, this is extra credit so I don't have to do it, but I am really interested in the answer.

Thanks for your help,
Dalton

dalton_holt said:

I am struggling with a an extra credit problem, and I was hoping to get some guidance. I've done alot of research and can't come up with anything. Anyone here have a place to look? The problem is this:

"Describe how the graph of f(x) = (ax+b)/(cx+d) can be obtained from the graph of f(x) = 1/x. Make sure to sure to describe the transformation(s) in terms of a, b, c, and/or d."


I'm at a complete loss. As I said, this is extra credit so I don't have to do it, but I am really interested in the answer.

Thanks for your help,
Dalton

Click to expand...

Have you considered Long Division?

tkhunny said:

Have you considered Long Division?

Click to expand...

Thank you for your idea, but I'm not tracking how long division applies. I'm trying to envision how 1/x applies to (ax+b)/(cx+d). Is the suggestion to divide one by the other?

I appreciate your ideas and help.

dalton_holt said:

Thank you for your idea, but I'm not tracking how long division applies. I'm trying to envision how 1/x applies to (ax+b)/(cx+d). Is the suggestion to divide one by the other?

I appreciate your ideas and help.

Click to expand...

Hint:

\(\displaystyle \displaystyle{ax + b \ = \frac{a}{c}*{(cx+d)} \ + \ (b - \ \frac{da}{c})}\) ... from long division

dalton_holt said:

Thank you for your idea, but I'm not tracking how long division applies. I'm trying to envision how 1/x applies to (ax+b)/(cx+d). Is the suggestion to divide one by the other?

Click to expand...

Yes, since you have an improper fraction, you can simplify (in a sense) by dividing the numerator by the denominator. The result, as Subhotosh Khan showed, is rather ugly, but that's what you get with lots of parameters.

You might find the idea easier to follow if you first try doing this with specific numbers, such as (4x + 3)/(2x + 1). Divide 4x + 3 by 2x + 1, and notice that the result will be a number plus a fraction more obviously related to 1/x. Then you can think about what transformations it takes to get this from 1/x. Once you've done that, you can more easily see what to do in the general case with parameters.