First, I appreciate your help, and this is very kind of all of you to help us.
Now, this problem we are both sick of: we are to use transformations of f(x)= 1/x or f(x)=1/x^2 to graph the rational function: f(x)= 1/x -4. I tried using the information you gave me, but my graphs look nothing like the examples.
I get confused too, with the various smooth lines and directions, and the dotted lines. Can you please explain?

1. This web site does not have a way of showing graphs or images, so I am not able to draw the graph for you. If the function really is (1/x) -4 and not 1/(x-4), then it will have a value of y = -4 at very large and very small values of x. As you approach x=0 from the left, the graph turns downwand sharply, going off the botton of the graph. The same thing happens in reverse when you approach x=0 from the right, namely, the curve goes quickly off the top of the graph. The set of two separate curves is called a hyperbola.

For a plotted example of what f(x)= y = 1/x looks like, see the third graph at this web site:
http://www.jimloy.com/geometry/analytiz.htm

posted by drwls
2. Hi,

You need to draw f(x) =1/x first and then transform it to f(x) = (1/x) - 4

To draw f(x) = 1/x, all you need to do is find several points, example:

let x=1, then f(1)=1, you get one point (1,1)
let x = 0.5, f(0.5)=1/0.5=2, another point (0.5,2).

Try to get 5-10 points and then connect then together using a smooth curve.

f(x)=(1/x) -4 is obtained by shifting the entire graph of f(x)=1/x down (negative y direction) by 4 units.(By the way this procedure is called transformation).

Contact me if you need further help, I can email you an attachment.

posted by Qun
3. drwls, thank you for the information. Yes the question is the first. I moved 4 points down the y-axis to the -4(?). Now, I need to graph it. I don't understand which way the curve goes (rt. or left).Thank you.

posted by Pam

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