
Michael W. answered 05/16/14
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Danielle,
Well, looks like you need to graph a log function. :) So let's start with the basic log function.
Question #1: What does the graph of f(x) = log3(x) look like? What is its domain (possible values of x), range (possible values of y), and asymptote? You're dealing with a log function, so there are limitations involved, either for x, y, or both. For the asymptote, it could be a vertical asymptote, where you plug in values of x that are closer and closer to some value, and f(x) either goes up or down forever. Or, it could be a horizontal asymptote, where you plug in values of x that are infinitely big in either direction (all the way to the right or left on your graph), and f(x) gets closer to some value but never reaches it. Which is it? Are there values of x that you can never plug into the log function because it's undefined? And if that's true, then are there values of y that you never can get back?
Question #2: Once you figure out Question 1, then we need to see what happens to it when we start messing around with the equation. What would f(x) = log3(x) + 3 look like? You're taking the old f(x) and adding 3 to it, so that's going to shift it somewhere. Which way? Try some values and see which way it went. For example, if log3(3) = 1, then what does log3(3) + 3 equal? That's the same value of x, but f(x) changed. Which way did it go?
Question #3: What would f(x) = log3(x-4) look like? Now, you've shifted the value of x that you're putting into the log function, so that's going to shift your graph, too. Which way? Do the same thing...try some values and see which way the graph moved. If log3(3) is 1, then when would log3(x-4) be 1? What would you have to put in for x to get 1 again? Compare those two points...which way did the graph move?
Question #4: If you can figure out questions 2 & 3, then your original question is just the combination of the two. What happens if you do both of those to f(x) at the same time? And then what are the resulting limitations on x and y?
Hope this helps,
-- Michael