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Graphing Logarithmic Functions: Examples (page 3 of 3)
This is similar to the graph of log2(x), but is shifted sideways. (Since the "+ 3" isn't outside of the log, the shift is not up or down; this means that the shift has to be to the left or to the right.) But which way? The way I keep track of this is to think of the basic point (1, 0). The log will be 0 when the argument, x + 3, is equal to 1. When is x + 3 equal to 1? When x = 2. Then the basic point (1, 0) is shifted to (2, 0); that is, the graph is shifted three units to the left. Among other things, this means that, since a log cannot have an argument of zero or less, then x + 3 > 0, so x must always be greater than 3. Remember that the standard log graph of y = log2(x) would crawl along the positive side of the y-axis. To remind myself that this log, y = log2(x + 3), will not go to the left of the line x = 3, I will insert a dashed line at x = 3:
A line like this, which marks off territory where the graph shouldn't go, is called a "vertical asymptote", or simply an "asymptote". It is not necessary to add this to the graph, but can be very helpful, and might convince your teacher on the test that you know what you're doing. After I dash in the asymptote, I plot some points: 20
= 1, so log2(1)
= 0; x
+ 3 = 1 for x
= 2: (2, 0) Then, working in the other direction: Copyright Š Elizabeth Stapel 2006-2008 All Rights Reserved 21
= 0.5, so log2(0.5)
= 1; (Note that, to find each of these points, I did not start with an x-value and then puzzle my way to a y-value; that would be too hard, and I'm too lazy. Instead, I started with a simple exponential statement, switched it around to the corresponding logarithmic statement, and then figured out, for that exponent (which is also my y-value), what the x-value needed to be. This method is, in my view, much the simpler way to work these problems.)
If you check this in your calculator, first, remember to put the "x + 3" inside parentheses, or your calculator will think you mean "log2(x) + 3", and you'll get the wrong answer; and, second, remember that your calculator can only follow its programming it can't think so the graph it displays will likely be incorrect, even if you enter the function correctly.
Now, you know full well that the log doesn't just "end" there, hanging in space like this. This isn't a square root, after all; it's a log. So why is the calculator doing it wrong? Because it's just a machine, and it's doing the best it can. The calculator graphs in a manner similar to how you do: it picks x-values, computes y-values, plots the points, and connects the dots. But, whereas you know that the log graph continues downward forever, getting infinitesimally close to the y-axis (or whatever the vertical asymptote happens to be), the calculator only knows that it tried one x-value on its list, got "ERROR" for an answer, tried the next x-value on its list, and got a valid y-value. Since it has no other dots before that first one, and because it can't think, it starts the graph with that first dot. This is another instance of "student smart; calculator stupid". Don't assume, just because the calculator displays a graph a certain way, that this is what the graph actually looks like. Use your head! To review: below are some different variations on the same basic logarithmic function, with the associated graph below each equation. Note that, even if the graph is moved left or right, or up or down, or is flipped upside-down, it still displays the same curve:
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Copyright Š 2006-2008 Elizabeth Stapel | About | Terms of Use |
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