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$y=-2\log(x-2)$
asymptotes $x=2$
lets $x=0$
$y=2\log(0-2)$
$y=2\log(-2)$

lets $y=0$
$0=2\log(x-2)$

I think but I don't know how to draw the graph. The answer of the graph is below

enter image description here

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Any questions about my answer? –  Gerry Myerson May 25 '12 at 6:26
    
i'm very appreciate your answer but I still don't idea, how to get (3,0). thx –  Sb Sangpi May 25 '12 at 6:57
    
OK, I've added a bit to explain that. Better? –  Gerry Myerson May 25 '12 at 7:17
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1 Answer

up vote 2 down vote accepted

$x=3$ is not an asymptote. $x=0$ is not an asymptote. I can't understand $y=\log(0-2)$, nor $y=2\log(-3)$; I don't know how you arrived at these equations, and I don't know what purpose you mean for them to serve, and they are nonsensical because they involve evaluating the logarithm of a negative number, which is a no-no. I don't know the purpose of $0=2\log(x-3)$.

Lastly, I don't know why you've tagged this "trigonometry" --- I'll edit that out when I've written this up.

Logarithm, as noted, is only defined for positive values of the argument, so if we are dealing with the function $y=-2\log(x-2)$ then we need to have $x\gt2$. This suggests that $x=2$ will be a vertical asymptote --- indeed, the only vertical asymptote. So, start by drawing a vertical line at $x=2$. That's not part of the graph of the function, but it is an asymptote, and helps you to sketch the graph of the function.

For what follows, it will be helpful to have a sketch of the graph of $y=\log x$ in front of you. Can you make or find one of those?

Now, what happens as $x\to2$ from the right? Well, that implies $x-2\to0$, and from that sketch of the graph of $y=\log x$, this implies $\log(x-2)\to-\infty$, which is to say, $\log(x-2)$ becomes larger and larger negative, without any bound. It follows that $-2\log(x-2)$ becomes larger and larger positive without any bound. This should enable you to sketch the part of the graph of $y=-2\log(x-2)$ near the vertical asymptote.

Now what happens when $x$ is large? Well, $x-2$ is also large, right? So what does the graph of $y=\log x$ tell you about $\log(x-2)$ when $x$ is large? And what does that tell you about $-2\log(x-2)$? So now you should know something about the part of the graph of $y=-2\log(x-2)$ for large values of $x$.

Now just connect the two parts you have, smoothly.

Here's another method: Start with that graph of $y=\log x$. You get the graph of $y=\log(x-2)$ by sliding it 2 units to the right --- make sure you understand why! Then you get the graph of $y=-\log(x-2)$ by reflecting the graph of $y=\log(x-2)$ in the $x$-axis --- make sure you understand why! Then you get the graph of $y=-2\log(x-2)$ by stretching the graph of $y=-\log(x-2)$ vertically by a factor of 2 --- make sure you understand why!

EDIT: in response to comment by OP asking how to get $(3,0)$, we have the function $y=-2\log(x-2)$, and we want to know what value of $x$ goes along with $y$ being zero, so we let $y$ be zero, giving us $$0=-2\log(x-2)$$ Now divide both sides by $-2$ to get $$\log(x-2)=0$$ Now there is only one number whose logarithm is zero, namely, $\log1=0$ (do you understand this?), so $x-2$ must be $1$. That is, $x-2=1$, so $x=3$ is the $x$-value we are looking for to go with the $y$-value of zero. That's where the point $(3,0)$ comes from.

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fantastic answer thanks you very much! –  Sb Sangpi May 25 '12 at 11:32
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