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I was surprised by the graph of $y=3+\ln(x+2)$:

Graph of y=3+ln(x+2)

I understand that $x=0 \implies y=3+\ln(2)$ and that $y=0 \implies x= e^{-3} -2$ and I derived this without problem. I was expecting the results to be different though. Considering the graph of $y=\ln(x)$ as a starting point, I was expecting the graph to translate 2 units to the left on the x-axis and 3 units up on the y-axis, sort of like with $f(x) = x^2$:

Graphs of y=x^2 and y=3+(x+2)^2

So my questions is, why does it translate up the extra $\ln(2)$ on the y-axis and less the $e^{-2}$ on the x-axis?

Thanks!

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    @joriki, you're right, of course.2011-04-06

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The graph did exactly what you expect. It now goes toward $y=-\infty$ at $x=-2$ instead of $0$, showing it moved $2$ units left. It shifted up by $3$ units, so the place where it crosses $y=0$ should be the place it used to cross $y=-3$ (shifted left by $2$). In the original graph, if $y=-3, x=e^{-3}$, so the final passage through $(e^{-3}-2,0)$ is to be expected. The point where it now crosses $y=0$ is where it used to cross $y=2$ (shifted up by $3$ and it used to cross $y=2$ at $(\ln 2,2)$ so you would expect $(0,3+\ln 2)$. All is well.

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    Ah thank you, yes that makes sense! For the other newbies: I was not looking at the right points on the graphs for where I expected the translation to be. It was not to do with how far the points that cross the axes move, but rather how far the the extremities moved. Just like in $y=x^2$ I was looking at the minimum point, not where the curve crosses the x-axis (although coincidentally it touches the x-axis at this min, which is why I was confused).2011-04-06