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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?


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$y=3+(x+2)^2$ moves $y=x^2$ by $7=3+2^2$ units up as the graph shows. – lhf Apr 6 '11 at 13:18
@Danny: You need to enclose compound exponents in braces. Also, function names like ln are usually not italicized; $\TeX$ has commands for the common ones (like "\ln"). I also inserted "x=" where I think you intended it. – joriki Apr 6 '11 at 13:19
Thanks Joriki, couldn't figure that one out. Will do from now on! – Danny King Apr 6 '11 at 13:21
@Danny: The graph behaves exactly as you expected it to. The problem is just that you're not taking into account that these two movements both change both the intersection with the $x$-axis and the intersection with the $y$-axis. So there's no contradiction between the movements you expected and the axis intersections you found. – joriki Apr 6 '11 at 13:22
@lhf: I think your comment contributes to the confusion more than resolving it. The problem is that Danny confused "moves the graph in direction $d$" and "changes the graph's intersection with the $d$-axis". The graph of $x^2$ doesn't move up 7 units; its intersection with the $y$-axis changes by 7 units if you move it 2 units to the left and 3 units up. – joriki Apr 6 '11 at 13:26
up vote 2 down vote accepted

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). – Danny King Apr 6 '11 at 13:27

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