# Sketching graph Exponential and logarithmic functions.

Sketch the graph of each of the following (labelling asymptotes), and state the range of each: $y=3^x+1$
is below graph. I have no idea, how can it get $(0,2)$ i know that the $+1$ is asymptotes, Isn't it $(0,2)$ suppose to be $(0,3)$accroding to the equation.

Appreciate your explaination! thanks

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It’s right: $3^0+1=1+1=2$, so $y=2$ when $x=0$. –  Brian M. Scott May 20 '12 at 9:34
i understand now! thx alot, can I delete my question. it seem very easy! –  Sb Sangpi May 20 '12 at 9:36
You could delete it, or in a few hours $-$ I forget how long you have to wait $-$ you could answer it yourself and accept your own answer. This is perfectly acceptable; in fact, it’s encouraged. –  Brian M. Scott May 20 '12 at 9:38
You can delete your own question if it's got no answers yet. –  Gigili May 20 '12 at 9:43
thank you for the honest respone! God bless! –  Sb Sangpi May 20 '12 at 9:56

## 1 Answer

yes, the graph plotted above is correct because
$y=3^x+1$, $x=0$
$y=3^0+1$
$y=1+1$
$y=2$ when $x=0$
so, the graph plotted it correct!

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