Determine the function which corresponds to the given graph. Hints: The graph is of a log The asymptote is x = -3 Picture will be posted below
@zepdrix @marcelie @.Sam.
@amorfide @Awolflover1
my guess was y = log (x+ 2)
"2"? Count that again.
@Zarkon
It looks like the graph is shifted two units to the left because it intersects 2 on the x axis, right?
-2*
Without the shift, where would it intersect the x-axis?
0,0
The origin, right?
Not quite. (1,0).
Oh .-. , sooo I guess it should be x + 0.xxx?
How do you solve this kind of problem, and where do you write in the asymptote?
log(x) has an asymptote at x=0 log(x+1) has an asymptote at x+1=0, x=-1 so your asymptote is at x=-3 and you know it is a log function so rearrange x=-3, to make it equal 0 and that is your expression part of your logarithm
if you draw the graph log(x) and compare it to the graph you are given you can figure out the translation
so its log(x+3) and brb ill graph and compare
That's the spirit.
:/ In the hour I spent graphing different things to try and figure it out myself, I tried this, but I was stumped when I compared log(x+3) to the graph because I noticed that it doesn't intersect -2 on the x axis nor does it cross the y axis at 1 or the vertical asymptote one could asume between y = 2.5 and 3
I think that there is something I dont understand
@amorfide Assuming that log(x+3) is not the final answer, what other steps can I take towards the correct answer?
@karim728
Draw log(x) on an accurate scale (As accurate as you can) then find its asymptote now draw on the asymptote the question gives you (x=-3) and see what translation puts you there next, draw that translated graph and it should be your final answer if not maybe it is also translated parallel to the y axis, that is, if you had f(x)=log(x) g(x)=log(x)+a is a translation a units, parallel to the y
but I think log(x+3) is your final answer show your graphs
y=log(x+3)
Okay I will post it:
So I refreshed my graph and now it looks perfect.. xD I guess it wasn't reloading properly when I checked it the first time, my bad.
Thank you!
no problem! happy to help!
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