# Graphing question

#### Death

I need some help please. I'm not that good at sketching graphs out.

Here's the problem. k(x) = log3 (x+9)
(the number three is the base)

What steps do I have to do in order to graph this problem? Thank you.

#### HallsofIvy

Homework Helper
Didn't you just post a similar question about an exponential?

General rule: If you already know the graph of y= f(x) and have a new graph that involves changing x before applying f, that changes the graph horizontally. That is, adding or subtracting a number from x moves the graph. Multiplying or dividing x stretches or shrinks the graph. If you change the value AFTER applying f, that's a change in y and changes the graph vertically.

For example, you know, I presume, that the graph of y= x2 is a parabola with vertex at (0,0) passing through (1,1) and (-1,1). The graph of y= (x-2)2 is that exact same graph moved to the right 2 places (the vertex of the "base" graph, y= x2 is at x=0 and we have replaced x by x-2: x-2= 0 when x= 2).
The vertex of y= (x-2)2 has vertex at (2,0) and passes through (1+2,1)=(3,1) and (-1+2,1)= (1,1) but looks exactly like y= x2 otherwise.
The graph of y= x2-2 is also a parabola but it is moved DOWN 2 because, since we have already done the squaring before subtracting 2, that is a change in y: the vertex is at (0,-2) and the graph passes through (-1,1-2)= (-1,-1) and (1,1-2)= (1,-1).

The graph of y= log3(x) (the "base" function here), like any logarithm, is undefined for x<=0, is asymptotic to the y-axis, passes through (1,0) and increases to infinity as x goes to infinity.
Adding 9 to x "moves" the graph 9 places to the left. The asymptote y-axis is, of course, x=0. Replacing x by x+ 9 means the asymptote will be where x+9= 0 which is x= -9. Also, any logarithm graph passes through (1,0) because log(1)= 0. x+9= 1 when x= -8 so
log3(x+9) passes through (-8,0).
The graph of y= log3(x+9) is asymptotic to the line x= -9 and passes through (-8,0). In shape, it is exactly the same as
y= log3(x).

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