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Graphing question

  1. Oct 26, 2003 #1
    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.
  2. jcsd
  3. Oct 26, 2003 #2


    User Avatar
    Science Advisor

    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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