How would you graph #y= -lnx# ?

1 Answer
Jul 28, 2017

Take the graph of #y=e^x#, reflect it in the line #y=x#, then in the #x# axis.

Explanation:

Do you know what the graph of #y = e^x# looks like?

graph{y=e^x [-10, 10, -5, 5]}

  • It is monotonically increasing.
  • It is always greater than #0#, so lies completely above the #x# axis.
  • It is rapidly asymptotic to the #x# axis for negative values of #x#.
  • It intersects the #y# axis at #(0, 1)#.
  • It grows very rapidly for positive values of #x#.

Next note that #ln x# is the inverse function of #e^x#.

So the graph of #y = ln x# can be found by swapping #x# and #y#, that is by reflecting the above graph in the diagonal line #y=x#, to get:

graph{y=ln x [-10, 10, -5, 5]}

Note that:

  • It is monotonically increasing.
  • It is only defined for #x > 0#, so the graph lies entirely to the right of the #y# axis.
  • It has a vertical asymptote at #x=0#.
  • It intersects the #x# axis at #(1, 0)#.
  • It grows very slowly for positive values of #x#.

Finally, to get the graph of #y = -ln x# we just have to reflect the above graph in the #x# axis to get:

graph{y=-ln x [-10, 10, -5, 5]}

Note that:

  • It is monotonically decreasing.
  • It is only defined for #x > 0#, so the graph lies entirely to the right of the #y# axis.
  • It has a vertical asymptote at #x=0#.
  • It intersects the #x# axis at #(1, 0)#.
  • It grows more negative very slowly for positive values of #x#.