Question

How would you graph `y= -lnx` ?

Answer 1

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`.