Question

How do you graph `f(X)=ln(2x-6)`?

Answer 1

Find the key points of a logarithm function:

`(x_1,0)`

`(x_2,1)`

`ln(g(x))->g(x)=0` (vertical asymptote)

Keep in mind that:

`ln(x)->`increasing and concave
`ln(-x)->`decreasing and concave

Explanation:

`f(x)=0`

`ln(2x-6)=0`

`ln(2x-6)=ln1`

`lnx` is `1-1`

`2x-6=1`

`x=7/2`

• So you have one point `(x,y)=(7/2,0)=(3.5,0)`

`f(x)=1`

`ln(2x-6)=1`

`ln(2x-6)=lne`

`lnx` is `1-1`

`2x-6=e`

`x=3+e/2~=4.36`

• So you have a second point `(x,y)=(1,4.36)`

Now to find the vertical line that `f(x)` never touches, but tends to, because of its logarithmic nature. This is when we try to estimate `ln0` so:

`ln(2x-6)`

`2x-6=0`

`x=3`

• Vertical asymptote for `x=3`
• Finally, since the function is logarithmic, it will be increasing and concave .

Therefore, the function will:

• Increase but curve downwards.
• Pass through `(3.5,0)` and `(1,4.36)`
• Tend to touch `x=3`

Here is the graph:

graph{ln(2x-6) [0.989, 6.464, -1.215, 1.523]}