Question

A example in my book asks to find any asymptotes and axes intercepts for this function. It also states that: "x approaches 1 from the right, y approaches negative infinity so the vertical asymptote is x=1"?

I don't understand how the y approaches negative infinity. Is there a way to find this without graphing the function?
The function is `f` `(x)` = `log_2` `(x-1)` `+1`

Answer 1

See explanation...

Explanation:

Let's find the inverse function...

Let:

`y = f(x) = log_2(x-1)+1`

Subtract `1` from both ends to get:

`y - 1 = log_2(x-1)`

Take the exponent with base `2` of both sides to get:

`2^(y-1) = x-1`

Add `1` to both sides and transpose to get:

`x = 2^(y-1)+1`

Swapping `x` and `y`, we can write the inverse function as:

`f^(-1)(x) = 2^(x-1)+1`

Now how does an exponential function behave?

Note that for any real value of `x`, `2^(x-1) > 0`, so `f^(-1)(x) > 1`

As `x->oo` we find `2^(x-1)->oo` very quickly.

As `x->-oo` we have `2^(x-1)->0` very quickly.

So:

`f^(-1)(x) -> 0+1 = 1" "` as `x->-oo`

That is: `f^(-1)(x)` has a horizontal asymptote `y=1` as `x->-oo`.

Note that reflecting the graph of `f^(-1)(x)` in the diagonal line `y=x` effectively swaps `x` and `y`, giving us the graph of `f(x)` and doing so we get a vertical asymptote at `x=1` from the right.

So without actually having to graph the function `f(x)` we have found the vertical asymptote.

Incidentally, note that `f^(-1)(0) = 2^(0-1)+1 = 3/2`. So `f^(-1)(x)` has a `y` intercept at `(0, 3/2)` and `f(x)` has an `x` intercept at `(3/2, 0)`.

Now let's graph `f(x)`, `f^(-1)(x)` and `y=x` together to make it a little clearer...

graph{(y-1-2^(x-1))(x-1-2^(y-1))(x-y) = 0 [-11.29, 8.71, -4.36, 5.64]}