Question

Graph then fill in the blanks and the table for the points on the graph

Answer 1

6) One of the main characteristics of the logarithmic function is that

`\text{domain}(\log _b(x))=(0,\infty)`

For b>0. Then, in our case, x-2 has to be always greater than zero; thus,

`\begin{gathered} x-2>0 \\ \Rightarrow x>2 \end{gathered}`

The domain of f(x) is

`\text{domain}(f(x))=x\in(2,\infty)`

Given the domain, we can calculate the range by evaluating the function in the extremes of the interval

`\lim _(x\to2)f(x)=\lim _(x\to2)\log _5(x-2)-1=-\infty-1=-\infty`

Any function of the form logb(x) ->-infinite when x->0

Similarly,

`\lim _(x\to\infty)f(x)=\lim _(x\to\infty)\log _5(x-2)-1=\infty-1=\infty`

Then, the range of the function is

`\text{range}(f(x))=(-\infty,\infty)`

In order to calculate the range of the function, we answer the third part of the question:

As x->infinite, f(x)->infinite

And

As x->2, f(x)->-infinite

To calculate the x-intercept, set f(x)=0 and solve for x, as follows

The x-intercept is x=7

The only asymptote is when x->2 because, in that case, f(x)->-infinite.

Then, the asymptote is x=2

It's not possible to exactly draw a logarithmic function by hand; however, we can use the information about the asymptote and the x-intercept to be more precise.

The graph of the function is

Table of values of f(x) (3 different values of x)

We already found that f(7)=0, the first value of the table is then

`\begin{gathered} x\to y \\ 7\to0 \end{gathered}`

(7,0)

Then,

`\begin{gathered} x=3 \\ \Rightarrow f(x)=\log _5(3-2)-1=\log _5(1)-1=0-1=-1 \\ \Rightarrow(3,-1) \end{gathered}`

(3,-1)

And

`\begin{gathered} x=4 \\ \Rightarrow f(x)=\log _5(4-2)-1=\log _5(2)-1=(\ln(2))/(\ln(5))-1\approx-0.56932\ldots \\ \Rightarrow(4,-0.56932\ldots) \end{gathered}`

(4, -0.56932...)