Question

Graph the function then fill in the blanks then fill in the table with three points from the graph

Answer 1

we have the function

`f(x)=\log _{(1)/(4)}(x-1)+1`

Applying property of log, change the base

Remember that

`\log _bM=(\log _nM)/(\log _nb)`

therefore

`\log _{(1)/(4)}(x-1)=(\log _(10)(x-1))/(\log _(10)(1)/(4))`

substitute

`f(x)=(\log_(10)(x-1))/(\log_(10)(1)/(4))+1`

using a graphing tool

The domain is the interval ------> (1, infinite)

The range is the interval ------> (-infinite, infinite)

end behavior

as x→+∞ ------> f(x)→-∞

as x→-∞ ------> f(x)→+∞

x-intercept ------> (5,0)

y-intercept -----> DNE

Asymptote: vertical asymptote at x=1

Construct the table

For x=2

substitute

`\begin{gathered} f(2)=(\log _(10)(2-1))/(\log _(10)(1)/(4))+1 \\ f(2)=1 \end{gathered}`

For x=10

`\begin{gathered} f(10)=(\log _(10)(10-1))/(\log _(10)(1)/(4))+1 \\ f(10)=-0.58 \end{gathered}`

For x=15

`\begin{gathered} f(15)=(\log _(10)(15-1))/(\log _(10)(1)/(4))+1 \\ f(15)=-0.90 \end{gathered}`