we have the function
[tex]f(x)=\log _{\frac{1}{4}}(x-1)+1[/tex]Applying property of log, change the base
Remember that
[tex]\log _bM=\frac{\log _nM}{\log _nb}[/tex]therefore
[tex]\log _{\frac{1}{4}}(x-1)=\frac{\log _{10}(x-1)}{\log _{10}\frac{1}{4}}[/tex]substitute
[tex]f(x)=\frac{\log_{10}(x-1)}{\log_{10}\frac{1}{4}}+1[/tex]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
[tex]\begin{gathered} f(2)=\frac{\log _{10}(2-1)}{\log _{10}\frac{1}{4}}+1 \\ f(2)=1 \end{gathered}[/tex]For x=10
[tex]\begin{gathered} f(10)=\frac{\log _{10}(10-1)}{\log _{10}\frac{1}{4}}+1 \\ f(10)=-0.58 \end{gathered}[/tex]For x=15
[tex]\begin{gathered} f(15)=\frac{\log _{10}(15-1)}{\log _{10}\frac{1}{4}}+1 \\ f(15)=-0.90 \end{gathered}[/tex]
