Graph y=-3 log base 3 of 1-x+6

Find where the expression ${\displaystyle -{\log }_3\left({(1-x)}^3\right)+6}$ is undefined.

Since ${\displaystyle -{\log }_3\left({(1-x)}^3\right)+6}$${\displaystyle \to }$${\displaystyle \infty }$ as ${\displaystyle x}$${\displaystyle \to }$${\displaystyle 1}$ from the left and ${\displaystyle -{\log }_3\left({(1-x)}^3\right)+6}$${\displaystyle \to }$${\displaystyle \infty }$ as ${\displaystyle x}$${\displaystyle \to }$${\displaystyle 1}$ from the right, then ${\displaystyle x=1}$ is a vertical asymptote.

Ignoring the logarithm, consider the rational function ${\displaystyle R\left(x\right)=\frac{a{x}^n}{b{x}^m}}$ where ${\displaystyle n}$ is the degree of the numerator and ${\displaystyle m}$ is the degree of the denominator.

1. If ${\displaystyle n<m}$, then the x-axis, ${\displaystyle y=0}$, is the horizontal asymptote.

2. If ${\displaystyle n=m}$, then the horizontal asymptote is the line ${\displaystyle y=\frac{a}{b}}$.

3. If ${\displaystyle n>m}$, then there is no horizontal asymptote (there is an oblique asymptote).

There are no horizontal asymptotes because ${\displaystyle Q\left(x\right)}$ is ${\displaystyle 1}$.

No Horizontal Asymptotes

No oblique asymptotes are present for logarithmic and trigonometric functions.

No Oblique Asymptotes

This is the set of all asymptotes.

Vertical Asymptotes: ${\displaystyle x=1}$

No Horizontal Asymptotes

Replace the variable ${\displaystyle x}$ with ${\displaystyle 0}$ in the expression.

Simplify the result.

Convert ${\displaystyle 6}$ to decimal.

Replace the variable ${\displaystyle x}$ with ${\displaystyle -1}$ in the expression.

Simplify the result.

Convert ${\displaystyle -{\log }_3\left(8\right)+6}$ to decimal.

Replace the variable ${\displaystyle x}$ with ${\displaystyle -2}$ in the expression.

Simplify the result.

Convert ${\displaystyle 3}$ to decimal.