QuestionQuestion 1 4
Use the function $y=-\log _{8}(x)-5$ to answer the following: (a) Determine the equation of the asymptote of $f(x)$. $\square$ (b) Determine the domain of $f(x)$ in interval notation. $\square$ (c) Determine the range of $f(x)$ in interval notation. $\square$
Question 2 4 pt
Find the domain and range of the following exponential function. $$y=-\ln (x-1)+6$$ a) Record the vertical asymptote below. Be sure to record you answer as $x=\#$. $\square$ b) Record the domain below using interval notation. $\square$ c) Record the range below using interval notation. $\square$

STEP 1

1. The function $y = -\log_{8}(x) - 5$ is a logarithmic function.
2. The function $y = -\ln(x-1) + 6$ is a natural logarithmic function.
3. Logarithmic functions have vertical asymptotes where the argument of the logarithm is zero.
4. The domain of a logarithmic function is determined by where the argument is positive.
5. The range of a logarithmic function is all real numbers.
_HIGH_LEVEL_APPROACH_ for Question 1:
1. Determine the equation of the asymptote for $y = -\log_{8}(x) - 5$.
2. Determine the domain of $y = -\log_{8}(x) - 5$.
3. Determine the range of $y = -\log_{8}(x) - 5$.
_HIGH_LEVEL_APPROACH_ for Question 2:
1. Determine the vertical asymptote for $y = -\ln(x-1) + 6$.
2. Determine the domain of $y = -\ln(x-1) + 6$.
3. Determine the range of $y = -\ln(x-1) + 6$.
**Question 1:**

STEP 2

STEP 3

For the function $y = -\log_{8}(x) - 5$, the asymptote occurs where the argument of the logarithm is zero.
$$x = 0$$
Thus, the equation of the vertical asymptote is:
$$x = 0$$

STEP 4

The domain of $y = -\log_{8}(x) - 5$ is where the argument of the logarithm is positive:
$$x > 0$$
In interval notation, the domain is:
$$(0, \infty)$$

STEP 5

The range of any logarithmic function is all real numbers. Therefore, the range of $y = -\log_{8}(x) - 5$ is:
$$(-\infty, \infty)$$
**Question 2:**
STEP_1: For the function $y = -\ln(x-1) + 6$, the vertical asymptote occurs where the argument of the logarithm is zero:
$$x - 1 = 0$$ $$x = 1$$
Thus, the equation of the vertical asymptote is:
$$x = 1$$
STEP_2: The domain of $y = -\ln(x-1) + 6$ is where the argument of the logarithm is positive:
$$x - 1 > 0$$ $$x > 1$$
In interval notation, the domain is:
$$(1, \infty)$$
STEP_3: The range of any logarithmic function is all real numbers. Therefore, the range of $y = -\ln(x-1) + 6$ is:
$$(-\infty, \infty)$$