Question Identify the domain and range of the logarithm functions $y=\log_2(x+6)-1$ and $y=\log_2(x-1)-2$ . Also state the asymptotes and $x$ -intercepts.

Studdy Solution

STEP 1

1. The logarithm function $\log_{b}(x)$ is defined for all $x > 0$ and $b > 0$ , $b \neq 1$ .
2. The domain of a function is the set of all possible input values (x-values) for which the function is defined.
3. The range of a function is the set of all possible output values (y-values) that the function can produce.
4. The vertical asymptote of a logarithmic function $\log_{b}(x - h) + k$ occurs where the argument $x - h$ is zero, i.e., $x = h$ .
5. The $x$ -intercept of a function is the point(s) where the function crosses the x-axis, i.e., where $y = 0$ .

STEP 2

1. Determine the domain of each function.
2. Determine the range of each function.
3. Identify the vertical asymptote of each function.
4. Find the $x$ -intercept of each function.
For function 4) $y=\log _{2}(x+6)-1$ :

STEP 3

Identify the domain of $y=\log _{2}(x+6)-1$ .
The argument of the logarithm, $x+6$ , must be positive. Therefore, we solve $x+6 > 0$ .

STEP 4

Find the solution to the inequality $x+6 > 0$ to determine the domain.
$x > -6$

STEP 5

State the range of $y=\log _{2}(x+6)-1$ .
The range of a logarithmic function is all real numbers, so the range is $(-\infty, \infty)$ .

STEP 6

Identify the vertical asymptote of $y=\log _{2}(x+6)-1$ .
The vertical asymptote occurs where $x+6 = 0$ , which is $x = -6$ .

STEP 7

Find the $x$ -intercept of $y=\log _{2}(x+6)-1$ by setting $y = 0$ and solving for $x$ .
$0 = \log _{2}(x+6)-1$

STEP 8

Solve the equation $\log _{2}(x+6)-1 = 0$ for $x$ to find the $x$ -intercept.
$\log _{2}(x+6) = 1$

STEP 9

Convert the logarithmic equation to exponential form to solve for $x$ .
$2^1 = x+6$

STEP 10

Solve for $x$ to find the $x$ -intercept.
$x = 2 - 6$ $x = -4$
For function 5) $y=\log _{2}(x-1)-2$ :

STEP 11

Identify the domain of $y=\log _{2}(x-1)-2$ .
The argument of the logarithm, $x-1$ , must be positive. Therefore, we solve $x-1 > 0$ .

STEP 12

Find the solution to the inequality $x-1 > 0$ to determine the domain.
$x > 1$

STEP 13

State the range of $y=\log _{2}(x-1)-2$ .
The range of a logarithmic function is all real numbers, so the range is $(-\infty, \infty)$ .

STEP 14

Identify the vertical asymptote of $y=\log _{2}(x-1)-2$ .
The vertical asymptote occurs where $x-1 = 0$ , which is $x = 1$ .

STEP 15

Find the $x$ -intercept of $y=\log _{2}(x-1)-2$ by setting $y = 0$ and solving for $x$ .
$0 = \log _{2}(x-1)-2$

STEP 16

Solve the equation $\log _{2}(x-1)-2 = 0$ for $x$ to find the $x$ -intercept.
$\log _{2}(x-1) = 2$

STEP 17

Convert the logarithmic equation to exponential form to solve for $x$ .
$2^2 = x-1$

STEP 18

Solve for $x$ to find the $x$ -intercept.
$x = 4 + 1$ $x = 5$
_SOLUTION_: For function 4) $y=\log _{2}(x+6)-1$ : - Domain: $x > -6$ - Range: $(-\infty, \infty)$ - Vertical Asymptote: $x = -6$ - $x$ -intercept: $x = -4$
For function 5) $y=\log _{2}(x-1)-2$ : - Domain: $x > 1$ - Range: $(-\infty, \infty)$ - Vertical Asymptote: $x = 1$ - $x$ -intercept: $x = 5$

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