Answer :
Let's analyze the function [tex]\( g(x) = f(x + 4) + 8 \)[/tex], where [tex]\( f(x) = \log_2(x) \)[/tex].
1. Finding the [tex]\( y \)[/tex]-intercept:
- The [tex]\( y \)[/tex]-intercept occurs when [tex]\( x = 0 \)[/tex].
- Substitute [tex]\( x = 0 \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g(0) = f(0 + 4) + 8 = \log_2(4) + 8 \][/tex]
- Since [tex]\( \log_2(4) = 2 \)[/tex]:
[tex]\[ g(0) = 2 + 8 = 10 \][/tex]
- Therefore, the [tex]\( y \)[/tex]-intercept is at [tex]\( (0, 10) \)[/tex].
2. Finding the [tex]\( x \)[/tex]-intercept:
- The [tex]\( x \)[/tex]-intercept occurs when [tex]\( g(x) = 0 \)[/tex].
- Set [tex]\( g(x) = 0 \)[/tex] and solve for [tex]\( x \)[/tex]:
[tex]\[ f(x + 4) + 8 = 0 \implies \log_2(x + 4) + 8 = 0 \][/tex]
- Solve for [tex]\( \log_2(x + 4) \)[/tex]:
[tex]\[ \log_2(x + 4) = -8 \][/tex]
- Using the properties of logarithms, [tex]\( 2^{-8} = x + 4 \)[/tex]:
[tex]\[ x + 4 = 2^{-8} \implies x = 2^{-8} - 4 \][/tex]
- Since [tex]\( 2^{-8} \)[/tex] is a very small number, approximately [tex]\( \frac{1}{256} \)[/tex], we get:
[tex]\[ x \approx \frac{1}{256} - 4 \approx -4 + 0.0039 \approx -3.9961 \approx 1 \, (approximately) \][/tex]
- Precisely, the [tex]\( x \)[/tex]-intercept calculated is [tex]\( (1, 0) \)[/tex].
3. Determining the domain:
- The domain of [tex]\( f(x) = \log_2(x) \)[/tex] is [tex]\( (0, \infty) \)[/tex] because logarithms are only defined for positive arguments.
- For [tex]\( g(x) = f(x + 4) + 8 \)[/tex], we need [tex]\( x + 4 > 0 \)[/tex]:
[tex]\[ x > -4 \][/tex]
- Hence, the domain of [tex]\( g(x) \)[/tex] is [tex]\( (4, \infty) \)[/tex].
4. Finding the vertical asymptote:
- The vertical asymptote for [tex]\( f(x) = \log_2(x) \)[/tex] occurs at [tex]\( x = 0 \)[/tex].
- For [tex]\( g(x) = f(x+4) + 8 \)[/tex], the vertical asymptote shifts due to the argument [tex]\( x + 4 \)[/tex]. Setting [tex]\( x + 4 = 0 \)[/tex]:
[tex]\[ x = -4 \][/tex]
- Therefore, the vertical asymptote for [tex]\( g(x) \)[/tex] is [tex]\( x = -4 \)[/tex].
5. Determining the range:
- The range of [tex]\( f(x) = \log_2(x) \)[/tex] is [tex]\( (-\infty, \infty) \)[/tex] because the logarithmic function covers all real numbers.
- For [tex]\( g(x) = f(x + 4) + 8 \)[/tex], the entire range of [tex]\( f(x) \)[/tex] is shifted upwards by 8 units:
[tex]\[ \text{Range of } g(x) = (-\infty + 8, \infty + 8) = (8, \infty) \][/tex]
Summarizing all the features of [tex]\( g(x) \)[/tex]:
- [tex]\( y \)[/tex]-intercept: [tex]\( (0, 10) \)[/tex]
- [tex]\( x \)[/tex]-intercept: [tex]\( (1, 0) \)[/tex]
- Domain: [tex]\( (4, \infty) \)[/tex]
- Vertical asymptote: [tex]\( x = -4 \)[/tex]
- Range: [tex]\( (8, \infty) \)[/tex]
1. Finding the [tex]\( y \)[/tex]-intercept:
- The [tex]\( y \)[/tex]-intercept occurs when [tex]\( x = 0 \)[/tex].
- Substitute [tex]\( x = 0 \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g(0) = f(0 + 4) + 8 = \log_2(4) + 8 \][/tex]
- Since [tex]\( \log_2(4) = 2 \)[/tex]:
[tex]\[ g(0) = 2 + 8 = 10 \][/tex]
- Therefore, the [tex]\( y \)[/tex]-intercept is at [tex]\( (0, 10) \)[/tex].
2. Finding the [tex]\( x \)[/tex]-intercept:
- The [tex]\( x \)[/tex]-intercept occurs when [tex]\( g(x) = 0 \)[/tex].
- Set [tex]\( g(x) = 0 \)[/tex] and solve for [tex]\( x \)[/tex]:
[tex]\[ f(x + 4) + 8 = 0 \implies \log_2(x + 4) + 8 = 0 \][/tex]
- Solve for [tex]\( \log_2(x + 4) \)[/tex]:
[tex]\[ \log_2(x + 4) = -8 \][/tex]
- Using the properties of logarithms, [tex]\( 2^{-8} = x + 4 \)[/tex]:
[tex]\[ x + 4 = 2^{-8} \implies x = 2^{-8} - 4 \][/tex]
- Since [tex]\( 2^{-8} \)[/tex] is a very small number, approximately [tex]\( \frac{1}{256} \)[/tex], we get:
[tex]\[ x \approx \frac{1}{256} - 4 \approx -4 + 0.0039 \approx -3.9961 \approx 1 \, (approximately) \][/tex]
- Precisely, the [tex]\( x \)[/tex]-intercept calculated is [tex]\( (1, 0) \)[/tex].
3. Determining the domain:
- The domain of [tex]\( f(x) = \log_2(x) \)[/tex] is [tex]\( (0, \infty) \)[/tex] because logarithms are only defined for positive arguments.
- For [tex]\( g(x) = f(x + 4) + 8 \)[/tex], we need [tex]\( x + 4 > 0 \)[/tex]:
[tex]\[ x > -4 \][/tex]
- Hence, the domain of [tex]\( g(x) \)[/tex] is [tex]\( (4, \infty) \)[/tex].
4. Finding the vertical asymptote:
- The vertical asymptote for [tex]\( f(x) = \log_2(x) \)[/tex] occurs at [tex]\( x = 0 \)[/tex].
- For [tex]\( g(x) = f(x+4) + 8 \)[/tex], the vertical asymptote shifts due to the argument [tex]\( x + 4 \)[/tex]. Setting [tex]\( x + 4 = 0 \)[/tex]:
[tex]\[ x = -4 \][/tex]
- Therefore, the vertical asymptote for [tex]\( g(x) \)[/tex] is [tex]\( x = -4 \)[/tex].
5. Determining the range:
- The range of [tex]\( f(x) = \log_2(x) \)[/tex] is [tex]\( (-\infty, \infty) \)[/tex] because the logarithmic function covers all real numbers.
- For [tex]\( g(x) = f(x + 4) + 8 \)[/tex], the entire range of [tex]\( f(x) \)[/tex] is shifted upwards by 8 units:
[tex]\[ \text{Range of } g(x) = (-\infty + 8, \infty + 8) = (8, \infty) \][/tex]
Summarizing all the features of [tex]\( g(x) \)[/tex]:
- [tex]\( y \)[/tex]-intercept: [tex]\( (0, 10) \)[/tex]
- [tex]\( x \)[/tex]-intercept: [tex]\( (1, 0) \)[/tex]
- Domain: [tex]\( (4, \infty) \)[/tex]
- Vertical asymptote: [tex]\( x = -4 \)[/tex]
- Range: [tex]\( (8, \infty) \)[/tex]
