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To determine which graph represents the function [tex]\( f(x) = \frac{2}{x-1} + 4 \)[/tex], let's analyze several key characteristics of the function:
1. Vertical Asymptote:
- A vertical asymptote occurs where the denominator of the function is zero, causing the function to approach infinity.
- For [tex]\( f(x) = \frac{2}{x-1} + 4 \)[/tex], set the denominator to zero: [tex]\( x - 1 = 0 \)[/tex].
- Solving for [tex]\( x \)[/tex] gives [tex]\( x = 1 \)[/tex].
- Therefore, there is a vertical asymptote at [tex]\( x = 1 \)[/tex].
2. Horizontal Asymptote:
- A horizontal asymptote is determined by the behavior of the function as [tex]\( x \)[/tex] approaches positive or negative infinity.
- As [tex]\( x \)[/tex] approaches [tex]\(\pm\infty\)[/tex], the term [tex]\( \frac{2}{x-1} \)[/tex] approaches 0.
- Hence, [tex]\( f(x) = \frac{2}{x-1} + 4 \)[/tex] approaches [tex]\( 4 \)[/tex].
- Therefore, there is a horizontal asymptote at [tex]\( y = 4 \)[/tex].
3. X-intercept:
- To find the x-intercept, set [tex]\( f(x) \)[/tex] to zero and solve for [tex]\( x \)[/tex]: [tex]\( f(x) = 0 \)[/tex].
- [tex]\( 0 = \frac{2}{x-1} + 4 \)[/tex].
- Rearrange to solve for [tex]\( x \)[/tex]:
[tex]\[ 0 = \frac{2}{x-1} + 4 \implies \frac{2}{x-1} = -4 \implies 2 = -4(x-1) \implies 2 = -4x + 4 \implies -4x = -2 \implies x = \frac{1}{2} \][/tex]
- Therefore, the x-intercept is [tex]\( x = \frac{1}{2} \)[/tex], or [tex]\( (0.5, 0) \)[/tex].
4. Y-intercept:
- To find the y-intercept, evaluate the function at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{2}{0-1} + 4 = \frac{2}{-1} + 4 = -2 + 4 = 2 \][/tex]
- Therefore, the y-intercept is [tex]\( y = 2 \)[/tex], or [tex]\( (0, 2) \)[/tex].
To summarize, the graph of the function [tex]\( f(x) = \frac{2}{x-1} + 4 \)[/tex] will have the following features:
- A vertical asymptote at [tex]\( x = 1 \)[/tex]
- A horizontal asymptote at [tex]\( y = 4 \)[/tex]
- An x-intercept at [tex]\( x = 0.5 \)[/tex]
- A y-intercept at [tex]\( y = 2 \)[/tex]
These characteristics will help identify the correct graph for the function [tex]\( f(x) = \frac{2}{x-1} + 4 \)[/tex].
1. Vertical Asymptote:
- A vertical asymptote occurs where the denominator of the function is zero, causing the function to approach infinity.
- For [tex]\( f(x) = \frac{2}{x-1} + 4 \)[/tex], set the denominator to zero: [tex]\( x - 1 = 0 \)[/tex].
- Solving for [tex]\( x \)[/tex] gives [tex]\( x = 1 \)[/tex].
- Therefore, there is a vertical asymptote at [tex]\( x = 1 \)[/tex].
2. Horizontal Asymptote:
- A horizontal asymptote is determined by the behavior of the function as [tex]\( x \)[/tex] approaches positive or negative infinity.
- As [tex]\( x \)[/tex] approaches [tex]\(\pm\infty\)[/tex], the term [tex]\( \frac{2}{x-1} \)[/tex] approaches 0.
- Hence, [tex]\( f(x) = \frac{2}{x-1} + 4 \)[/tex] approaches [tex]\( 4 \)[/tex].
- Therefore, there is a horizontal asymptote at [tex]\( y = 4 \)[/tex].
3. X-intercept:
- To find the x-intercept, set [tex]\( f(x) \)[/tex] to zero and solve for [tex]\( x \)[/tex]: [tex]\( f(x) = 0 \)[/tex].
- [tex]\( 0 = \frac{2}{x-1} + 4 \)[/tex].
- Rearrange to solve for [tex]\( x \)[/tex]:
[tex]\[ 0 = \frac{2}{x-1} + 4 \implies \frac{2}{x-1} = -4 \implies 2 = -4(x-1) \implies 2 = -4x + 4 \implies -4x = -2 \implies x = \frac{1}{2} \][/tex]
- Therefore, the x-intercept is [tex]\( x = \frac{1}{2} \)[/tex], or [tex]\( (0.5, 0) \)[/tex].
4. Y-intercept:
- To find the y-intercept, evaluate the function at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{2}{0-1} + 4 = \frac{2}{-1} + 4 = -2 + 4 = 2 \][/tex]
- Therefore, the y-intercept is [tex]\( y = 2 \)[/tex], or [tex]\( (0, 2) \)[/tex].
To summarize, the graph of the function [tex]\( f(x) = \frac{2}{x-1} + 4 \)[/tex] will have the following features:
- A vertical asymptote at [tex]\( x = 1 \)[/tex]
- A horizontal asymptote at [tex]\( y = 4 \)[/tex]
- An x-intercept at [tex]\( x = 0.5 \)[/tex]
- A y-intercept at [tex]\( y = 2 \)[/tex]
These characteristics will help identify the correct graph for the function [tex]\( f(x) = \frac{2}{x-1} + 4 \)[/tex].
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