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To graph the function [tex]\( r(x) = \frac{1}{x+4} \)[/tex] by using transformations of the graph of [tex]\( f(x) = \frac{1}{x} \)[/tex], we need to understand how the transformation will modify the original graph. Let's go through the process step-by-step.
1. Identify the base function and its basic properties:
The base function is [tex]\( f(x) = \frac{1}{x} \)[/tex].
- This function has a vertical asymptote at [tex]\( x = 0 \)[/tex].
- It also has a horizontal asymptote at [tex]\( y = 0 \)[/tex].
- The graph of [tex]\( f(x) = \frac{1}{x} \)[/tex] is hyperbolic and has two separate branches:
- For [tex]\( x > 0 \)[/tex], [tex]\( f(x) \)[/tex] is in the first quadrant and approaches zero as [tex]\( x \)[/tex] increases.
- For [tex]\( x < 0 \)[/tex], [tex]\( f(x) \)[/tex] is in the third quadrant and also approaches zero as [tex]\( x \)[/tex] becomes more negative.
2. Apply the transformation:
The function [tex]\( r(x) = \frac{1}{x+4} \)[/tex] can be seen as a horizontal shift of the base function [tex]\( f(x) = \frac{1}{x} \)[/tex].
- The transformation [tex]\( x \to x + 4 \)[/tex] shifts the graph 4 units to the left. This is because in the denominator, [tex]\( x + 4 \)[/tex] implies that we are replacing [tex]\( x \)[/tex] with [tex]\( x + 4 \)[/tex].
3. Determine the new asymptotes:
- Vertical Asymptote: For [tex]\( f(x) = \frac{1}{x} \)[/tex], the vertical asymptote is at [tex]\( x = 0 \)[/tex]. After shifting left by 4 units, the new vertical asymptote for [tex]\( r(x) = \frac{1}{x+4} \)[/tex] is at [tex]\( x = -4 \)[/tex].
- Horizontal Asymptote: The horizontal asymptote for [tex]\( f(x) = \frac{1}{x} \)[/tex] is at [tex]\( y = 0 \)[/tex]. This asymptote does not change with horizontal shifts. Therefore, the horizontal asymptote for [tex]\( r(x) = \frac{1}{x+4} \)[/tex] remains at [tex]\( y = 0 \)[/tex].
4. Compute key points:
Let's calculate some specific points on the graph to help us plot it accurately.
- For [tex]\( x = -6 \)[/tex], [tex]\( r(-6) = \frac{1}{-6+4} = \frac{1}{-2} = -0.5 \)[/tex].
- For [tex]\( x = -5 \)[/tex], [tex]\( r(-5) = \frac{1}{-5+4} = \frac{1}{-1} = -1.0 \)[/tex].
- For [tex]\( x = -4.5 \)[/tex], [tex]\( r(-4.5) = \frac{1}{-4.5+4} = \frac{1}{-0.5} = -2.0 \)[/tex].
- For [tex]\( x = -3.5 \)[/tex], [tex]\( r(-3.5) = \frac{1}{-3.5+4} = \frac{1}{0.5} = 2.0 \)[/tex].
- For [tex]\( x = -3 \)[/tex], [tex]\( r(-3) = \frac{1}{-3+4} = \frac{1}{1} = 1.0 \)[/tex].
These key points are:
- [tex]\( (-6, -0.5) \)[/tex]
- [tex]\( (-5, -1.0) \)[/tex]
- [tex]\( (-4.5, -2.0) \)[/tex]
- [tex]\( (-3.5, 2.0) \)[/tex]
- [tex]\( (-3, 1.0) \)[/tex]
5. Sketch the graph:
Using the asymptotes and the key points, we can plot the graph.
- The graph will have a vertical asymptote at [tex]\( x = -4 \)[/tex] and a horizontal asymptote at [tex]\( y = 0 \)[/tex].
- For values of [tex]\( x \)[/tex] less than [tex]\(-4\)[/tex], the function values will be negative, approaching zero as [tex]\( x \)[/tex] decreases.
- For values of [tex]\( x \)[/tex] greater than [tex]\(-4\)[/tex], the function values will be positive, approaching zero as [tex]\( x \)[/tex] increases.
- Plot the key points to shape the hyperbolic branches. One branch will be in the second quadrant (lower left), and the other will be in the first quadrant (upper right) adjacent to the vertical asymptote.
By following these steps, you should have an accurate graph of [tex]\( r(x) = \frac{1}{x+4} \)[/tex], showing the behavior of the function clearly around the asymptotes and through the plotted points.
1. Identify the base function and its basic properties:
The base function is [tex]\( f(x) = \frac{1}{x} \)[/tex].
- This function has a vertical asymptote at [tex]\( x = 0 \)[/tex].
- It also has a horizontal asymptote at [tex]\( y = 0 \)[/tex].
- The graph of [tex]\( f(x) = \frac{1}{x} \)[/tex] is hyperbolic and has two separate branches:
- For [tex]\( x > 0 \)[/tex], [tex]\( f(x) \)[/tex] is in the first quadrant and approaches zero as [tex]\( x \)[/tex] increases.
- For [tex]\( x < 0 \)[/tex], [tex]\( f(x) \)[/tex] is in the third quadrant and also approaches zero as [tex]\( x \)[/tex] becomes more negative.
2. Apply the transformation:
The function [tex]\( r(x) = \frac{1}{x+4} \)[/tex] can be seen as a horizontal shift of the base function [tex]\( f(x) = \frac{1}{x} \)[/tex].
- The transformation [tex]\( x \to x + 4 \)[/tex] shifts the graph 4 units to the left. This is because in the denominator, [tex]\( x + 4 \)[/tex] implies that we are replacing [tex]\( x \)[/tex] with [tex]\( x + 4 \)[/tex].
3. Determine the new asymptotes:
- Vertical Asymptote: For [tex]\( f(x) = \frac{1}{x} \)[/tex], the vertical asymptote is at [tex]\( x = 0 \)[/tex]. After shifting left by 4 units, the new vertical asymptote for [tex]\( r(x) = \frac{1}{x+4} \)[/tex] is at [tex]\( x = -4 \)[/tex].
- Horizontal Asymptote: The horizontal asymptote for [tex]\( f(x) = \frac{1}{x} \)[/tex] is at [tex]\( y = 0 \)[/tex]. This asymptote does not change with horizontal shifts. Therefore, the horizontal asymptote for [tex]\( r(x) = \frac{1}{x+4} \)[/tex] remains at [tex]\( y = 0 \)[/tex].
4. Compute key points:
Let's calculate some specific points on the graph to help us plot it accurately.
- For [tex]\( x = -6 \)[/tex], [tex]\( r(-6) = \frac{1}{-6+4} = \frac{1}{-2} = -0.5 \)[/tex].
- For [tex]\( x = -5 \)[/tex], [tex]\( r(-5) = \frac{1}{-5+4} = \frac{1}{-1} = -1.0 \)[/tex].
- For [tex]\( x = -4.5 \)[/tex], [tex]\( r(-4.5) = \frac{1}{-4.5+4} = \frac{1}{-0.5} = -2.0 \)[/tex].
- For [tex]\( x = -3.5 \)[/tex], [tex]\( r(-3.5) = \frac{1}{-3.5+4} = \frac{1}{0.5} = 2.0 \)[/tex].
- For [tex]\( x = -3 \)[/tex], [tex]\( r(-3) = \frac{1}{-3+4} = \frac{1}{1} = 1.0 \)[/tex].
These key points are:
- [tex]\( (-6, -0.5) \)[/tex]
- [tex]\( (-5, -1.0) \)[/tex]
- [tex]\( (-4.5, -2.0) \)[/tex]
- [tex]\( (-3.5, 2.0) \)[/tex]
- [tex]\( (-3, 1.0) \)[/tex]
5. Sketch the graph:
Using the asymptotes and the key points, we can plot the graph.
- The graph will have a vertical asymptote at [tex]\( x = -4 \)[/tex] and a horizontal asymptote at [tex]\( y = 0 \)[/tex].
- For values of [tex]\( x \)[/tex] less than [tex]\(-4\)[/tex], the function values will be negative, approaching zero as [tex]\( x \)[/tex] decreases.
- For values of [tex]\( x \)[/tex] greater than [tex]\(-4\)[/tex], the function values will be positive, approaching zero as [tex]\( x \)[/tex] increases.
- Plot the key points to shape the hyperbolic branches. One branch will be in the second quadrant (lower left), and the other will be in the first quadrant (upper right) adjacent to the vertical asymptote.
By following these steps, you should have an accurate graph of [tex]\( r(x) = \frac{1}{x+4} \)[/tex], showing the behavior of the function clearly around the asymptotes and through the plotted points.
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