Certainly! Let's analyze the given functions and their transformation:
1. The original function is given by:
[tex]\[
f(x) = \frac{1}{x}
\][/tex]
2. The transformed function is:
[tex]\[
g(x) = \frac{1}{x} + 9
\][/tex]
To determine how the graph of [tex]\(f(x) = \frac{1}{x}\)[/tex] is transformed to become [tex]\(g(x) = \frac{1}{x} + 9\)[/tex], we need to examine what happens when we add 9 to the original function [tex]\(f(x)\)[/tex].
### Step-by-Step Analysis:
- Start with the function [tex]\(f(x)\)[/tex]:
[tex]\[
f(x) = \frac{1}{x}
\][/tex]
This is the basic reciprocal function. Its graph is a hyperbola with two branches, one in the first quadrant and one in the third quadrant. It is asymptotic to both the x-axis (y = 0) and the y-axis (x = 0).
- Define the transformation:
To obtain [tex]\(g(x)\)[/tex], we add 9 to [tex]\(f(x)\)[/tex]:
[tex]\[
g(x) = \frac{1}{x} + 9
\][/tex]
Adding a constant value to a function [tex]\(f(x)\)[/tex] results in a vertical shift of the graph. Specifically, when we add a positive constant [tex]\(c\)[/tex] to [tex]\(f(x)\)[/tex], the graph of [tex]\(f(x)\)[/tex] shifts upward by [tex]\(c\)[/tex] units.
- Determine the direction of the shift:
In this case, the constant added to [tex]\(f(x)\)[/tex] is 9. Therefore, the entire graph of [tex]\(f(x) = \frac{1}{x}\)[/tex] will be shifted upward by 9 units.
### Conclusion:
The effect on the graph of [tex]\(f(x) = \frac{1}{x}\)[/tex] when transformed to [tex]\(g(x) = \frac{1}{x} + 9\)[/tex] is that the graph of [tex]\(f(x)\)[/tex] is shifted 9 units up.
So, the correct answer is:
[tex]\[ \boxed{D. \text{The graph of } f(x) \text{ is shifted 9 units up.}} \][/tex]
