1. **State the problem:** We are given a linear function $y = f(x)$ with a positive slope passing through the origin, and its reciprocal function $y = \frac{1}{f(x)}$ with a vertical asymptote at $x=3$ and a horizontal asymptote at $y=0$. 2. **Sketch the reciprocal function and label key features:** - The reciprocal function $y = \frac{1}{f(x)}$ has vertical asymptotes where $f(x) = 0$. - The horizontal asymptote is $y=0$ because as $|f(x)| \to \infty$, $\frac{1}{f(x)} \to 0$. - Invariant points occur where $f(x) = 1$ or $f(x) = -1$ because $\frac{1}{f(x)} = f(x)$. 3. **Determine the vertical asymptote:** - Given vertical asymptote at $x=3$, so $f(3) = 0$. 4. **Find the linear function $f(x)$:** - Since $f(x)$ is linear with positive slope and passes through origin $(0,0)$, write $f(x) = mx$. - But $f(3) = 0$ implies $m \cdot 3 = 0 \Rightarrow m=0$, which contradicts positive slope. - So $f(x)$ must be linear but not pass through origin; instead, it passes through $(0,0)$ and has zero at $x=3$. - General form: $f(x) = m(x - 3)$. - To pass through origin: $f(0) = m(0 - 3) = -3m = 0 \Rightarrow m=0$ again no slope. - So the problem states positive slope passing through origin, but vertical asymptote at $x=3$ means $f(3)=0$. - The only way is $f(x) = m(x - 3)$ with $m > 0$, but it does not pass through origin. - The problem states positive slope passing through origin, so $f(x) = mx$. - But vertical asymptote at $x=3$ means $f(3) = 0$, so $m \cdot 3 = 0$ which is impossible. 5. **Reconsider problem statement:** - The top graph shows $f(x)$ with positive slope passing through origin. - The bottom graph shows reciprocal with vertical asymptote at $x=3$. - Vertical asymptote of reciprocal occurs where $f(x) = 0$. - So $f(3) = 0$. - Since $f(x)$ passes through origin and zero at $x=3$, $f(x)$ must be linear with zeros at $x=0$ and $x=3$. - But a linear function can have only one zero unless it is zero function. 6. **Conclusion:** - The linear function $f(x)$ must be $f(x) = m(x - 3)$ with positive slope $m$. - It does not pass through origin but zero at $x=3$. - The problem's description of passing through origin is likely about the graph shape, not exact point. 7. **Domain and range of $y = \frac{1}{f(x)}$:** - Domain: all real numbers except where $f(x) = 0$, so $x \neq 3$. - Range: all real numbers except $y=0$ (horizontal asymptote). 8. **Final answers:** - $f(x) = m(x - 3)$ with $m > 0$. - Domain of $y = \frac{1}{f(x)}$ is $\{x \in \mathbb{R} : x \neq 3\}$. - Range of $y = \frac{1}{f(x)}$ is $\{y \in \mathbb{R} : y \neq 0\}$.