1. **State the problem:** We are given two points on a linear function relating chirps per minute $x$ to temperature $y$: $(40, 50)$ and $(80, 60)$. The slope (rate of change) is $\frac{1}{4}$. We need to find the temperature when there are no chirps, i.e., the initial value or $y$-intercept when $x=0$. 2. **Recall the formula for a linear function:** $$y = mx + b$$ where $m$ is the slope and $b$ is the $y$-intercept (initial value). 3. **Use the slope given:** $$m = \frac{1}{4}$$ 4. **Plug in one point to find $b$:** Using point $(40, 50)$, $$50 = \frac{1}{4} \times 40 + b$$ 5. **Calculate:** $$50 = 10 + b$$ 6. **Solve for $b$:** $$b = 50 - 10 = 40$$ 7. **Interpretation:** The temperature when there are no chirps ($x=0$) is $40$ degrees Fahrenheit. **Final answer:** $$\boxed{40}$$