We are given the following:

• The months February through June correspond to $x$-values ranging from 2 (for February) to 6 (for June).
• The temperatures for each month (based on the graph) seem to be approximately:
  • February (x=2): $10^\circ F$
  • March (x=3): $30^\circ F$
  • April (x=4): $45^\circ F$
  • May (x=5): $60^\circ F$
  • June (x=6): $70^\circ F$

We are tasked with finding the linear function $f(x)$ that best models this data. To do so, we should:

1. Calculate the slope $m$ using two points on the graph. Let’s use the points for February $(2, 10)$ and June $(6, 70)$: $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{70 - 10}{6 - 2} = \frac{60}{4} = 15$ So, the slope is $m = 15$.

2. Use the slope-intercept form $y = mx + b$ and substitute one of the points to solve for the y-intercept $b$. Using the point $(2, 10)$: $10 = 15(2) + b \implies 10 = 30 + b \implies b = 10 - 30 = -20$

Thus, the equation of the line is $f(x) = 15x - 20$. However, none of the provided options match this exactly. The closest match among the choices is $f(x) = 13x - 12$, which suggests that this option may have been selected by estimating the data points differently or using a slightly less steep slope.

Would you like further clarification, or do you have any additional questions?

Here are 5 related questions:

1. How can we confirm the accuracy of a linear regression model?
2. How would this problem change if we used Celsius instead of Fahrenheit?
3. What other forms of functions (besides linear) could model temperature data over time?
4. How would increasing the data set (more months) affect the model?
5. What role does the slope play in understanding temperature changes?

Tip: When creating models based on data, it's important to check if a linear approximation is appropriate or if a different function (e.g., quadratic) fits better.