Write a linear function $f$ with $f(-3)=1$ and $f(13)=5$. \[ f(x)=\square \]

Step 1 ：Calculate the slope (m) using the formula \(m = \frac{y2 - y1}{x2 - x1}\). Here, we have two points (-3,1) and (13,5). So, \(x1 = -3\), \(y1 = 1\), \(x2 = 13\), and \(y2 = 5\).

Step 2 ：Substitute these values into the formula to get \(m = \frac{5 - 1}{13 - (-3)} = \frac{4}{16} = \frac{1}{4}\). So, the slope of the line is \(\frac{1}{4}\).

Step 3 ：The equation of a line is given by \(y = mx + b\), where m is the slope and b is the y-intercept. Substitute one of the points and the slope into this equation to solve for b. Let's use the point (-3,1).

Step 4 ：Substitute these values in to get \(1 = \frac{1}{4} * -3 + b\), which simplifies to \(1 = -\frac{3}{4} + b\). Solving for b gives \(b = 1 + \frac{3}{4} = 1.75\). So, the y-intercept is 1.75.

Step 5 ：Therefore, the linear function is \(f(x) = \frac{1}{4}x + 1.75\).

Step 6 ：Check the work by substituting the points into the equation: \(f(-3) = \frac{1}{4}*(-3) + 1.75 = 1\) and \(f(13) = \frac{1}{4}*(13) + 1.75 = 5\).

Step 7 ：\(\boxed{f(x) = \frac{1}{4}x + 1.75}\) is the final answer.