Write a linear function $f$ with $f(-3)=1$ and $f(13)=5$.
$$f(x)=◻$$

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 $1=\frac{1}{4}\ast -3+b$, 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: $f(-3)=\frac{1}{4}\ast (-3)+1.75=1$ and $f(13)=\frac{1}{4}\ast (13)+1.75=5$.

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