From the table given
We can see that
when x=3, f(x)=2.5
when x=6, f(x) =4
when x=10, f(x)=6
when x=15, f(x)=8.5
If two coordinates are selected, then we can find the relationship by using the equation below
[tex]\frac{y_2-y_1}{x_2-x_1}=\frac{y_{}-y_1}{x_{}-x_1}[/tex]If we select the points
x=6 and f(x)=4
x=10 and f(x)=6
Therefore, If we substitute these values
[tex]\begin{gathered} \frac{6-4}{10-6}=\frac{y-4}{x-6} \\ \\ \frac{2}{4}=\frac{y-4}{x-6} \\ \\ \frac{1}{2}=\frac{y-4}{x-6} \\ \end{gathered}[/tex][tex]\begin{gathered} \text{cross multiply} \\ y-4=\frac{1}{2}(x-6) \\ \\ y-4=\frac{1}{2}x-3 \\ \\ y=\frac{1}{2}x-3+4 \\ \\ y=\frac{1}{2}x+1 \end{gathered}[/tex]Answer=>
y=1/2x+1
