Your picture unfortunately cuts off the upper limit of the interval, so let's set it to [tex]b>2[/tex]. Then the average rate of change of a function [tex]y=f(x)[/tex] over the interval [tex][2,b][/tex] is
[tex]\dfrac{f(b)-f(2)}{b-2}[/tex]
We have
[tex]y=3x+5\implies\dfrac{(3b+5)-(3\cdot2+5)}{b-2}=\dfrac{3b-6}{b-2}=\dfrac{3(b-2)}{b-2}=3[/tex]
(where we can cancel the factors of [tex]b-2[/tex] because we assume [tex]b>2[/tex])
[tex]y=3x^2+1\implies\dfrac{(3b^2+1)-(3\cdot2^2+1)}{b-2}=\dfrac{3b^2-12}{b-2}=\dfrac{3(b-2)(b+2)}{b-2}=3(b+2)[/tex]
[tex]y=3^x\implies\dfrac{3^b-3^2}{b-2}=\dfrac{3^b-9}{b-2}[/tex]
