Given the function:
[tex]y\text{ = -5x + 3}[/tex]The rate of change (slope) is expressed as
[tex]\begin{gathered} \text{slope = }\frac{rise}{run} \\ =\frac{y_2-y_1}{x_2-x_1} \end{gathered}[/tex]The slope is also evaluated from the general equation of the line function expressed as
[tex]\begin{gathered} y\text{ = mx + c} \\ \text{where} \\ m\Rightarrow slope \\ c\Rightarrow y-intercept \end{gathered}[/tex]In the given function,
[tex]y\text{ = -5x + 3}[/tex]This implies that in comparison with the general equation of the line function, the rate of change (slope) is evaluated to be -5.
In option A,
taking any two points for (x₁, y₁) and (x₂, y₂),
[tex]\begin{gathered} (x_1,y_1)\Rightarrow(4,\text{ -10)} \\ (x_2,y_2)\Rightarrow(8,\text{ -15)} \\ \end{gathered}[/tex]the slope is evaluated to be
[tex]\begin{gathered} \text{slope = }\frac{-15--10}{8-4}=\frac{-15+10}{8-4} \\ =-\frac{5}{4} \\ \text{slope = -}\frac{5}{4} \end{gathered}[/tex]In option B,
[tex]\begin{gathered} (x_{1,}y_1)\Rightarrow(-1,0) \\ (x_2,y_2)\Rightarrow(0,\text{ -5)} \end{gathered}[/tex]the slope is evaluated to be
[tex]\begin{gathered} \text{slope = }\frac{-5-0}{0--1}=-\frac{5}{1} \\ \text{slope =-5} \end{gathered}[/tex]In option C,
[tex]\begin{gathered} (x_{1,}y_1)\Rightarrow(0,4) \\ (x_2,y_2)\Rightarrow(3,22\text{)} \end{gathered}[/tex]the slope is evaluated to be
[tex]\begin{gathered} \text{slope = }\frac{22-4}{3-0}=\frac{18}{3} \\ \text{slope = 6} \end{gathered}[/tex]In option D,
[tex]\begin{gathered} (x_{1,}y_1)\Rightarrow(-1,0) \\ (x_2,y_2)\Rightarrow(0,4\text{)} \end{gathered}[/tex]the slope is thus evaluated as
[tex]\begin{gathered} \text{slope =}\frac{4-0}{0--1}=\frac{4}{1} \\ \text{slope = 4} \end{gathered}[/tex]Since the slope in option B is evaluated to be -5 which is equivalent to the slope of the function in question, the correct option is B.