Given:
A linear function starting at point C at (3, –2) and ending at point D at (–2, 3).
To find:
The rate of change and the initial value.
Solution:
If a linear function passes through two points [tex](x_1,y_1)\text{ and }(x_2,y_2)[/tex], then the equation of linear function is
[tex]y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)[/tex]
The given linear function passes through (3,-2) and (-2,3). So, the equation is
[tex]y-(-2)=\dfrac{3-(-2)}{-2-3}(x-3)[/tex]
[tex]y+2=\dfrac{3+2}{-5}(x-3)[/tex]
[tex]y+2=\dfrac{5}{-5}(x-3)[/tex]
[tex]y+2=-(x-3)[/tex]
Subtracting 2 on both sides, we get
[tex]y=-x+3-2[/tex]
[tex]y=-x+1[/tex]
On comparing this equation with slope intercept form [tex]y=mx+b[/tex], where, m is slope and b is y-intercept, we get
[tex]m=-1[/tex] and [tex]b=1[/tex]
Therefore, the rate of change is -1 and the initial value is 1.
