Question

A relation is plotted as a linear function on a coordinate plane starting at point C at (3, –2) and ending at point D at (–2, 3). What is the rate of change for the linear function and what is its initial value?The rate of change is ______ and the initial value is ______.

Answer 1

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
`(x_1,y_1)\text{ and }(x_2,y_2)`, then the equation of linear function is

`y-y_1=(y_2-y_1)/(x_2-x_1)(x-x_1)`

The given linear function passes through (3,-2) and (-2,3). So, the equation is

`y-(-2)=(3-(-2))/(-2-3)(x-3)`

`y+2=(3+2)/(-5)(x-3)`

`y+2=(5)/(-5)(x-3)`

`y+2=-(x-3)`

Subtracting 2 on both sides, we get

`y=-x+3-2`

`y=-x+1`

On comparing this equation with slope intercept form
`y=mx+b`, where, m is slope and b is y-intercept, we get

`m=-1` and
`b=1`

Therefore, the rate of change is -1 and the initial value is 1.