Rate of Change and Initial Value

The initial value is the value of the dependent variable when the independent variable is $0$: the starting amount, a flat fee, the water already in a pool. The rate of change tells you how much the dependent variable changes for each increase of 1 in the independent variable. In a real situation it comes with units that use the word "per": dollars per kilometre, litres per minute. To calculate it from any two rows of a table or two points on the graph, divide the change in the dependent variable by the change in the independent variable.

Look at how the same two numbers show up in each representation of the taxi fare. In the story they are "4 dollars to start" and "2 dollars per km". In the table, the cost in the row where $d=0$ is the initial value, 4, and the first difference (with $d$ going up by 1) is the rate of change, 2. On the graph, the line meets the vertical axis at $(0,4)$, and it rises 2 dollars for every 1 km to the right. In the equation $C=2d+4$, the number multiplying $d$ is the rate of change and the number added on is the initial value. In general, a linear relation can be written as $y=(\text{rate of change})x+(\text{initial value})\text{.}$

Rate of Change and Initial Value | MTH1W Grade 9 Math (Ontario Curriculum)

For a linear relation whose rate of change is not $0$: when the initial value is $0$, the relation is a direct variation. Gas is an example: at 1.60 dollars per litre, $C=1.6V$. Zero litres cost zero dollars, so the line passes through the origin $(0,0)$, and doubling the amount doubles the cost: 20 L cost 32 dollars and 40 L cost 64 dollars. When the initial value is not $0$, the relation is a partial variation, like the taxi fare $C=2d+4$. Part of the cost is fixed and part varies. Doubling the distance does not double the cost: a 3 km ride costs $2(3)+4=10$ dollars, but a 6 km ride costs $2(6)+4=16$ dollars, not 20.

5.1/5.2 Direct vs Partial Variation (Grade 9 Academic MPM1D)

A backyard pool already holds 2000 L of water, and a hose adds 25 L every minute. The initial value is 2000 L and the rate of change is $+25$ L per minute, so $V=25t+2000$, where $t$ is the time in minutes. After 1 hour ($t=60$) the pool holds $25(60)+2000=3500$ L. Now picture a hot tub holding 1200 L that drains at 40 L per minute. The amount decreases, so the rate of change is negative: $V=-40t+1200$, often written $V=1200-40t$. The tub is empty when $V=0$: $1200-40t=0$ gives $t=30$ minutes. On a graph, a positive rate makes the line rise from left to right, and a negative rate makes it fall.