Many real situations change at a constant rate: a bill that adds the same charge each hour, a tank that drains steadily. With slope-intercept form you can turn such a story into an equation, then use it to predict.
Why it works
A constant rate means equal steps in $x$ add equal amounts to $y$, so the data lie on a line. Two points pin down a line: their rise over run is the rate, and walking back along the line to $x = 0$ gives the starting value.
Key idea
Rate: $m = \dfrac{y_2 - y_1}{x_2 - x_1}$. Starting value: $b = y - mx$, using either point. Model: $y = mx + b$. In context, $m$ is the change per unit of $x$ and $b$ is the value when $x = 0$.
Worked examples
Worked example: Two points in a story
A plumber’s bill is \$165 for 3 hours and \$245 for 5 hours, with a constant charge per hour. Write a model.
- Rate: $m = \frac{245 - 165}{5 - 3} = \frac{80}{2} = 40$ dollars per hour.
- Start: $b = 165 - 40 \cdot 3 = 165 - 120 = 45$ dollars.
- Model: $y = 40x + 45$. The 45 is a fixed charge for coming out; the 40 is the charge per hour.
Answer: $y = 40x + 45$
Worked example: A table that starts late
$$\begin{array}{c|ccc} x & 2 & 4 & 6 \\ \hline y & 31 & 43 & 55 \end{array}$$ Find the rate and the starting value.
- Rate: $\frac{43 - 31}{4 - 2} = 6$.
- The table starts at $x = 2$, so 31 is not the start. Go back 2 steps: $b = 31 - 6 \cdot 2 = 19$.
Answer: rate 6, starting value 19: $y = 6x + 19$
Worked example: Predict
Using $y = 40x + 45$, what would 7 hours cost?
- $y = 40 \cdot 7 + 45 = 280 + 45 = 325$.
Answer: \$325
Common mistake
Using the first value in a table as the starting value when the table does not begin at $x = 0$. In Example 2, 31 is the value at $x = 2$. The starting value is $b$, the value at $x = 0$: work backward with the rate.