Gradient and intercept in context

Finding the gradient of a line as rise over run, reading it as a rate with units, and interpreting the vertical intercept as the starting value or fixed cost.

Every straight-line model answers two questions. Where does it start? And how fast does it change? The vertical intercept answers the first. The gradient answers the second.

Gradient is a rate

Gradient

$$m = \frac{\text{rise}}{\text{run}} = \frac{\text{change in } y}{\text{change in } x}$$

In context the gradient always has units: the $y$ unit per $x$ unit. Dollars per hour, litres per minute, kilometres per litre. A positive gradient means the quantity goes up as $x$ increases. A negative gradient means it goes down.

Example 1●

A plumber's charges are graphed against time. The line passes through $(0, 90)$ and $(4, 390)$. Find the gradient and say what it means.

$m = \dfrac{390 - 90}{4 - 0} = \dfrac{300}{4} = 75$

Rise: from 90 to 390 dollars. Run: from 0 to 4 hours.

The plumber charges $\$75$ per hour.

The units are dollars per hour.

Example 2●

A tank's volume falls along a line from $(0, 500)$ to $(20, 0)$, with time in minutes and volume in litres. Find the gradient and interpret it.

$m = \dfrac{0 - 500}{20 - 0} = -25$

The volume drops 500 L over 20 minutes.

The tank loses $25$ litres per minute.

Negative means decreasing.

InteractiveRise over run, with units

Try this

Pick the candle. Drag the two points to different places on the line. Does the gradient change? Write a sentence about what it means. Then pick the hire car and explain the intercept in words.

The intercept is the starting value

The vertical intercept is where the line crosses the vertical axis, where $x = 0$. In context that is the starting amount, or a fixed cost you pay before any usage.

Plumber: the intercept $90$ is the $\$90$ call-out fee, paid even for a zero-hour job.
Tank: the intercept $500$ is the volume at the start.

Example 3●●

A hire car costs $\$65$ a day plus a charge per kilometre. A $240$ km trip costs $\$161$ for one day. Find the gradient and intercept of the cost line and interpret them.

Intercept $= 65$: the $\$65$ daily fee.

When the distance is 0 km you still pay the daily fee.

Gradient $= \dfrac{161 - 65}{240 - 0} = \dfrac{96}{240} = 0.4$

The extra 96 dollars covers 240 km.

The car costs $40$ cents per kilometre on top of the daily fee.

Units: dollars per kilometre.

Your turn

A taxi fare is graphed against distance. The line passes through $(0, 4.20)$ and $(10, 27.20)$. Find the gradient and interpret both the gradient and the intercept.

Show the working

$m = \dfrac{27.20 - 4.20}{10} = 2.3$. The fare rises by $\$2.30$ per kilometre. The intercept $4.20$ is the $\$4.20$ flagfall charged before the taxi moves.

Flashcards

Gradient from two points

$\dfrac{y_2 - y_1}{x_2 - x_1}$, as a rate in $y$ units per $x$ unit

Vertical intercept in context

The value when the input is zero: a starting amount or fixed cost

Negative gradient

The quantity decreases as $x$ increases

Practice

Find the gradient of the line through $(1, 2)$ and $(5, 14)$.
Find the gradient of the line through $(2, 10)$ and $(6, 4)$.
A line shows cost (dollars) against hours. Its gradient is $45$. This means:
A line shows the balance of a savings account against weeks. Its vertical intercept is $200$. This means:
A tank holds $500$ L at the start and is empty after $20$ minutes, draining steadily. What is the gradient of the volume–time graph?
A hire car costs $\$65$ a day, and $\$161$ for a day with $240$ km driven. What is the cost per km, in dollars?
Which of these has a negative gradient when graphed against time?
A taxi fare line passes through $(0, 4.20)$ and $(10, 27.20)$, distance in km and fare in dollars. Find the gradient.
Exam style: A mobile phone's battery charge (%) is graphed against hours of video playback. The line passes through $(0, 100)$ and $(8, 20)$. Which is the correct interpretation of the gradient?
Exam style: For the same phone, how many hours of video would take the battery from $100$% to $0$%?