The Lumenary · Grades 6–8

Graphs of change

Name: ______________________

Date: ______________________

You will learn to

Interpret distance–time and other real-world graphs, describing where a quantity increases, decreases or stays constant
Compute average rates of change between two points of a graph or table, with units, and interpret them
Prove that a graph is increasing exactly when its average rates of change are positive, and that the rate over a whole interval lies between the rates of its parts
Sketch a graph from a verbal description, and describe a situation from a graph
Explain why two situations can share an average rate yet behave differently, and why a round trip's average speed is not the mean of the two speeds

1.
On the graph of Keza's trip (segments A–E in the lesson), during which segment was she riding fastest?
- A.A
- B.B
- C.C
- D.D
- E.E
2.
A village water tank holds $1200$ litres at 8 a.m. and $300$ litres at 11 a.m. What is the average rate of change of the amount of water, in litres per hour?
3.
A bamboo shoot in a garden in Kyoto is $40$ cm tall on day $0$ , $130$ cm on day $3$ and $180$ cm on day $5$ . What is its average rate of growth from day $0$ to day $5$ , in cm per day?
4.
A cyclist rides slowly up a steep hill, then quickly down the other side, each at a steady speed. Which describes the graph of distance travelled against time?
- A.It rises gently, then rises steeply.
- B.It rises, then falls, like the hill.
- C.It rises steeply, then rises gently.
- D.It is flat, then rises.
5.
Over the same week, the graph of plant P's height is steeper than the graph of plant Q's height. Which statement must be true?
- A.P grew more than Q during that week.
- B.P is taller than Q.
- C.P was shorter than Q at the start.
- D.Q did not grow at all.
6.
A bus travels $90$ km at $45$ km/h and then returns the same $90$ km at $90$ km/h. What is its average speed for the whole round trip, in km/h?
7.
Amira cycles for a while at an average of $12$ km/h, then for a while at an average of $20$ km/h (she spends some time on each leg). Which of these could be her average speed for the whole ride? (Choose all that apply.)
- A. $10$ km/h
- B. $15$ km/h
- C. $16$ km/h
- D. $19$ km/h
- E. $25$ km/h
8.
Which Lean statement says that f : Rat → Rat is increasing (y gets larger whenever x gets larger)?
- A.∀ x y : Rat, x < y → f x < f y
- B.∀ x y : Rat, f x < f y
- C.∃ x y : Rat, x < y ∧ f x < f y
- D.∀ x : Rat, f x < f (x + 1)
9.
Write average_rates(points) for a list of points (x, y) with increasing $x$ -values. Return the list of average rates of change between consecutive points, as Fractions. For Keza's journey [(0, 0), (20, 4), (25, 4), (35, 6), (55, 6), (75, 0)] (minutes, km) it should return the rates in km per minute.
Computational — Practice
10.
In the model $d(t) = 4.9t^2$ , the average speed of the falling stone over the interval from $t = 3$ to $t = 3 + h$ (with $h \neq 0$ ) is …
- A. $29.4 + 4.9h$ m/s
- B. $44.1$ m/s
- C. $19.6 + 4.9h$ m/s
- D. $4.9h$ m/s

Answer key — Graphs of change

1.
The steepest segment is E: $6$ km in $20$ min $= 0.3$ km/min $= 18$ km/h. A and C are both $12$ km/h; B and D are flat (stopped). Going down only means getting closer to home.
2.
$\dfrac{300 - 1200}{11 - 8} = \dfrac{-900}{3} = -300$ litres per hour. On average the tank lost $300$ litres every hour.
3.
$\dfrac{180 - 40}{5} = \dfrac{140}{5} = 28$ cm per day. Notice that $28$ lies between the rates of the two stretches ( $30$ and $25$ ) but is not their plain average $27.5$ : it is the weighted average $\frac{3 \cdot 30 + 2 \cdot 25}{5} = 28$ .
4.
Distance travelled keeps increasing, so the graph always rises. Uphill she is slow (gentle slope), downhill she is fast (steep slope). The graph does not look like the hill at all.
5.
A steeper graph means a larger average rate of change, and over the same week that means a larger change in height: P grew more. It says nothing about which plant is taller — that is read from the heights of the graphs, not their slopes.
6.
The trip out takes $\frac{90}{45} = 2$ hours and the trip back $\frac{90}{90} = 1$ hour. Average speed $= \frac{180}{3} = 60$ km/h, which matches the formula $\frac{2uv}{u + v} = \frac{2 \cdot 45 \cdot 90}{135} = 60$ . It is less than the mean $67.5$ km/h.
7.
With $t_1$ hours at $12$ km/h and $t_2$ hours at $20$ km/h, the average speed is $\frac{12 t_1 + 20 t_2}{t_1 + t_2}$ , a weighted average of $12$ and $20$ . With $t_1, t_2 > 0$ it lies strictly between $12$ and $20$ , and by choosing the times every such value is possible. So $15$ , $16$ and $19$ are possible; $10$ and $25$ are not.
8.
∀ x y : Rat, x < y → f x < f y. It is the left-hand side of increasing_iff_avg_rate_pos in the lesson, which proves it equivalent to "every average rate of change is positive".

from fractions import Fraction

def average_rates(points):
    rates = []
    for (x1, y1), (x2, y2) in zip(points, points[1:]):
        rates.append(Fraction(y2 - y1, x2 - x1))
    return rates

zip(points, points[1:]) pairs each point with the next one. Exact fractions keep rates such as $\frac{1}{5}$ km/min exact; multiplying by $60$ converts them to km/h.

10.
$\dfrac{4.9(3 + h)^2 - 4.9 \cdot 3^2}{h} = \dfrac{4.9(6h + h^2)}{h} = 4.9(6 + h) = 29.4 + 4.9h$ . As $h \to 0$ this approaches $29.4$ m/s, the instantaneous speed at $t = 3$ — exactly $9.8 \times 3$ , matching the velocity formula $v = 9.8t$ from the physics viewpoint.