Reading a graph

A rain barrel graph rises from 20 to 60 liters in two hours, so the water came in at an average of 20 liters an hour.

When did the barrel hold 40 liters?

Try a question before reading on. A garden rain barrel, made up for this unit, fills with rain, is emptied onto the garden and fills again. The graph below records liters against hours. When did it hold exactly 40 liters? One answer jumps out. There is a second. Find both, and then check them in the section on the line at 40.

The rain barrel, hour by hour

Many functions reach us as graphs rather than formulas: temperature through a day, the price of fuel over a year, the water level in a tank. Take an example made up for this unit. A garden rain barrel holds 20 liters at noon (hour 0). Rain fills it to 60 liters by hour 2. The rain stops and the level stays at 60 until hour 4. Then someone waters the garden and the barrel empties by hour 7. A second shower brings it back to 30 liters by hour 9, where it stays until hour 10. The graph of this function W(t), liters against hours, is a chain of straight pieces.

The line at 40 meets the graph twice

To find W(1), go to 1 on the horizontal axis, up to the graph, and across to the vertical axis: 40 liters. Halfway between hour 0 and hour 2 the level is halfway between 20 and 60. To answer the reverse question, when was there 40 liters, draw the horizontal line at 40 and see where it meets the graph. It does so twice, at t = 1 while filling and at t = 5 while emptying. So W(t) = 40 has two solutions. The line at 70 meets nothing, so the equation W(t) = 70 has no solution; the barrel never held that much.

Increasing and decreasing intervals are written in hours

A function is increasing on an interval if its graph rises as you move to the right there, decreasing if it falls, and constant if it stays level. The barrel is increasing on (0, 2) and (7, 9), constant on (2, 4) and (9, 10), and decreasing on (4, 7). Intervals are written with x-values, the times, not with the heights. The highest value is 60, held over the whole stretch from 2 to 4; the lowest is 0, at t = 7. Where a graph switches from rising to falling there is a local maximum, and where it switches from falling to rising a local minimum.

20 liters an hour in, 20 liters an hour out

How fast did the water come in? The average rate of change of a function between a and b is the change in output divided by the change in input: (f(b) − f(a)) / (b − a). From hour 0 to hour 2 it is (60 − 20)/(2 − 0) = 20 liters per hour. From hour 4 to hour 7 it is (0 − 60)/(7 − 4) = −20 liters per hour; the negative sign says the water was leaving. On a graph this is the slope of the straight line joining the two points, so the rate of change of a line is the same everywhere, while a curved graph has different average rates on different intervals.

In the lab, drag across the graph to read W(t) at any time, and switch on Solve f(x) = c to slide a horizontal line up and down. The purple marks are every solution of W(t) = c; at c = 60 or c = 30 you get a whole stretch of solutions instead of single points, because the graph is level there.