From a Graph to an Average Rate
Tables give values in rows and columns. A graph presents the same kind of information visually: input values lie along the horizontal axis, and the corresponding output values lie along the vertical axis. To estimate an average rate from a graph, identify the two input values of interest, read the graph’s output at each one, and compare the change in output with the change in input.
The result is an estimate because a plotted curve may pass between grid lines, a point may be thick or blurry, or the axis labels may require interpolation. The graph’s scale matters too: one grid square on the vertical axis need not represent the same numerical amount as one square on the horizontal axis.
The two graph points are \((x_1,f(x_1))\) and \((x_2,f(x_2))\). The line joining them is a secant line, and its slope is the same quotient. The rate’s units are output units per input unit. The order of the endpoints must be consistent: use the output change from the first point to the second and the input change in that same direction.
Read the Axis Scales Before Counting
A grid square has no fixed numerical value on its own. Read the labels and tick marks first, then determine the value represented by each square or subdivision. If adjacent labelled ticks are 2 units apart and there are four equal grid intervals between them, each interval represents \(2/4=0.5\) units.
Read the horizontal and vertical scales separately. For example, moving three squares to the right might represent an input change of 6, while moving three squares upward might represent an output change of 15. Equal visual distances do not necessarily mean equal numerical changes.
Use the horizontal axis labels and grid to identify \(x_1\) and \(x_2\).
At each input, follow vertically to the plotted function and estimate its height using the vertical-axis scale.
Find \(f(x_2)-f(x_1)\) and \(x_2-x_1\), keeping the endpoint order the same.
Compute output change divided by input change, then state the result as an estimate with appropriate units.
Why the Secant Slope Gives the Average Rate
Suppose the graph passes through the two points \((x_1,f(x_1))\) and \((x_2,f(x_2))\), where \(x_1\ne x_2\). A line through two points has slope equal to its vertical change divided by its horizontal change. For the line joining these graph points, those changes are \(f(x_2)-f(x_1)\) and \(x_2-x_1\).
The average rate of change is defined by that same change quotient. Therefore, the secant slope is the average rate over the interval. If the coordinates are estimated from a graph, the quotient is an estimate of that rate. This reasoning uses endpoint changes only; it does not require the function to be a straight line between the endpoints.
Worked Example: Different Scales on the Two Axes
A graph has horizontal-axis labels \(0,2,4,6,8\), with one grid interval between successive labels. Its vertical-axis labels \(0,5,10,15,20\) are also one grid interval apart. The plotted function passes through the readable points \((2,5)\) and \((8,15)\). Estimate the average rate from \(x=2\) to \(x=8\).
The input change is
The output change is
Thus the estimated average rate is
The estimate is about \(1.67\) output units per input unit. To check the calculation, multiply the rate by the input change: \(\frac{5}{3}\cdot 6=10\), which recovers the output change from the two graph readings.
Signs, Units, and Visual Appearance
If the second endpoint lies higher than the first, the output change is positive. If the second endpoint lies lower, it is negative. The sign of the average rate therefore describes the net direction of change over the chosen interval. A negative rate does not mean that every point between the endpoints decreases; it says only that the ending output is below the starting output.
Units also help check whether the calculation makes sense. If the vertical axis records distance in metres and the horizontal axis records time in seconds, the quotient has units of metres per second. Write units as output units divided by input units, not as a unitless number.
Do not estimate a numerical rate by judging the angle of the curve or secant line alone. A graph’s visual steepness depends on how the horizontal and vertical scales are drawn, including the size of the plotted area. The numerical rate comes from coordinate differences measured with the axis scales.
Worked Example: A Decreasing Graph with Unequal Axis Scales
On a graph, each horizontal grid interval represents \(0.5\) input units and each vertical grid interval represents \(3\) output units. At \(x=-1\), the curve is at \(y=6\). At \(x=2\), it is at \(y=-3\). Estimate the average rate over this interval.
The input change is
The output change is
Therefore, the average rate is
The negative sign is appropriate because the graph’s ending output is lower than its starting output. The rate is \(-3\) output units per input unit. Check by multiplying by the input change: \((-3)(3)=-9\), which agrees with the change from \(6\) to \(-3\). The stated grid scales help interpret the coordinates; once the coordinates are read, use their numerical values rather than counting squares again.
Reading Uncertainty
A graph may not show an endpoint value exactly. Instead of pretending that a reading is exact, state a reasonable estimate and, when useful, a range of plausible values. The grid spacing, thickness of the plotted curve, and clarity of the image all affect how precisely an output can be read.
Suppose the input endpoints are known, and the vertical readings have bounded uncertainty: \(f(x_1)\) is within \(e_1\) of its estimated value and \(f(x_2)\) is within \(e_2\) of its estimated value. Then the uncertainty in the output difference can be as large as \(e_1+e_2\). To see why, write the two actual values as their estimates plus errors, where the absolute values of the errors are at most \(e_1\) and \(e_2\). The error in their difference is the second error minus the first, whose absolute value is at most the sum of their absolute values.
When \(x_2>x_1\), dividing by the positive interval length preserves the order of the lower and upper bounds. Thus a rate estimate can be reported with a corresponding interval. This bound assumes that the horizontal endpoint values are known; uncertainty in those inputs would also affect the denominator and requires separate consideration.
Worked Example: Reporting a Rate Range
A graph is read at \(x_1=1\) and \(x_2=7\). The estimated output values are \(4.2\) and \(10.1\). Because the plotted curve is thick, suppose the first reading is uncertain by at most \(0.3\) output units and the second by at most \(0.4\). Find the central rate estimate and a range consistent with these reading bounds.
The input change is
The estimated output change is
The central rate estimate is therefore
The maximum uncertainty in the output difference is \(0.3+0.4=0.7\). Therefore, the actual output change is between \(5.9-0.7=5.2\) and \(5.9+0.7=6.6\). Since the input change is the positive number \(6\), the rate lies between
To two decimal places, the central estimate is \(0.98\), and the bounded range is approximately \(0.87\) to \(1.10\) output units per input unit. As a check, multiplying the endpoints of the rate range by \(6\) gives \(5.2\) and \(6.6\), the lower and upper possible output changes. The range communicates more than the central estimate alone: it shows how the graph-reading uncertainty affects the result.
A Common Scale Mistake
A frequent error is to count grid squares as though each square represented one unit on both axes. Suppose the graph’s horizontal scale is \(2\) input units per square and its vertical scale is \(4\) output units per square. Between two points, imagine the graph moves 3 squares to the right and 2 squares upward. The incorrect calculation treats those movements as changes of \(3\) and \(2\), giving a rate of \(2/3\).
That calculation ignores the labels. The actual input change is \(3\cdot2=6\) units, and the actual output change is \(2\cdot4=8\) units. The correct rate is
Checking by multiplication gives \(\frac{4}{3}\cdot6=8\), the correct vertical change. A reliable habit is to convert each graph movement into coordinate units before dividing.
Putting the Estimate in Context
A graph-based average rate describes the net output change across a specified input interval. The interval should be stated, particularly when the curve bends: different endpoint choices can produce different secant slopes. A good report includes the interval, the estimated rate, its units, and a brief indication of uncertainty if the graph does not support an exact reading.
When reporting a decimal, retain enough digits to reflect the quality of the graph rather than suggesting false precision. If the endpoints can only be read to about a tenth of an output unit, many extra decimal places in the quotient do not make the estimate more accurate. Keep unrounded values during the calculation and round the final estimate to a sensible precision.
Check Your Understanding
Use the graph scales and endpoint readings stated in each question. Show the changes before dividing.
- A curve is at output \(3\) when the input is \(1\), and at output \(11\) when the input is \(5\). What is the average-rate estimate over this interval?
- Each horizontal grid square represents \(2\) input units and each vertical square represents \(5\) output units. Between two points the curve moves 4 squares right and 3 squares down. What is the average rate?
- Why can two visually equal grid movements represent different numerical changes?
- A graph-based output-change estimate is \(12\), with uncertainty at most \(1\), and the known input change is \(4\). Give a central rate estimate and a range consistent with the bound.
- What does a negative average rate tell you about the two endpoint outputs, and what does it not necessarily tell you about every point between them?
