Fit linear, quadratic and exponential functions to data on a scatter plot and use the fitted function to solve problems in context

Learning goal: Represent data on two quantitative variables on a scatter plot and describe how the variables are related; fit a linear, quadratic or exponential function to the data, either using a given function or choosing the function suggested by the context; and use the fitted function to solve problems in the context of the data, saying where its answers can and cannot be trusted.

From a scatter plot to a description. Data on two quantitative variables come in pairs, one pair of numbers for each thing measured. A scatter plot shows each pair as one point. The variable thought to explain or predict the other (the explanatory variable) goes on the horizontal axis, and the variable being predicted (the response variable) goes on the vertical axis. Describing how the variables are related means reporting four things: the direction (as x increases, does y tend to rise, to fall, or to rise and then fall?), the form (do the points follow a straight line, a curve that bends one way, or a curve with a peak?), the strength (how closely do the points follow that form?) and any unusual points far from the pattern. The description comes before any fitting, because the form decides which kind of function is worth fitting.

What it means to fit a function to the data. A fitted function is a formula y = f(x) whose graph passes close to the points and so summarizes the relationship. It will usually not pass through the points, because measured data vary. At each data point there is a vertical gap between the point and the graph: the error the function makes there. When you ask technology for a regression, you name a family of functions and the technology returns one member of it. For linear and quadratic regression it returns the member for which the sum of the squares of those vertical gaps is as small as possible. This is the method of least squares, and it is what the linear and quadratic regression commands of a graphing calculator, or a spreadsheet trendline, carry out (exponential regression is usually done differently, as explained below). You choose the family; the technology chooses the numbers. Choosing the family is the step that needs judgment, and two things guide it: the form seen in the scatter plot and what the context says about how the quantities behave.

Three families and what suggests each. A linear function y = mx + b is suggested when the context describes a constant rate (the same amount gained or lost for each unit of x) and the points lie near a straight line. In a table with equally spaced x-values, the differences between successive y-values are roughly equal. A quadratic function y = ax² + bx + c, with a ≠ 0, is suggested when a quantity rises to a maximum and then falls (or falls to a minimum and rises), when it is the product of two quantities that each change at a constant rate, or when it describes an area as a length changes or the height of a thrown object; the plot bends one way throughout, and when the data reach past the turning point it shows a peak or a valley with a rough line of symmetry (data that stop short of the turning point show only one arm of the curve). An exponential function y = a·bˣ, with b > 0 and b ≠ 1 (and a > 0 for the positive quantities modeled here), is suggested when the context describes change by a constant percent or a constant factor in each equal step; the plot curves, growing ever steeper when b > 1 or leveling off toward zero when b < 1, and in a table with equally spaced x-values the ratios of successive y-values are roughly equal. With measured data the differences or ratios are only roughly equal, never exactly.

A linear model worked in full. A hypothetical candle is lit and its height is measured each hour. The numbers are invented for teaching and chosen so that the fitted coefficients come out round. The pairs (hours burned, height in cm) are (0, 24.2), (1, 21.2), (2, 19.1), (3, 16.3), (4, 14.3), (5, 11.6), (6, 8.6), (7, 6.7). On a scatter plot with time on the horizontal axis and height on the vertical axis, the direction is negative, the form is linear, the association is strong and no point is unusual. The context agrees: a candle of even thickness that burns steadily loses about the same height each hour. The hourly differences are −3.0, −2.1, −2.8, −2.0, −2.7, −3.0 and −1.9 cm, roughly equal, with mean −2.5. Linear regression returns h = −2.5t + 24. Fitting a linear function for a scatter plot that suggests a linear association, the Algebra 1 and Math 1 part of this standard, can also be done by hand: lay a line through two points that follow the trend, here the first and last, (0, 24.2) and (7, 6.7). Its slope is (6.7 − 24.2) ÷ 7 = −2.5, so the hand-fitted line is h = 24.2 − 2.5t. It passes through those two points and lies above five of the other six, so its residuals average −0.2 cm; lowering it by 0.2 cm gives the regression line, which has the same slope. The slope says the candle loses 2.5 cm of height per hour; the intercept, 24 cm, is the model's height at lighting, close to the measured 24.2 cm. At t = 3 the function gives 16.5 cm against a measured 16.3 cm: close, not equal.

Now the fitted function is used to solve problems in the context of the data. How tall is the candle after 4.5 hours? h = 24 − 2.5(4.5) = 12.75, about 12.8 cm. When will the candle be 10 cm tall? Solve 24 − 2.5t = 10: 2.5t = 14, t = 5.6 hours, which is 5 hours 36 minutes and sits sensibly between the measurements of 11.6 cm at 5 hours and 8.6 cm at 6 hours. When will the candle burn out? Solve 24 − 2.5t = 0: t = 9.6 hours. The first two answers are interpolation, predictions inside the range of the data (0 to 7 hours). The third is extrapolation, a prediction outside that range. It assumes the steady rate continues for 2.6 hours past the last measurement, which the data cannot confirm: the candle may widen at its base, or the flame may drown in the holder. It is reported as "about 9.6 hours if the burning rate stays the same". Push the function further and it gives nonsense: at 12 hours it predicts 24 − 30 = −6 cm.

How technology fits an exponential function. Exponential regression usually works differently from the other two. Taking the natural logarithm of y = a·bˣ gives ln y = ln a + (ln b)·x, which is a linear function of x. Many calculators and spreadsheets therefore fit a least-squares line to the points (x, ln y) and convert back: a is e raised to that line's intercept, and b is e raised to its slope. This method has two conditions. Every y-value must be positive, because the logarithm of zero or of a negative number is undefined; data with a zero in the y-column make the command fail. And the method makes the squared gaps in ln y smallest, not the squared gaps in y, so a tool that works on y directly returns different coefficients: close to these when the points lie near an exponential curve, noticeably different when they do not. Either is acceptable if you say which method was used. The exponential fits in this lesson use the logarithm method.

Given functions. Sometimes the function is supplied: by a manufacturer, by a scientific law or by an earlier study. Then nothing has to be fitted, but the given function is still compared with the data before it is used, because a formula that misses the points cannot be trusted to answer questions about them.

The misconception: the closest curve is always the right one. It is tempting to try every regression command and keep whichever curve passes nearest the points. Closeness matters, but inside a short range of data two different families can both pass close to the points and then disagree completely beyond it. A function can pass close to seven points without being a good description of the situation. The worked examples below show a quadratic function that fits a bouncing ball's data to within 2 cm and then predicts that the ball bounces higher and higher. The choice of family is made from the form of the plot together with the context; closeness then confirms or questions that choice, and the lesson on residuals makes that check sharper.

Worked quadratic model: A hypothetical school club sells T-shirts at seven different prices, one week each, and records the week's profit. The pairs (price in dollars, profit in dollars) are (6, 142), (8, 236), (10, 300), (12, 323), (14, 301), (16, 237), (18, 141). The scatter plot rises to a peak near $12 and falls again, roughly symmetrically, so the form is not linear; the least-squares line for these points is the horizontal line P = 240, which misses the middle point by 83 and the end points by 98 and 99. The context suggests a quadratic function: profit is the profit on each shirt times the number sold; raising the price raises the first factor and lowers the second, and if each factor changes by a constant amount per dollar, their product is a quadratic function of the price. Quadratic regression returns P = −5p² + 120p − 400. Check it against the data: P(10) = −500 + 1,200 − 400 = 300 (observed 300) and P(8) = −320 + 960 − 400 = 240 (observed 236). Now solve problems. Best price: the vertex is at p = −120 ÷ (2 × −5) = 12, where P = −720 + 1,440 − 400 = 320, so the model says a $12 price gives about $320 a week (observed 323). Prices earning at least $275: write P = −5(p − 12)² + 320; then −5(p − 12)² + 320 ≥ 275 gives (p − 12)² ≤ 9, so 9 ≤ p ≤ 15. Both answers are inside the tested prices of $6 to $18. The model's zeros, p = 4 and p = 20, lie outside them, so "the club breaks even at $4 and $20" is an extrapolation and is reported as one.

Worked exponential model: A hypothetical ball is dropped from 100 cm and the height of each rebound is read from a video. The pairs (bounce number, height in cm), with bounce 0 as the release, are (0, 100), (1, 78), (2, 64), (3, 51), (4, 40), (5, 33), (6, 26). The plot falls steeply at first and then levels off: negative, curved, strong. The differences are −22, −14, −13, −11, −7 and −7, far from equal, so a linear function is not suggested. The ratios of successive heights are 0.780, 0.821, 0.797, 0.784, 0.825 and 0.788, roughly equal, and the context agrees: a ball tends to return to about the same fraction of its previous height on each bounce. Exponential regression by the logarithm method returns a = 99.06 and b = 0.8005, so the fitted function, rounded, is h = 99.1(0.800)ⁿ. Each bounce reaches about 80% of the one before, a loss of about 20% per bounce, and 99.1 cm is the model's value at bounce 0 (the true release height was 100 cm). Check: at n = 3 the function gives 99.1 × 0.512 = 50.7 cm against a measured 51 cm. Solve problems. Height of bounce 8: 99.1 × 0.8⁸ = 99.1 × 0.1678 = 16.6 cm. First bounce lower than 10 cm: 99.1(0.8)ⁿ < 10 means 0.8ⁿ < 0.1009, so n > ln 0.1009 ÷ ln 0.8 = 10.28, and the first whole bounce is bounce 11. A table confirms it: 10.6 cm at bounce 10 and 8.5 cm at bounce 11. (These values use the rounded function; a calculator that keeps the unrounded coefficients gives 16.7, 10.7 and 8.6 cm, and the same answer, bounce 11.) Both answers extrapolate beyond bounce 6, and a real ball eventually stops bouncing, so they are stated as what the model predicts.

Worked use of a given function: Suppose the ball's packaging states that it "returns to 80% of its previous height". From a 100 cm drop that statement gives the function h = 100(0.8)ⁿ with no fitting at all. Compare it with the data first: it predicts 100, 80, 64, 51.2, 41.0, 32.8 and 26.2 cm against the measured 100, 78, 64, 51, 40, 33 and 26 cm; the largest miss is 2 cm, at bounce 1, against 1.3 cm for the fitted function. The given function describes these data nearly as well as the fitted one, so it can be used. Bounce 8: 100 × 0.8⁸ = 16.8 cm (the fitted function said 16.6 cm). First bounce under 10 cm: 0.8ⁿ < 0.1 gives n > 10.32, so bounce 11, where h = 8.6 cm; at bounce 10 it is still 10.7 cm. The two functions agree on every conclusion. Had the given function missed the points by a wide margin or in a steady pattern, the fitted function would have been preferred.

Worked choice between two families: Quadratic regression on the same bounce data returns h = 1.357n² − 20.14n + 98.79, and inside the data it also fits well: at n = 3 it gives 12.21 − 60.42 + 98.79 = 50.6 cm, and its largest miss is 2 cm. But a quadratic function with a positive leading coefficient has a minimum and then rises. Its vertex is at n = 20.14 ÷ (2 × 1.357) = 7.42, where h is about 24 cm; after that the predicted heights climb, to about 33 cm at bounce 10 and about 52.5 cm at bounce 12. A ball that bounces higher each time is impossible. Within bounces 0 to 6 both functions stay within about 2 cm of every point, so closeness alone would not rule the quadratic function out, but the context does: a constant fraction per bounce means an exponential function, and the exponential function's predictions keep shrinking toward zero as the situation requires.

Represent and fit: Enter the candle, T-shirt and bounce data sets into a spreadsheet or graphing calculator. For each one, (1) draw the scatter plot with labeled axes and units and write one sentence giving direction, form, strength and unusual points; (2) compute the successive differences and, for the bounce data, the successive ratios, and say which family they suggest; (3) run the regression for that family, write the fitted function with its coefficients, and graph it over the points; (4) for the bounce data, also fit by hand: use the release height for a and the mean of the six ratios for b, and compare your function with the one from technology.

Independent performance: Collect or look up at least eight pairs of measurements of two quantitative variables for which you can argue from the context what kind of function to expect (for example, the height of water in a straight-sided container as equal cups are added, the area covered by a shadow as an object moves toward a lamp, or the temperature difference between a warm drink and the room every two minutes). (1) Make the scatter plot and describe how the variables are related. (2) State the function family suggested by the context and give the reason before you fit anything. (3) Fit the function with technology and name the method, including whether an exponential fit used logarithms. (4) Use the fitted function to solve two problems in context: one prediction of y from x, and one where you solve an equation or inequality for x. (5) Label each answer as interpolation or extrapolation and name one thing that could make the extrapolation fail. Success means the axes carry units, the family is justified from the context and the plot rather than from closeness alone, every answer has units and a sentence in context, and no prediction outside the data is reported as certain.