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A. Quadratic Functions · Quadratic Functions

Finding a Quadratic Model for Data

A3.2 A3.1About 75 minutes
Learning goals
  • I can find the equation of a quadratic model from a scatter plot using an estimated vertex or estimated zeros.
  • I can compare my model with a regression equation from technology.
  • I can judge a model using residuals, R² and the context.
  • I can use a model to interpolate and extrapolate, and explain the risks.

Words to know

quadratic regression
A technology method that finds the quadratic y = ax² + bx + c that best fits a set of data.
R² (coefficient of determination)
A number from 0 to 1 showing how closely data fits a regression curve; closer to 1 is closer.
residual
Measured value minus predicted value.
interpolation / extrapolation
Predicting inside / outside the range of the data.
curve of best fit
A smooth curve that models the trend of the data.

From dots to a formula

In Collecting and Graphing Data That Can Be Modelled by a Quadratic, you gathered data and saw it curve like a parabola. A graph is useful, but an equation is more powerful: it lets you predict values you did not measure, find the exact maximum, and compare experiments.

In this lesson you will find the equation of the quadratic function that best models a data set, by hand using what you know about the forms, and then compare it with the curve of best fit that technology produces.

Strategy 1: vertex form (when you can see the vertex)

  1. Draw a smooth curve of best fit on the scatter plot.
  2. Estimate the vertex (h, k) from the curve (not necessarily a data point).
  3. Write y = a(x − h)² + k.
  4. Choose another point on the curve (or a reliable data point far from the vertex), substitute it, and solve for a.

Strategy 2: factored form (when you can see the zeros)

  1. Estimate the zeros r and s where the curve meets the horizontal axis.
  2. Write y = a(x − r)(x − s).
  3. Substitute another point (often near the top or bottom) and solve for a.

Strategy 3: technology (quadratic regression)

Desmos, a graphing calculator or a spreadsheet can compute the quadratic regression: the curve y = ax² + bx + c that makes the total squared vertical distances from the points as small as possible. In Desmos, type the data into a table, then type y₁ ~ ax₁² + bx₁ + c. The output includes R², a number from 0 to 1; the closer to 1, the closer the points are to the curve.

Worked example

Example 1: A tossed ball (vertex form)

A ball was tossed upward and filmed. Its height h in metres after t seconds:

t: 0, 0.2, 0.4, 0.6, 0.8, 1.0, 1.2

h: 1.1, 2.1, 3.0, 3.4, 3.5, 3.0, 2.2

Find a quadratic model by hand.

  1. Plot the points and draw a curve of best fit. The highest points are at t = 0.6 and t = 0.8 (3.4 and 3.5), so the curve peaks between them.
  2. Estimate the vertex: about (0.7, 3.5). So h = a(t − 0.7)² + 3.5.
  3. Use a point far from the vertex, (0, 1.1): 1.1 = a(0 − 0.7)² + 3.5, so 0.49a = −2.4 and a ≈ −4.9.
  4. Model: h = −4.9(t − 0.7)² + 3.5.
  5. Test it at t = 1.2: −4.9(0.25) + 3.5 = 2.275 ≈ 2.3. Measured 2.2. Close. ✓

h ≈ −4.9(t − 0.7)² + 3.5.

t (s)h (m)00.20.40.60.811.21.401234estimated vertex (0.7, 3.5)solid: hand model h = −4.9(t − 0.7)² + 3.5dashed: regression h = −4.94t² + 6.93t + 1.03
The ball data with the hand model (solid) and the technology regression (dashed). The two curves are so close that they almost overlap.
Worked example

Example 2: Comparing with technology

Desmos gives the quadratic regression h = −4.94t² + 6.93t + 1.03 with R² ≈ 0.994 for the ball data. Compare it with the hand model h = −4.9(t − 0.7)² + 3.5.

  1. Write the hand model in standard form: −4.9(t² − 1.4t + 0.49) + 3.5 = −4.9t² + 6.86t + 1.10.
  2. Compare coefficients: a: −4.9 vs −4.94; b: 6.86 vs 6.93; c: 1.10 vs 1.03. Very close.
  3. Compare predictions at t = 0.5: hand −4.9(0.04) + 3.5 = 3.30; regression −1.235 + 3.465 + 1.03 = 3.26. They differ by only 0.04 m.
  4. R² ≈ 0.994 is very close to 1, so a quadratic is a very good model for this data.
  5. Notice that a ≈ −4.9 in both. For an object moving freely near Earth's surface, the t² coefficient is about −4.9 when heights are in metres and time is in seconds (half of the acceleration due to gravity, 9.8 m/s²). The data agrees with physics.

The hand model and the regression agree closely; both have a ≈ −4.9.

Worked example

Example 3: A water jet (factored form)

A garden-hose jet of water was photographed against a grid. The height y (cm) of the jet at horizontal distance x (cm) from the nozzle:

x: 0, 20, 40, 60, 80, 100, 120

y: 0, 22, 37, 41, 36, 23, 1

Find a model in factored form.

  1. The jet starts at height 0 at x = 0 and lands at about x = 120. Zeros ≈ 0 and 120.
  2. y = a(x − 0)(x − 120) = ax(x − 120).
  3. The axis is halfway, at x = 60, and the data point (60, 41) is at the top. 41 = a(60)(60 − 120) = −3600a, so a = −41 ÷ 3600 ≈ −0.0114.
  4. Model: y = −0.0114x(x − 120).
  5. Technology gives y = −0.01125x² + 1.357x − 0.071 (R² ≈ 0.999). Our model in standard form is −0.0114x² + 1.368x. Very close.
  6. The regression has c = −0.071, a tiny negative height at the nozzle. The hand model starts exactly at 0, which matches the situation better there.

y ≈ −0.0114x(x − 120).

x (cm)y (cm)02040608010012001020304050zeros ≈ 0 and 120(60, 41)y = −0.0114x(x − 120)
The water-jet data with the factored-form model. The zeros at about 0 and 120 and the top point (60, 41) are enough to find the equation.

Judging a model

  • Residuals: residual = measured value − predicted value. Small residuals, some positive and some negative, mean a good fit.
  • R² close to 1 means the points lie close to the curve. It does not prove the situation is truly quadratic.
  • Context: does the model make sense at the ends of the domain (a height of 0 at launch, no negative heights)?
  • Interpolation (predicting inside the data range) is usually reliable. Extrapolation (predicting beyond it) is riskier, because the situation may change: the ball hits the ground, the bottle empties.
Just for fun

Your hand model and the computer's regression disagree in the second decimal place. Congratulations: you are now arguing with a computer about a ball, which is basically what scientists do all day.

Hands-on activity

Model your own data

You need: the data you collected in the ramp activity of Collecting and Graphing Data That Can Be Modelled by a Quadratic (or the ramp data from that lesson: t = 0, 0.5, 1.0, 1.5, 2.0, 2.5 s and d = 0, 4, 15, 34, 61, 94 cm), graph paper, a calculator, and Desmos if you have a device.

  1. Plot the data and draw a curve of best fit.
  2. The ball starts from rest, so the vertex is (0, 0). Write d = at².
  3. Use a point far from the vertex to find a. Try two different points and compare.
  4. Use your model to predict the distance at a time you did not measure (interpolate), and at a time beyond your data (extrapolate). Which prediction do you trust more? Why?
  5. If you have Desmos, enter the data and find the quadratic regression. Compare a, b, c and R² with your model.

What to look for: with the sample data, (2, 61) gives a = 15.25 and (2.5, 94) gives a = 15.04, so d ≈ 15t². At t = 1.75 s this predicts about 46 cm. Extrapolating to t = 4 s predicts 240 cm, which only works if the ramp is at least 2.4 m long. A regression of the sample data gives a close to 15, with b and c close to 0.

Your turn

1

A curve of best fit has estimated vertex (2, 9) and passes through (0, 1). Find its equation.

2

A curve of best fit has zeros 1 and 7 and passes through (3, −8). Find its equation.

3

Find a model for x = 0, 1, 2, 3, 4, 5, 6 and y = 12, 7, 4, 3, 4, 7, 12.

4

A kicked ball leaves the ground at t = 0, lands at t = 3.0 s, and its highest measured height is 11 m at t = 1.5 s. Find a model in factored form.

5

A calculator's quadratic regression for the data in the third question is y = x² − 6x + 12. Is it the same as your model?

6

What does R² = 0.98 tell you about a quadratic regression?

More practice: warm-up

1

Which form uses an estimated vertex?

2

Which form uses estimated zeros?

3

A parabola has vertex (0, 5) and passes through (1, 3). Find a.

4

Name a free technology tool that can do quadratic regression.

5

A measured value is 10 and the model predicts 9.4. What is the residual?

More practice: on your own

1

An arch was measured: x = 0, 10, 20, 30, 40, 50 m and y = 0, 16, 24, 24, 16, 0 m. Find a model and the maximum height.

2

For the ramp data, use (2, 61) to find a in d = at². Then use (2.5, 94). Compare.

3

The draining-bottle data (t = 0, 20, 40, 60, 80, 100 s; h = 20, 15.2, 11.0, 7.4, 4.6, 2.4 cm) has regression h = 0.00083t² − 0.259t + 20.02 with R² ≈ 0.99998. Use it to predict the height at t = 50 s.

4

Use the ball data points (0, 1.1), (0.6, 3.4) and (1.2, 2.2). The hand model predicts 1.10, 3.45 and 2.28; the regression predicts 1.03, 3.41 and 2.23. Find the residuals for each model. Which fits these three points better?

5

Explain the difference between interpolation and extrapolation, and why extrapolation is riskier.

Challenge

1

Give a reason why a hand-made model might be more useful than a regression in some situations.

2

For the tossed-ball regression h = −4.94t² + 6.93t + 1.03, use the physics model h = −4.9t² + v₀t + h₀ to estimate the ball's initial upward speed v₀ and the release height h₀.

3

Use the bottle regression h = 0.00083t² − 0.259t + 20.02 to estimate when the water level reaches 0. Then explain why the model should not be used after that time.

Check your understanding

Quiz

1.

To find a in y = a(x − h)² + k after estimating the vertex, you…

2.

A curve of best fit has zeros 0 and 10 and passes through (5, 20). What is its equation?

3.

What does quadratic regression do?

4.

Which R² value shows the closest fit?

5.

Predicting a value beyond the range of the data is called…

6.

For heights in metres and time in seconds, a thrown object's model usually has a t² coefficient close to…

For parents and teachers

Big idea: expectation A3.2 asks students to determine, through investigation using a variety of strategies, the equation of the quadratic function that best models a suitable data set graphed on a scatter plot, and to compare this equation to the equation of a curve of best fit generated with technology.

How to help: if you have a computer, Desmos regression is worth ten minutes together: enter a table, type y₁ ~ ax₁² + bx₁ + c, and compare with the hand model. Ask "Which model makes more sense at the start? At the end?"

Common difficulties:

  • Using a data point that is very close to the estimated vertex to find a (small errors become big ones). Use a point far away.
  • Reading R² as "the probability the model is right".
  • Trusting extrapolations without checking the context.
  • Sign errors with h and with zeros.