Year 10 Extended · Unit 3 Functions, quadratics and transformations · Lesson 4 of 7

From measurements to an equation

Measurements of a basin floor, a cable or a thrown ball can be turned into one equation. This lesson asks how many measurements that takes, finds a, b and c from points, from a vertex or from zeros, and asks when data cannot be a parabola.

FunctionsGDC optional

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. From Year 9: solve the pair a + b = 7 and a − b = 1.
    Answera = 4 and b = 3. Add the equations to get 2a = 8.
  2. From this unit, Lesson 3: find the vertex of y = x² − 6x + 14.
    Answer(3, 5). Half of 6 is 3, and 14 − 9 = 5.
  3. From this unit, Lesson 2: expand 2(x − 1)(x − 6).
    Answer2x² − 14x + 12. (x − 1)(x − 6) = x² − 7x + 6, and then double.
Words used in this lesson
unknown
a letter whose value must be found. For example: a, b and c in ax² + bx + c.
independent facts
facts that each give new information. For example: the heights at x = 0 and at x = 2, not the height at x = 0 measured twice.
collinear
lying on one straight line. For example: (0, 1), (2, 5) and (4, 9).
fit
to find the equation that passes through given points. For example: a quadratic fitted to three measurements.
model
an equation that describes a real situation. For example: the height of a thrown ball.
check
put a point back in to test an answer. For example: substitute x = 3 and compare y.

The salt basin and its three measurements

A survey team in the Atacama measures the height of the floor of a salt basin above a marker level (a negative height is below the marker) at some known distances across it, and wants one equation y = ax² + bx + c for the whole floor. Every measurement they take gives one fact, and the equation has some unknown numbers in it.

Predict first: how many measurements are enough to pin down exactly one parabola, and why that many? Then use the figure: move the heights and watch the curve and its equation.

3
−1
3
y = x² − 1equation
(0, −1)vertex
1value of a

Heights 3, −1 and 3 at x = −2, 0 and 2 give a = (3 + 3 − 2 × (−1)) ÷ 8 = 1, b = (3 − 3) ÷ 4 = 0 and c = −1: y = x² − 1, with vertex (0, −1).

Show me the steps
  1. Set the heights to 5, 1 and 5. Then a = (5 + 5 − 2) ÷ 8 = 1, b = 0 and c = 1, so y = x² + 1 with vertex (0, 1). Equal end heights put the axis exactly halfway, at x = 0.
  2. Set the heights to 5, 1 and 9. Then a = (5 + 9 − 2) ÷ 8 = 1.5, b = (9 − 5) ÷ 4 = 1 and c = 1, so y = 1.5x² + x + 1. The higher right end drags the vertex to the left of the y-axis, at about (−0.33, 0.83).
  3. Set the heights to 1, 2 and 3. Then a = (1 + 3 − 4) ÷ 8 = 0, so the equation is y = 0.5x + 2, a straight line with no vertex. Three points in a line do not fix a parabola.
  4. Set the heights to −4, 2 and −4. Then a = (−8 − 4) ÷ 8 = −1.5, so y = −1.5x² + 2. It opens downward, and the highest point is (0, 2).

Count the unknowns. y = ax² + bx + c has three unknowns, so it needs three independent facts: three points with different x, or the vertex and one more point, or the two zeros and one more point.

Method for points. Put each point into y = ax² + bx + c to get an equation. A point on the y-axis gives c at once. Subtract equations to remove c (or b), solve, and check with a point you did not use.

Choose the form that makes the facts cheap. With the vertex (h, k), use y = a(x − h)² + k and find only a. With zeros p and q, use y = a(x − p)(x − q) and find only a.

If the points are in a straight line, the equations give a = 0: the data is linear, not quadratic.

Two points are not enough. y = x² and y = 0.5x² + x both pass through (0, 0) and (2, 4). And a model is only trusted near its data: a parabola fitted to a stretch of basin floor says nothing sensible far beyond it.

Worked examples

1. Find the quadratic through (0, 4), (1, 3) and (3, 13).

  1. (0, 4) gives c = 4. Then (1, 3): a + b + 4 = 3, so a + b = −1. And (3, 13): 9a + 3b + 4 = 13, so 3a + b = 3.
  2. Subtract: (3a + b) − (a + b) = 3 − (−1), so 2a = 4 and a = 2. Then b = −1 − 2 = −3.
  3. y = 2x² − 3x + 4. Check (3): 18 − 9 + 4 = 13.

2. The vertex is (2, −6) and the graph passes through (0, 2). Find the equation.

  1. Use vertex form: y = a(x − 2)² − 6.
  2. Put in (0, 2): 2 = 4a − 6, so 4a = 8 and a = 2.
  3. y = 2(x − 2)² − 6 = 2x² − 8x + 2.

3. The zeros are −2 and 4 and the graph passes through (0, 4). Find the equation and the vertex.

  1. y = a(x + 2)(x − 4). Put in (0, 4): 4 = a(2)(−4) = −8a, so a = −0.5.
  2. y = −0.5(x + 2)(x − 4) = −0.5x² + x + 4.
  3. The axis is halfway between the zeros: x = 1, and y = −0.5(3)(−3) = 4.5. The vertex is (1, 4.5).

Your go. The first step is given. Find the quadratic through (−1, 4), (0, −1) and (2, 7).

  1. (0, −1) gives c = −1.
  2. Now put in (−1, 4) and (2, 7), and solve for a and b.
Check the answer

(−1, 4): a − b − 1 = 4, so a − b = 5. (2, 7): 4a + 2b − 1 = 7, so 2a + b = 4. Adding: 3a = 9, a = 3 and b = −2. So y = 3x² − 2x − 1. Check (2): 12 − 4 − 1 = 7.

With a GDC (optional)

A GDC can fit a quadratic to a list of points. Enter the x values in one list and the y values in another, and choose quadratic regression from the statistics or calculate menu. It reports a, b and c. For exactly three points it gives the same quadratic you find by hand; for more points it gives the best fit, which need not pass through any of them.

Check the result by putting a point back in. A GDC reports rounded decimals, so give exact values from your hand working. Everything here can be done by hand.

Check before you go on

1. A parabola passes through (0, 5), (1, 2) and (2, 3). What is its equation?

Your turn

Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.

Level 1 · Getting started

2. y = ax² + bx + 4 passes through (1, 9) and (−1, 5). Find a.

3. A parabola has zeros at x = 1 and x = 5 and passes through (0, 10). What is its y value at x = 3?

Level 2 · Building

4. A parabola has its vertex at (4, −3) and passes through (6, 5). Find its y-intercept.

5. A walkway cable in Thimphu hangs in a parabola. Its height above the deck is 6 m at the left tower (x = 0), 1 m at x = 10 and 6 m at the right tower (x = 30). Find the lowest height of the cable, in metres.

Level 3 · Stretch

6. y = x² + bx + c passes through (1, 4) and (5, 4). Find c.

7. A parabola passes through (1, 2), (3, 0) and (5, 6). Find its y-intercept.

Something to think about

In the figure, find some sets of three heights that make a = 0. What do they have in common? Say it as a sentence about the middle height and the other two, and then explain, using the graph, why a straight line comes out.

One way to think about it

a = 0 exactly when p + r = 2q: the middle height is the average of the other two. The three points are then equally spaced along a straight line, so the only curve of the form y = ax² + bx + c through them is that line, and the x² term has nothing to do.

Exit ticket

8. Which three points do NOT lie on any graph of the form y = ax² + bx + c with a not equal to 0?