Quadratic functions and equations: a week of ready-to-teach maths

Five lessons for Year 9 Algebra. Print this pack and the week is prepared: each day has a one-page plan and a student worksheet with practice and a stretch section, plus table sort cards, parabola match cards, a mini-check and every answer with its working.

Start here: five minutes to Monday

1. Skim the week at a glance on the next page.
2. Print the five days. Each day is two A4 sheets: a plan and a worksheet.
3. Print the two cut-out sheets, one copy per pair, and cut them before Day 1: the table sort cards are for Day 1 and the parabola match cards for Day 3.
4. Free graphing software or a graphics calculator helps on Days 3 and 5, one device per pair; every task also works with the printed graphs and a calculator.
5. Open the free interactive unit on the board. Every plan says which picture to show and which buttons to press.

The idea in plain words

A quadratic function has x² as its highest power, such as y = x² − 2x − 3. Its graph is a parabola: a U-shape with a mirror line, the axis of symmetry, through its lowest point, the vertex. In a table the first differences change steadily and the second differences stay the same. Solving x² − 2x − 3 = 0 means finding where y = 0, which is where the parabola crosses the x-axis; a parabola can cross twice, touch once or miss altogether. When the equation factorises, the null factor law gives the roots exactly: (x + 1)(x − 3) = 0 makes x = −1 or x = 3. When it does not, a graph or a table of values traps each root between decimals.

One day, one lesson

The five lessons fill a week of maths, one lesson of about 60 minutes a day. Run them in order: each day stands on the one before. Every plan marks a split point for two shorter sessions, and every worksheet ends with a stretch section for students who finish early.

How we write these answers

A root is written x = 3, and a pair of roots x = −1 or x = 3. A vertex is a point, such as (1, −4); an axis of symmetry is a line, such as x = 1. Decimal roots from a graph or a table are given to the accuracy asked for, with ≈. In this pack a monic quadratic has x² with no number in front, and every letter stands for a real number.

The week at a glance

One lesson a day for a week. Each day stands on the day before, so run them in order.

How the week builds

Days 1 and 2 meet the parabola and its symmetry; Day 3 reads solutions from a graph; Day 4 finds them exactly by factorising; and Day 5 traps the ones that are not whole numbers. It builds on Year 8, where students graphed linear relations and solved linear equations graphically and algebraically, and on expanding and factorising monic quadratics earlier in Year 9. It prepares modelling with linear and quadratic functions and transforming graphs later this year, and the exponential relations of Year 10.

Materials for the week (one trip)

• From the classroom: rulers, sharp pencils, scissors, envelopes for the cards, calculators, this pack printed.
• Cut-out sheet 1, one per pair: 12 parabola match cards, six graphs and six equations (Day 3).
• Cut-out sheet 2, one per pair: 12 table sort cards (Day 1).
• If you have them: devices with free graphing software for Days 3 and 5. No other equipment to buy.

Words for the word wall

quadratic, parabola, vertex, turning point, axis of symmetry, minimum, maximum, y-intercept, root, solution, x-intercept, first difference, second difference, null factor law, factorise, substitute, sign change, approximate, graphing software

Dear families

This week in maths, Year 9 is learning about quadratic functions, such as y = x² − 2x − 3, whose graphs are U-shaped curves called parabolas. We find the lowest point of the curve, read off where it crosses the x-axis to solve equations, and solve some equations exactly by factorising.

Try this at home

• Throw a ball gently and watch its path: it traces a parabola, opening downward.
• Ask: if (x − 2)(x − 5) = 0, what can x be? (2 or 5: one of the brackets has to be 0.)
• Ask: does x² + 1 = 0 have a solution? (No real one: x² is never negative, so x² + 1 is at least 1.)
• Ask: what is special about the second differences of 1, 4, 9, 16, 25? (They are all 2.)

My quadratic week

Fill one row a day. Write a quadratic from class and one thing you found out about it.

Printed from the free seegongsik Quadratic functions and equations teaching pack · seegongsik.com/au/y9/algebra/AC9M9A04/pack

From a table to a parabola

A quadratic function has x² as its highest power. Its table has steady second differences, and its graph is a parabola with a vertex and a mirror line.

We are learning to

• complete a table of values for a quadratic and plot a smooth parabola,
• use first and second differences to tell a quadratic table from a linear one,
• decide from its equation whether a function is quadratic.

Success criteria

• I can plot y = x² − 4 from a table and mark its vertex.
• I can show from the differences that a table is quadratic, not linear.

You need

The worksheet; the 12 table sort cards (cut-out sheet 2), one set per pair; rulers and sharp pencils.

Lesson flow (about 60 minutes)

Two shorter sessions instead? Split after the table sort; start the second with Part A.

Watch for these ideas

• Joining the points with a ruler: the curve is smooth and rounded at the vertex, not a V.
• Squaring a negative wrongly: (−3)² = 9, so x = −3 gives y = 9 − 4 = 5.
• Stopping at first differences: a quadratic’s first differences change, but steadily.

Answers

• Part A: y = 5, 0, −3, −4, −3, 0, 5; vertex (0, −4), a minimum.
• Part B: y = x² + 1 gives 1, 2, 5, 10, 17, 26; first differences 1, 3, 5, 7, 9; second differences 2, 2, 2, 2: quadratic. y = 3x + 1 gives 1, 4, 7, 10, 13, 16; first differences 3, 3, 3, 3, 3: linear.
• Part C: 1 quadratic (the highest power of x is 2); 2 linear (the highest power of x is 1); 3 quadratic (it expands to x² − 4x); 4 neither (x is under a fraction bar, not a power of 1 or 2); 5 linear (it simplifies to 2x + 1: the x² terms cancel); 6 neither (the power is 3).
• Part D: (x + 1)² − x² = 2x + 1, so the first differences of y = x² are 1, 3, 5, ..., and those go up by 2 each time: the second difference is always 2.

From a table to a parabola

A. Table, then graph

Complete the table for y = x² − 4, plot the points and join them with a smooth curve. Mark and name the vertex.

B. Differences

Fill the y-values for y = x² + 1 and for y = 3x + 1, then the first and second differences. Which is quadratic?

C. Quadratic, linear or neither?

Simplify first if you need to, then decide and give a reason.

D. Stretch: why always 2?

Simplify (x + 1)² − x². Use it to explain why the second differences of y = x² are always 2.

Symmetry and the vertex

A parabola is symmetrical about a vertical line, its axis of symmetry, and the vertex sits on that line. Any two points at the same height are mirror images, so the axis is halfway between them.

We are learning to

• find the axis of symmetry halfway between two points at the same height,
• find the vertex by substituting the axis value, and say whether it is a minimum or a maximum,
• read the y-intercept straight from the equation, and use the pattern x = −b ÷ 2.

Success criteria

• I can find the axis and vertex of y = x² − 8x + 12 from its table.
• I can say where y = x² + bx + c crosses the y-axis, and why.

You need

The worksheet; calculators for checking. Graphing software is optional.

Lesson flow (about 60 minutes)

Two shorter sessions instead? Split after Part A; start the second with the demo parabola again.

Watch for these ideas

• Writing the axis as a point: it is a line, x = 1, while the vertex is a point, (1, −4).
• Sign of the axis: for x² + 6x + 5 the axis is x = −3, not 3.
• Forgetting to substitute: the axis gives the x of the vertex; the y needs the equation.

Answers

• Part A: x² − 4x + 3: y = 0 at 1 and 3, axis x = 2, vertex (2, −1); x² + 2x − 8: y = 0 at −4 and 2, axis x = −1, vertex (−1, −9); x² − 8x + 12: y = 0 at 2 and 6, axis x = 4, vertex (4, −4); x² + 6x + 5: y = 0 at −5 and −1, axis x = −3, vertex (−3, −4).
• Part B: (−1, 7) and (5, 7): x = 2; (0, 3) and (6, 3): x = 3; (−7, −2) and (1, −2): x = −3.
• Part C: b = −2, c = −15: axis x = 1, vertex (1, −16), y-intercept −15; b = 10, c = 21: axis x = −5, vertex (−5, −4), y-intercept 21; b = −12, c = 20: axis x = 6, vertex (6, −16), y-intercept 20; b = 4, c = 0: axis x = −2, vertex (−2, −4), y-intercept 0. The axis is always x = −b ÷ 2.
• Part D: y = −x² + 4x is 0 at x = 0 and x = 4, so the axis is x = 2 and the vertex (2, 4) is a maximum: the −x² turns the U upside down.

Symmetry and the vertex

A. Mirror points, axis and vertex

Make a short table to find two x-values with the same y. The axis is halfway between them. Substitute to find the vertex.

B. The axis from two points

Each pair of points lies on one parabola. Write the equation of its axis of symmetry.

C. The pattern

For each y = x² + bx + c, find the axis, the vertex and the y-intercept. Then write a rule for the axis using b.

Rule: the axis of y = x² + bx + c is x = ______________

D. Stretch: upside down

Make a table for y = −x² + 4x from x = −1 to x = 5. Find the axis and the vertex. Is the vertex a minimum or a maximum, and why?

Solving from the graph

Solving x² + bx + c = 0 means finding where y = 0: where the parabola meets the x-axis. It can cross twice, touch once at its vertex, or miss the axis altogether.

We are learning to

• read the solutions of a quadratic equation where its graph crosses the x-axis,
• explain why a quadratic equation can have two solutions, one or none,
• solve x² + bx + c = k with a horizontal line on the graph.

Success criteria

• I can read the solutions of an equation from its graph and check them by substituting.
• I can say how many solutions an equation has from where its vertex is.

You need

The worksheet; the 12 parabola match cards (cut-out sheet 1), one set per pair; rulers; graphing software if you have it.

Lesson flow (about 60 minutes)

Two shorter sessions instead? Split after parabola match; start the second with Part B.

Watch for these ideas

• Reading the y-intercept as a solution: the solutions are on the x-axis.
• Counting a touch as two: a parabola that touches the axis at its vertex has one solution.
• Trusting a sketch too far: a reading from a graph is checked by substituting.

Answers

• Part A: x² + 2x − 3 = 0: x = −3 or x = 1; x² − 4x + 4 = 0: x = 2; x² − 2x + 3 = 0: no solution.
• Part B: k = 5: x = −2 or x = 4; k = −3: x = 0 or x = 2; k = −4: x = 1; k = −5: no solution.
• Part C: x² − 9 = 0: 2 (x = −3 or x = 3); x² = 0: 1 (x = 0); x² + 4 = 0: 0 (no solution).
• Part D: the vertex of y = x² − 6x + k is at x = 3, height k − 9. One solution when the vertex is on the axis: k = 9, and then x = 3.

Solving from the graph

A. Read the solutions

Each graph is y = the quadratic shown. Solve quadratic = 0 from the graph, then check by substituting.

B. A horizontal line

C. How many solutions?

Sketch each graph in your head from its vertex. How many solutions does each equation have? Find them.

D. Stretch: exactly one

For which value of k does x² − 6x + k = 0 have exactly one solution? Explain using the vertex.

Solving by factorising

If two numbers multiply to zero, one of them is zero: the null factor law. So once a quadratic equation is factorised and equal to zero, each bracket gives a root exactly, with no graph needed.

We are learning to

• use the null factor law to solve an equation in factorised form,
• factorise a monic quadratic and solve it, including roots of 0 and differences of squares,
• rearrange an equation to = 0 before factorising, and check every root by substituting.

Success criteria

• I can solve x² + 2x − 24 = 0 and check both roots.
• I can explain why x² = 3x + 10 must be rearranged before it is factorised.

You need

The worksheet. No calculators.

Lesson flow (about 60 minutes)

Two shorter sessions instead? Split after why zero matters; start the second with Part B.

Watch for these ideas

• Taking the sign from the bracket: x + 7 = 0 gives x = −7.
• Dividing by x: x² + 5x = 0 divided by x loses the root x = 0; factorise x(x + 5) instead.
• Forgetting the second root of x² = 49: it is ±7.

Answers

• Part A: (x + 7)(x − 4) = 0: x = −7 or x = 4; (x + 9)(x + 2) = 0: x = −9 or x = −2; x(x − 6) = 0: x = 0 or x = 6.
• Part B: x² + 8x + 15 = 0, (x + 5)(x + 3) = 0: x = −5 or x = −3; x² − 9x + 20 = 0, (x − 4)(x − 5) = 0: x = 4 or x = 5; x² + 2x − 24 = 0, (x + 6)(x − 4) = 0: x = −6 or x = 4; x² − 49 = 0, (x + 7)(x − 7) = 0: x = −7 or x = 7; x² + 5x = 0, x(x + 5) = 0: x = −5 or x = 0.
• Part C: x² = 3x + 10 becomes x² − 3x − 10 = 0, (x + 2)(x − 5) = 0: x = −2 or x = 5; x² + 4x = 21 becomes x² + 4x − 21 = 0, (x + 7)(x − 3) = 0: x = −7 or x = 3; x(x + 1) = 30 becomes x² + x − 30 = 0, (x + 6)(x − 5) = 0: x = −6 or x = 5.
• Part D: w(w + 3) = 40, so w² + 3w − 40 = 0 and (w + 8)(w − 5) = 0: w = −8 or w = 5. A width cannot be negative, so the rectangle is 5 cm by 8 cm.

Solving by factorising

A. Already factorised

Use the null factor law. Write both solutions.

B. Factorise, then solve

Factorise, solve, and check one root of each by substituting.

C. Make it equal zero first

Rearrange so one side is 0, then factorise and solve.

D. Stretch: a rectangle

A rectangle is 3 cm longer than it is wide, and its area is 40 cm². Write an equation for its width w, solve it, and explain which solution makes sense.

Roots that are not whole numbers

Many quadratics do not factorise over the integers. Their roots can still be trapped: where the value of the quadratic changes sign, the graph has crossed the axis, and closer and closer decimals pin the root down.

We are learning to

• trap a root between two decimals by looking for a change of sign in a table of values,
• find decimal roots with graphing software or a calculator table,
• choose a method: factorise, solve numerically, or say there is no real solution.

Success criteria

• I can show that x² − 7 = 0 has a root between 2.6 and 2.7 and give it to one decimal place.
• I can say which equations factorise before I start, and why.

You need

The worksheet; calculators; graphing software if you have it; the unit’s self-check quiz for the game.

Lesson flow (about 60 minutes)

Two shorter sessions instead? Split after Part A; start the second with Part B and the quiz.

Watch for these ideas

• Stopping at the first sign change: a quadratic can have two roots, so look on both sides of the vertex.
• Rounding from the wrong end: a root between 2.6 and 2.65 rounds to 2.6.
• Taking only the positive root of x² = 7: it is ±√7.

Answers

• Part A: x = 2: −3 (negative); x = 3: 2 (positive); x = 2.6: −0.24 (negative); x = 2.7: 0.29 (positive); x = 2.65: 0.0225 (positive). The root is between 2.6 and 2.65, so x ≈ 2.6; the other root is −2.6. (Exactly ±√7 ≈ ±2.6458.)
• Part B: x² + x − 4 = 0: x ≈ −2.56 or x ≈ 1.56; x² − 4x + 1 = 0: x ≈ 0.27 or x ≈ 3.73.
• Part C: x² − 11x + 24 = 0: factorise, (x − 3)(x − 8); x² + 3x − 1 = 0: numerically, x ≈ −3.3 or 0.3; x² + 2x + 5 = 0: no real solutions; x² − 16 = 0: factorise, (x + 4)(x − 4); x² − 5x + 2 = 0: numerically, x ≈ 0.44 or 4.56.
• Part D: x² − 2x − 1 = (x − 1)² − 2 = 0, so (x − 1)² = 2, x − 1 = ±√2 and x = 1 ± √2: the board’s roots exactly.

Roots that are not whole numbers

A. Trap the root

Work out x² − 7 for each x and write whether it is negative or positive. Where does it change sign?

So the positive root is x ≈ ______ to one decimal place, and the other root is ______.

B. Decimal roots

Use a table of values or graphing software. Give both roots to two decimal places.

C. Choose a method

Circle the method, then solve.

D. Stretch: exact roots

Show that x² − 2x − 1 = (x − 1)² − 2. Use it to find the exact roots of x² − 2x − 1 = 0, and compare with the decimals on the board.

Parabola match cards

For Day 3, one set per pair: 12 cards

Match each graph to its equation. Use where the graph crosses or touches the x-axis, and where it crosses the y-axis.

Teacher note: the key is with the mini-check answers, so this sheet can go to pairs whole.

Table sort cards

For Day 1, one set per pair: 12 cards

Write the first differences under each table, and the second differences if you need them. Then sort the cards: linear, quadratic or neither.

Teacher note: the key is with the mini-check answers, so this sheet can go to pairs whole.

What we know: Quadratic functions and equations

Work on your own. Use a calculator only for Question 8. Show your working.

1. Complete a table for y = x² − 3 from x = −2 to x = 2. What is the vertex? ______________________
2. y = x² − 10x + 16 meets the x-axis at x = 2 and x = 8. Find the axis of symmetry and the vertex. ______________________
3. A parabola y = x² + bx + c crosses the x-axis at x = −1 and x = 6. Write its equation. ______________________
4. Solve x² − 7x + 10 = 0. ______________________
5. Solve x² + 3x − 18 = 0, and check one solution. ______________________
6. Solve x² = 4x + 12. ______________________
7. How many solutions does each have? x² − 6x + 9 = 0 and x² + 4 = 0. ______________________
8. x² + 2x − 5 = 0 has a positive root between 1 and 2. Find it to one decimal place with a table of values. ______________________

Answers and marking guide

Answers with working

1. y = 1, −2, −3, −2, 1; vertex (0, −3).
2. Axis x = (2 + 8) ÷ 2 = 5; vertex: 5² − 10 × 5 + 16 = −9, so (5, −9).
3. y = (x + 1)(x − 6) = x² − 5x − 6.
4. (x − 2)(x − 5) = 0, so x = 2 or x = 5.
5. (x + 6)(x − 3) = 0, so x = −6 or x = 3; check: 3² + 3 × 3 − 18 = 0.
6. x² − 4x − 12 = 0, (x + 2)(x − 6) = 0, so x = −2 or x = 6.
7. x² − 6x + 9 = (x − 3)²: one solution, x = 3. x² + 4 is at least 4: no real solutions.
8. x = 1.4 gives −0.24, x = 1.5 gives 0.25, x = 1.45 gives 0.0025: the sign changes between 1.4 and 1.45, so x ≈ 1.4 (1.449 to three places).

A quick three-level guide

Eight questions, four ideas. A student at standard answers most questions with working, and checks each root by substituting.

Card keys

Parabola match (Day 3): x² − 2x − 8 = (x + 2)(x − 4); x² + 6x + 8 = (x + 4)(x + 2); x² − 6x + 5 = (x − 1)(x − 5); x² + 4x − 5 = (x + 5)(x − 1); x² − 8x + 16 = (x − 4)²; x² − 1 = (x + 1)(x − 1).

Table sort (Day 1): linear: E (y = 2x + 3), G (y = 10 − 3x), H (y = 0.5x); quadratic: B (y = x² + 2), C (y = 2x²), I (y = x² − 4x), J (y = 9 − x²), L (y = x² + x); neither: A (y = 2^x), D (y = x³), F (y = 12 ÷ (x + 1)), K (y = 10 − 2^x).

Weekly class record

Jot a tick as you move around the room; the mini-check fills any gaps. A tick a day is plenty.

The five columns are the five days: tables and parabolas, symmetry, solving from graphs, factorising, and decimal roots.