Mathematics · Simulation and virtual lab Free
Quadratic functions: roots, vertex and discriminant
Move a, b and c and watch the parabola change. Discover how the discriminant tells you the number of roots.
- You will learn to
- See how a, b and c change the graph of y = ax² + bx + c.
- Find roots, vertex and axis of symmetry from the graph and by calculation.
- Use the discriminant to predict the number of real roots.
- You will measure
- Roots, vertex, y-intercept and the discriminant Δ for each curve.
- Time needed
- About 25–35 min
- Equipment
- Nothing extra. A notebook to check the answers by calculation.
- Price
- Free: no payment, no account.
Virtual lab
Follow the steps with the simulation above (go back to the simulation ↑). Each time you tap Record, a row is added to the table and a point to the graph.
Aim
To investigate how a, b and c change the graph of y = ax² + bx + c, and how the discriminant decides the number of roots.
What to do
- Start with y = x² − 2x − 3. Read the roots, the vertex and the y-intercept. Tap Record.
- Change only a (try 2, 0.5, −1). What happens to the shape and the direction of opening?
- Change only c. What happens to the y-intercept and to the roots?
- Make three curves with Δ > 0, Δ = 0 and Δ < 0, and record each one. Count the roots.
- Tap Example: a kicked ball. Find where the ball lands and its maximum height.
- Tap Challenge and match the dashed curve. Then try another.
Results
Questions
What does the sign of a tell you?
If a > 0 the parabola opens upwards and has a minimum point. If a < 0 it opens downwards and has a maximum point.
Find the roots and the vertex of y = x² − 2x − 3.
x² − 2x − 3 = (x − 3)(x + 1) = 0, so x = 3 or x = −1. Vertex: x = −b/2a = 1, y = 1 − 2 − 3 = −4, so (1, −4).
How many real roots does y = x² + 2x + 5 have? Explain with the discriminant.
Δ = b² − 4ac = 4 − 20 = −16 < 0, so there are no real roots: the curve does not cut the x-axis.
The height of a ball is h = −0.2x² + 2x. Where does it land and how high does it go?
h = 0 when x(−0.2x + 2) = 0, so x = 0 or x = 10 m: it lands 10 m away. The vertex is at x = 5 m, h = −5 + 10 = 5 m.
What is the equation of the axis of symmetry of y = ax² + bx + c?
x = −b/(2a).
Which value tells you where the curve cuts the y-axis?
c, because when x = 0, y = c.
For teachers: expected results and how the model works
Tap or drag on the graph to trace any point. The challenge curves always have integer roots, so learners can find them from the graph and write y = a(x − p)(x − q), then expand to get b and c. Use the table to collect evidence for the discriminant rule.
Exam practice Exam set
National-exam style questions on this simulation (20 marks). Try each one on paper first, then open the marking scheme.
For y = x² − 6x + 8: (a) find the roots, (b) find the vertex, (c) find the axis of symmetry, (d) sketch the graph. (7 marks)
Marking scheme: (a) (x − 2)(x − 4) = 0, x = 2 or x = 4 (2). (b) x = 3, y = 9 − 18 + 8 = −1: vertex (3, −1) (2). (c) x = 3 (1). (d) U-shaped curve through (2, 0), (4, 0), (0, 8) with minimum at (3, −1) (2).
Use the discriminant to find the number of real roots of (a) 2x² + 3x + 5 = 0, (b) x² − 4x + 4 = 0, (c) x² + x − 6 = 0. (6 marks)
Marking scheme: (a) Δ = 9 − 40 = −31 < 0: no real roots (2). (b) Δ = 16 − 16 = 0: one repeated root (2). (c) Δ = 1 + 24 = 25 > 0: two real roots (2).
Find the value of k for which x² + kx + 9 = 0 has equal roots. (3 marks)
Marking scheme: Equal roots when Δ = 0: k² − 4 × 1 × 9 = 0 (1), k² = 36 (1), k = 6 or k = −6 (1).
A ball is thrown so that its height is h = 20t − 5t² metres after t seconds. Find the maximum height and when the ball lands. (4 marks)
Marking scheme: Vertex at t = −20 ÷ (2 × −5) = 2 s (1); h = 40 − 20 = 20 m (1). Lands when 20t − 5t² = 0, t(20 − 5t) = 0 (1), t = 4 s (1).
Donat Sciences and MathsMathematics · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Quadratic functions: roots, vertex and discriminant
Aim: To investigate how a, b and c change the graph of y = ax² + bx + c, and how the discriminant decides the number of roots.
Apparatus (real laboratory): Nothing extra. A notebook to check the answers by calculation.
What to do (with the simulation on www.donatstem.co.rw)
- Start with y = x² − 2x − 3. Read the roots, the vertex and the y-intercept. Tap Record.
- Change only a (try 2, 0.5, −1). What happens to the shape and the direction of opening?
- Change only c. What happens to the y-intercept and to the roots?
- Make three curves with Δ > 0, Δ = 0 and Δ < 0, and record each one. Count the roots.
- Tap Example: a kicked ball. Find where the ball lands and its maximum height.
- Tap Challenge and match the dashed curve. Then try another.
Results
| # | Equation | Δ = b² − 4ac | Number of roots | Roots | Vertex |
|---|
| 1 | | | | | |
| 2 | | | | | |
| 3 | | | | | |
| 4 | | | | | |
| 5 | | | | | |
| 6 | | | | | |
| 7 | | | | | |
| 8 | | | | | |
Graph
Sketch each parabola you investigate on the grid. Mark the roots, the vertex and the axis of symmetry.
Questions
- What does the sign of a tell you? [2]
- Find the roots and the vertex of y = x² − 2x − 3. [2]
- How many real roots does y = x² + 2x + 5 have? Explain with the discriminant. [2]
- The height of a ball is h = −0.2x² + 2x. Where does it land and how high does it go? [2]
- What is the equation of the axis of symmetry of y = ax² + bx + c? [2]
- Which value tells you where the curve cuts the y-axis? [2]
- For y = x² − 6x + 8: (a) find the roots, (b) find the vertex, (c) find the axis of symmetry, (d) sketch the graph. [7]
- Use the discriminant to find the number of real roots of (a) 2x² + 3x + 5 = 0, (b) x² − 4x + 4 = 0, (c) x² + x − 6 = 0. [6]
- Find the value of k for which x² + kx + 9 = 0 has equal roots. [3]
- A ball is thrown so that its height is h = 20t − 5t² metres after t seconds. Find the maximum height and when the ball lands. [4]
Donat Sciences and MathsMathematics · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Marking scheme: Quadratic functions: roots, vertex and discriminant
Expected results
Tap or drag on the graph to trace any point. The challenge curves always have integer roots, so learners can find them from the graph and write y = a(x − p)(x − q), then expand to get b and c. Use the table to collect evidence for the discriminant rule.
Answers
- What does the sign of a tell you? [2]
If a > 0 the parabola opens upwards and has a minimum point. If a < 0 it opens downwards and has a maximum point.
- Find the roots and the vertex of y = x² − 2x − 3. [2]
x² − 2x − 3 = (x − 3)(x + 1) = 0, so x = 3 or x = −1. Vertex: x = −b/2a = 1, y = 1 − 2 − 3 = −4, so (1, −4).
- How many real roots does y = x² + 2x + 5 have? Explain with the discriminant. [2]
Δ = b² − 4ac = 4 − 20 = −16 < 0, so there are no real roots: the curve does not cut the x-axis.
- The height of a ball is h = −0.2x² + 2x. Where does it land and how high does it go? [2]
h = 0 when x(−0.2x + 2) = 0, so x = 0 or x = 10 m: it lands 10 m away. The vertex is at x = 5 m, h = −5 + 10 = 5 m.
- What is the equation of the axis of symmetry of y = ax² + bx + c? [2]
x = −b/(2a).
- Which value tells you where the curve cuts the y-axis? [2]
c, because when x = 0, y = c.
- For y = x² − 6x + 8: (a) find the roots, (b) find the vertex, (c) find the axis of symmetry, (d) sketch the graph. [7]
(a) (x − 2)(x − 4) = 0, x = 2 or x = 4 (2). (b) x = 3, y = 9 − 18 + 8 = −1: vertex (3, −1) (2). (c) x = 3 (1). (d) U-shaped curve through (2, 0), (4, 0), (0, 8) with minimum at (3, −1) (2).
- Use the discriminant to find the number of real roots of (a) 2x² + 3x + 5 = 0, (b) x² − 4x + 4 = 0, (c) x² + x − 6 = 0. [6]
(a) Δ = 9 − 40 = −31 < 0: no real roots (2). (b) Δ = 16 − 16 = 0: one repeated root (2). (c) Δ = 1 + 24 = 25 > 0: two real roots (2).
- Find the value of k for which x² + kx + 9 = 0 has equal roots. [3]
Equal roots when Δ = 0: k² − 4 × 1 × 9 = 0 (1), k² = 36 (1), k = 6 or k = −6 (1).
- A ball is thrown so that its height is h = 20t − 5t² metres after t seconds. Find the maximum height and when the ball lands. [4]
Vertex at t = −20 ÷ (2 × −5) = 2 s (1); h = 40 − 20 = 20 m (1). Lands when 20t − 5t² = 0, t(20 − 5t) = 0 (1), t = 4 s (1).