Maths4U
« Back to Subchapter

Maths 9 – Quadratic functions

📘 Notes

Quadratic Functions

In these questions, you need to draw the graph of a quadratic function and then read important information from the graph.

The graph of a quadratic function is a parabola. It can open upwards or downwards.

1. General form

A quadratic function has the form:

y = ax2 + bx + c, where a ≠ 0

  • If a > 0, the parabola opens upwards.
  • If a < 0, the parabola opens downwards.
  • The graph is symmetric.

2. What to find from the graph

  • Which quadrants the graph passes through
  • Points where the graph meets the x-axis and y-axis
  • Domain
  • Range
  • Intervals where the function is increasing and decreasing

3. Key facts

Vertex

The vertex is the highest or lowest point of the parabola.

x-coordinate of vertex:

x = -b / 2a

Then substitute this value into the function to find y.

If the parabola opens upwards, the vertex is a minimum point.
If the parabola opens downwards, the vertex is a maximum point.

Axis of symmetry

The axis of symmetry is the vertical line x = -b / 2a.

Intersection with the y-axis

Put x = 0. Then y = c, so the graph crosses the y-axis at (0, c).

Intersection with the x-axis

Put y = 0 and solve:

ax2 + bx + c = 0

4. Domain and range

Domain

The domain of every quadratic function is all real numbers.

Range

The range depends on the y-coordinate of the vertex.

  • If the parabola opens upwards and the vertex is (p, q), then range: y ≥ q
  • If the parabola opens downwards and the vertex is (p, q), then range: y ≤ q

5. Increasing and decreasing

Use the x-coordinate of the vertex.

  • If the parabola opens upwards, it is decreasing before the vertex and increasing after the vertex.
  • If the parabola opens downwards, it is increasing before the vertex and decreasing after the vertex.

6. Standard method for these questions

  1. Write the function clearly.
  2. Find the direction of opening from the sign of a.
  3. Find the vertex using x = -b / 2a.
  4. Find the y-intercept by putting x = 0.
  5. Find the x-intercepts by solving ax2 + bx + c = 0.
  6. Draw a neat parabola using the key points.
  7. State the quadrants, domain, range, and intervals of increase and decrease.

7. Worked example 1

Function: y = x2 + 2x - 8

Step 1: Direction

a = 1, so the parabola opens upwards.

Step 2: Vertex

x = -b / 2a = -2 / 2 = -1

y = (-1)2 + 2(-1) - 8 = 1 - 2 - 8 = -9

Vertex: (-1, -9)

Step 3: Intersections with axes

y-axis: x = 0, so y = -8

Point: (0, -8)

x-axis: x2 + 2x - 8 = 0

(x + 4)(x - 2) = 0

Points: (-4, 0) and (2, 0)

Step 4: Domain and range

Domain: all real numbers

Range: y ≥ -9

Step 5: Increasing and decreasing

Decreasing for x < -1

Increasing for x > -1

Step 6: Quadrants

The graph passes through Quadrants I, II, III and IV.

8. Worked example 2

Function: y = -x2 + 4x + 5

Step 1: Direction

a = -1, so the parabola opens downwards.

Step 2: Vertex

x = -b / 2a = -4 / -2 = 2

y = -(2)2 + 4(2) + 5 = -4 + 8 + 5 = 9

Vertex: (2, 9)

Step 3: Intersections with axes

y-axis: x = 0, so y = 5

Point: (0, 5)

x-axis: -x2 + 4x + 5 = 0

x2 - 4x - 5 = 0

(x - 5)(x + 1) = 0

Points: (5, 0) and (-1, 0)

Step 4: Domain and range

Domain: all real numbers

Range: y ≤ 9

Step 5: Increasing and decreasing

Increasing for x < 2

Decreasing for x > 2

Step 6: Quadrants

The graph passes through Quadrants I, II, III and IV.

9. Quick notes about quadrants

  • Quadrant I: x > 0, y > 0
  • Quadrant II: x < 0, y > 0
  • Quadrant III: x < 0, y < 0
  • Quadrant IV: x > 0, y < 0

10. Common mistakes

  • Using the wrong formula for the x-coordinate of the vertex.
  • Forgetting that the domain is always all real numbers.
  • Writing the wrong range sign. Use y ≥ ... for a minimum and y ≤ ... for a maximum.
  • Mixing up increasing and decreasing intervals.
  • Giving incorrect quadrants without checking where the graph really lies.