Math 20-1 · Quadratic functions
The parabola, in two forms
One curve, written two ways. Vertex form tells you where it is and how wide; standard form tells you where it crosses the y-axis. Most of this unit is moving between them and reading what each one is good for.
- 1. The words, first
- 2. Reading vertex form
- 3. Completing the square
- 4. Using the vertex to answer a question
- 5. What costs marks
The words, first
The idea: Every symbol in y = a(x − p)² + q controls something you can see on the graph.
| Word | What it means |
|---|---|
| Parabola | The U-shaped curve of a quadratic function. Every quadratic graphs to one; only its position, width and direction change. |
| Vertex | The turning point — the highest point if the parabola opens downward, the lowest if it opens upward. |
| Axis of symmetry | The vertical line through the vertex. The curve is a mirror image across it, so it is written as an equation, x = p. |
| Vertex form | y = a(x − p)² + q. The vertex is (p, q). Note the minus sign is part of the form, so x − 3 means p = 3. |
| Standard form | y = ax² + bx + c. The y-intercept is c, and the axis of symmetry is x = −b/(2a). |
| a | Sets the direction and the width. Positive opens up, negative opens down. |a| > 1 is narrower than y = x²; |a| < 1 is wider. |
| Domain | The x-values allowed. For any quadratic function it is all real numbers. |
| Range | The y-values reached. It always starts or stops at q: y ≥ q if a > 0, y ≤ q if a < 0. |
| Maximum / minimum value | The y-coordinate of the vertex — the value, not the point, and not the x that produces it. |
| Completing the square | Rewriting ax² + bx + c as a(x − p)² + q by building the perfect square trinomial that is hiding inside it. |
Reading vertex form
The idea: Everything the graph does is visible in the four symbols, without plotting a single point.
For y = −2(x + 1)² + 4: a = −2, so it opens downward and is narrower than y = x². Writing x + 1 as x − (−1) gives p = −1, so the vertex is (−1, 4), the axis of symmetry is x = −1, and because it opens down, 4 is a maximum. Domain: all real numbers. Range: y ≤ 4.
No x-intercepts? That happens when the vertex is already on the far side of the axis from the arms — a and q with the same sign. An upward parabola whose lowest point is above the x-axis never reaches it.
Completing the square
The idea: Take half the coefficient of x, square it, add it and subtract it again. You have changed nothing and gained a perfect square.
When a = 1. For y = x² + 6x + 5: half of 6 is 3, and 3² = 9.
y = (x² + 6x + 9) − 9 + 5 = (x + 3)² − 4
So the vertex is (−3, −4). Check by putting x = 0 into both forms: the original gives 5, and (3)² − 4 = 5 ✓.
When a ≠ 1, factor a out of the first two terms only:
y = 2x² − 8x + 3 = 2(x² − 4x) + 3 = 2(x² − 4x + 4 − 4) + 3 = 2(x − 2)² − 8 + 3 = 2(x − 2)² − 5
Notice the −4 gets multiplied by the 2 on its way out of the bracket. That is where most errors happen.
The shortcut. If you only want the vertex, use x = −b/(2a) = 8/4 = 2, then substitute: y = 2(4) − 16 + 3 = −5. Same answer, less writing — but you still need completing the square when the question asks for vertex form.
Using the vertex to answer a question
The idea: Maximum height, maximum area, minimum cost: every one of them is the vertex of a parabola, and the question decides whether you report p or q.
Projectile. h = −5t² + 20t + 1. The vertex is at t = −b/(2a) = −20/(−10) = 2 s, and h = −5(4) + 40 + 1 = 21 m. If the question asks when, answer 2 seconds; if it asks how high, answer 21 metres. Reporting the wrong coordinate is a whole-mark error.
Maximum area. With 40 m of fence, the sides are x and 20 − x, so A = x(20 − x) = −x² + 20x. The vertex is at x = 10, giving 100 m². For a fixed perimeter the square always wins, which is a useful sanity check.
Building the equation from a vertex and a point. Vertex (−1, 4) through (1, 0): start at y = a(x + 1)² + 4, substitute the point, 0 = 4a + 4, so a = −1 and y = −(x + 1)² + 4.
What costs marks
The idea: Sign errors and answering the wrong question account for most of them.
- Reading p with the wrong sign. y = (x + 3)² has p = −3, not 3. The form contains a minus sign.
- Forgetting to multiply the subtracted square by a when completing the square with a ≠ 1.
- Giving the x-value when the question asked for the maximum. The maximum value is q.
- Writing the axis of symmetry as a number. It is a line, so it is x = 4, not 4.