In Grade 10, the parabola was $y = ax^2 + q$ — it could only move up or down. In Grade 11, we add the horizontal shift and unlock the full turning point form. This is one of the most important functions in matric.
The Two Forms#
| Form | Formula | Best used when… |
|---|---|---|
| Turning Point | $y = a(x - p)^2 + q$ | You know the turning point |
| Standard | $y = ax^2 + bx + c$ | You know the y-intercept and x-intercepts |
You can convert between them by expanding or completing the square.
1. Understanding the Parameters#
$$ y = a(x - p)^2 + q $$| Parameter | What it controls | Effect |
|---|---|---|
| $a > 0$ | Opens upward (smile) | Minimum turning point |
| $a < 0$ | Opens downward (frown) | Maximum turning point |
| $\|a\| > 1$ | Narrower (steeper) | |
| $\|a\| < 1$ | Wider (flatter) | |
| $p$ | Horizontal shift | $y = (x - 3)^2$ shifts RIGHT to $x = 3$ |
| $q$ | Vertical shift | Moves the turning point up/down |
⚠️ The Sign Trap#
In $y = a(x - p)^2 + q$, there is a minus before $p$ built into the formula.
- $y = (x - 3)^2$ → turning point at $x = +3$ (shifts RIGHT)
- $y = (x + 2)^2 = (x - (-2))^2$ → turning point at $x = -2$ (shifts LEFT)
The sign you see is the OPPOSITE of the direction.
2. Key Features to Identify#
For any parabola, you should be able to state:
| Feature | How to find it |
|---|---|
| Turning point | $(p; q)$ from the equation |
| Axis of symmetry | $x = p$ |
| y-intercept | Let $x = 0$, solve for $y$ |
| x-intercepts | Let $y = 0$, solve for $x$ |
| Domain | Always $x \in \mathbb{R}$ (all real numbers) |
| Range | $y \geq q$ if $a > 0$, or $y \leq q$ if $a < 0$ |
| Increasing/Decreasing | Increasing for $x > p$ (if $a > 0$), decreasing for $x < p$ |
3. Sketching from the Equation — Step by Step#
Worked Example: Sketch $y = 2(x - 1)^2 - 8$#
Step 1 — Identify parameters: $a = 2 > 0$ (opens up), $p = 1$, $q = -8$
Step 2 — Turning point: $(1; -8)$
Step 3 — Axis of symmetry: $x = 1$
Step 4 — y-intercept (let $x = 0$):
$y = 2(0-1)^2 - 8 = 2(1) - 8 = -6$
y-intercept: $(0; -6)$
Step 5 — x-intercepts (let $y = 0$):
$0 = 2(x-1)^2 - 8$
$2(x-1)^2 = 8$
$(x-1)^2 = 4$
$x - 1 = \pm 2$
$x = 3$ or $x = -1$
x-intercepts: $(-1; 0)$ and $(3; 0)$
Step 6 — Domain and Range:
Domain: $x \in \mathbb{R}$
Range: $y \geq -8$ (minimum at turning point)
Step 7 — Sketch: Plot the turning point, intercepts, draw a smooth U-shape opening upward.
4. Finding the Equation from a Graph#
Method 1: Given the turning point + one other point#
Use the turning point form $y = a(x - p)^2 + q$.
Example: Turning point $(2; -3)$, passes through $(0; 5)$.
$y = a(x - 2)^2 - 3$
Substitute $(0; 5)$: $5 = a(0-2)^2 - 3 = 4a - 3$
$4a = 8 \Rightarrow a = 2$
$y = 2(x-2)^2 - 3$
Method 2: Given the x-intercepts + one other point#
If the x-intercepts are $x_1$ and $x_2$:
$y = a(x - x_1)(x - x_2)$
Example: x-intercepts at $-1$ and $3$, y-intercept is $6$.
$y = a(x + 1)(x - 3)$
Substitute $(0; 6)$: $6 = a(1)(-3) = -3a$
$a = -2$
$y = -2(x + 1)(x - 3)$
Method 3: Given three points (use $y = ax^2 + bx + c$)#
Substitute each point to get 3 equations in $a$, $b$, $c$. Solve simultaneously.
5. Completing the Square (Converting Standard → Turning Point)#
To convert $y = ax^2 + bx + c$ to $y = a(x - p)^2 + q$:
Worked Example: $y = 2x^2 - 12x + 13$#
Step 1: Factor out $a$ from the first two terms:
$y = 2(x^2 - 6x) + 13$
Step 2: Complete the square inside the bracket. Take half the coefficient of $x$ and square it: $(\frac{-6}{2})^2 = 9$
$y = 2(x^2 - 6x + 9 - 9) + 13$
Step 3: Take the extra term out (multiply by $a$):
$y = 2(x - 3)^2 - 18 + 13$
$y = 2(x - 3)^2 - 5$
Turning point: $(3; -5)$
Quick formula#
$p = -\frac{b}{2a}$ and $q = f(p)$ (substitute $p$ back into the original equation)
For $y = 2x^2 - 12x + 13$: $p = -\frac{-12}{4} = 3$, $q = 2(9) - 36 + 13 = -5$ ✓
6. The Symmetry Shortcut#
A parabola is perfectly symmetric about its axis of symmetry ($x = p$).
This means: if one x-intercept is at $x = -1$ and the axis of symmetry is at $x = 3$, the other x-intercept must be at $x = 7$ (same distance of 4 units on the other side).
Also: the axis of symmetry is always halfway between the x-intercepts:
$p = \frac{x_1 + x_2}{2}$
🚨 Common Mistakes#
- The sign of $p$: $y = (x + 2)^2$ means $p = -2$, NOT $p = +2$. The turning point is at $(-2; q)$. This is the #1 error in tests.
- Expanding too early: If you’re given the turning point, use $y = a(x-p)^2 + q$ directly. Don’t expand to standard form — you’ll make more errors.
- y-intercept ≠ $q$: The y-intercept is found by letting $x = 0$. It only equals $q$ when $p = 0$.
- x-intercepts don’t always exist: If $a > 0$ and $q > 0$ (smile above x-axis), or $a < 0$ and $q < 0$ (frown below x-axis), there are NO x-intercepts. The discriminant tells you: $\Delta < 0$ → no x-intercepts.
- Domain vs Range: Domain is ALWAYS $x \in \mathbb{R}$ for a parabola. Range depends on the turning point and whether it opens up or down.
💡 Pro Tip: Reading a Graph Backwards#
In exams, you’re often given the graph and asked to find the equation. Always ask yourself:
- Can I see the turning point? → Use turning point form.
- Can I see both x-intercepts? → Use intercept form $y = a(x - x_1)(x - x_2)$.
- Then use one more point (usually the y-intercept) to find $a$.
🔗 Related Grade 11 topics:
- Quadratic Equations — the x-intercepts of a parabola come from solving $ax^2 + bx + c = 0$
- Quadratic Patterns — $T_n = an^2 + bn + c$ IS a parabola (with $n$ as the input)
- The Hyperbola & The Exponential — the other two functions you must sketch
📌 Grade 10 foundation: Sketching Graphs
📌 Grade 12 extension: Quadratic Function & Inverse — domain restriction and inverse graphs.
🏠 Back to Functions | ⏭️ Hyperbola