Why the Parabola Matters#
The parabola is arguably the most important graph in Grade 12. It appears in Functions, Calculus (as the derivative of a cubic), and even in Finance (compound growth curves). Understanding every parameter here sets you up for the rest of the year.
1. The Three Forms#
Standard Form#
$$ y = ax^2 + bx + c $$Useful for finding the y-intercept ($c$) and using the quadratic formula.
Turning Point Form#
$$ y = a(x - p)^2 + q $$The most powerful form. You can read the turning point directly: Turning Point = $(p; q)$.
Factored (Root) Form#
$$ y = a(x - x_1)(x - x_2) $$Useful when you know the x-intercepts ($x_1$ and $x_2$).
Converting between forms: You can always expand Turning Point Form to get Standard Form. To go from Standard Form to Turning Point Form, use Completing the Square.
2. The Parameters: What Each One Does#
The “$a$” Value — Shape, Width, and Direction#
The $a$ value is the most important parameter. It controls three things at once:
| Value of $a$ | Effect |
|---|---|
| $a > 0$ | Parabola opens upward (“happy face” $\smile$) — has a minimum turning point |
| $a < 0$ | Parabola opens downward (“sad face” $\frown$) — has a maximum turning point |
| $ | a |
| $0 < | a |
| $a = 1$ | Standard width parabola |
Think of $a$ as the “zoom” control. A large $|a|$ zooms in (making the parabola look narrow), and a small $|a|$ zooms out (making it look wide).
The “$p$” Value — Horizontal Shift#
In $y = a(x - p)^2 + q$:
| Value of $p$ | Effect |
|---|---|
| $p > 0$ | Graph shifts right by $p$ units |
| $p < 0$ | Graph shifts left by $ |
| $p = 0$ | No horizontal shift (axis of symmetry is the y-axis) |
The “opposite sign” trap: In $y = (x - 3)^2$, the shift is right 3 (not left). In $y = (x + 2)^2 = (x - (-2))^2$, the shift is left 2. Always look at the sign inside the bracket.
Axis of Symmetry: The vertical line $x = p$ is the axis of symmetry. The parabola is a perfect mirror image on either side of this line.
The “$q$” Value — Vertical Shift#
| Value of $q$ | Effect |
|---|---|
| $q > 0$ | Graph shifts up by $q$ units |
| $q < 0$ | Graph shifts down by $ |
| $q = 0$ | No vertical shift |
$q$ is the y-coordinate of the turning point. Combined with $p$, the turning point is always at $(p; q)$.
3. Key Properties#
| Property | Formula / Value |
|---|---|
| Turning Point | $(p; q)$ |
| Axis of Symmetry | $x = p$ |
| y-intercept | Set $x = 0$: $y = a(0-p)^2 + q = ap^2 + q$ |
| x-intercepts | Set $y = 0$ and solve $a(x-p)^2 + q = 0$, or use $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ |
| Domain | $x \in \mathbb{R}$ |
| Range | If $a > 0$: $y \ge q$ (i.e. $y \in [q; \infty)$) |
| If $a < 0$: $y \le q$ (i.e. $y \in (-\infty; q]$) |
The Discriminant and x-intercepts#
The discriminant $\Delta = b^2 - 4ac$ tells you how many x-intercepts the parabola has:
| Discriminant | Meaning |
|---|---|
| $\Delta > 0$ | Two x-intercepts (graph cuts the x-axis twice) |
| $\Delta = 0$ | One x-intercept (graph touches the x-axis at the turning point) |
| $\Delta < 0$ | No x-intercepts (graph floats above or below the x-axis) |
4. Finding the Equation#
Given the Turning Point and one other point#
Example: Turning point $(2; -3)$, passes through $(4; 5)$.
- Write the form: $y = a(x - 2)^2 + (-3)$
- Substitute the point $(4; 5)$: $$ 5 = a(4 - 2)^2 - 3 $$ $$ 5 = 4a - 3 $$ $$ 4a = 8 $$ $$ a = 2 $$
- Final equation: $y = 2(x - 2)^2 - 3$
Given the x-intercepts and one other point#
Example: x-intercepts at $x = -1$ and $x = 3$, passes through $(0; 6)$.
- Write the factored form: $y = a(x + 1)(x - 3)$
- Substitute $(0; 6)$: $$ 6 = a(0 + 1)(0 - 3) $$ $$ 6 = -3a $$ $$ a = -2 $$
- Final equation: $y = -2(x + 1)(x - 3)$
Completing the Square (Standard → Turning Point Form)#
Convert $y = 2x^2 - 12x + 22$ to turning point form:
- Factor out $a$ from the first two terms: $y = 2(x^2 - 6x) + 22$
- Half the coefficient of $x$ and square it: $(\frac{-6}{2})^2 = 9$
- Add and subtract inside the bracket: $$ y = 2(x^2 - 6x + 9 - 9) + 22 $$ $$ y = 2((x - 3)^2 - 9) + 22 $$ $$ y = 2(x - 3)^2 - 18 + 22 $$ $$ y = 2(x - 3)^2 + 4 $$
Turning point: $(3; 4)$
5. The Inverse of a Quadratic Function#
This is where Grade 12 gets interesting. A parabola is Many-to-One — two different $x$-values can give the same $y$-value (e.g. $(-3)^2 = 9$ and $(3)^2 = 9$).
The Problem#
If you swap $x$ and $y$ in $y = x^2$:
$$ x = y^2 $$$$ y = \pm\sqrt{x} $$The “$\pm$” means for every $x$, there are two $y$-values. This fails the Vertical Line Test — it is not a function.
The Solution: Domain Restriction#
To make the inverse a function, we only use half of the original parabola.
For $y = x^2$:
| Restriction | Inverse |
|---|---|
| $x \ge 0$ (right half) | $f^{-1}(x) = \sqrt{x}$ (positive root only) |
| $x \le 0$ (left half) | $f^{-1}(x) = -\sqrt{x}$ (negative root only) |
The restriction is always at the turning point. For $y = (x - 3)^2$, the turning point is at $x = 3$, so the restriction is $x \ge 3$ or $x \le 3$.
Full Worked Example#
Find the inverse of $f(x) = 2(x - 1)^2 + 3$ for $x \ge 1$.
- Swap $x$ and $y$: $$ x = 2(y - 1)^2 + 3 $$
- Solve for $y$: $$ x - 3 = 2(y - 1)^2 $$ $$ \frac{x - 3}{2} = (y - 1)^2 $$ $$ y - 1 = \pm\sqrt{\frac{x - 3}{2}} $$ $$ y = 1 \pm \sqrt{\frac{x - 3}{2}} $$
- Choose the correct sign: Since we restricted $x \ge 1$ (right half), the inverse must give $y \ge 1$, so we take the positive root: $$ f^{-1}(x) = 1 + \sqrt{\frac{x - 3}{2}} $$
Domain of $f^{-1}$: $x \ge 3$ (the range of the original restricted $f$). Range of $f^{-1}$: $y \ge 1$ (the domain of the original restricted $f$).
The Horizontal Line Test#
Use your ruler horizontally on the original graph. If it hits the graph in two places, the inverse will not be a function without restriction.
6. Reading the Graph in an Exam#
When given a parabola graph and asked for the equation:
- Read the turning point $(p; q)$ directly from the graph.
- Determine the sign of $a$: Opens up = positive, opens down = negative.
- Find $a$: Use any other point on the graph (often the y-intercept) and substitute into $y = a(x - p)^2 + q$.
- Write the final equation.
Worked Example: Combining Functions#
Given: $f(x) = -(x - 2)^2 + 9$
(a) Write down the turning point, axis of symmetry, and range.
- Turning point: $(2; 9)$
- Axis of symmetry: $x = 2$
- Range: $y \le 9$ (since $a = -1 < 0$)
(b) Find the x-intercepts.
$$ 0 = -(x - 2)^2 + 9 $$$$ (x - 2)^2 = 9 $$$$ x - 2 = \pm 3 $$$$ x = 5 \text{ or } x = -1 $$x-intercepts: $(-1; 0)$ and $(5; 0)$
(c) For which values of $x$ is $f(x) > 0$? The parabola is above the x-axis between the roots:
$$ -1 < x < 5 $$(d) Determine $f^{-1}$ if the domain of $f$ is restricted to $x \ge 2$. Swap: $x = -(y - 2)^2 + 9$
$$ (y - 2)^2 = 9 - x $$$$ y - 2 = -\sqrt{9 - x} $$(We take the negative root because $x \ge 2$ maps to the right half, and the inverse must give $y \ge 2$… wait — the original function is decreasing for $x \ge 2$ since $a < 0$. For $x \ge 2$, $f(x) \le 9$, and the range of $f$ restricted to $x \ge 2$ is $y \le 9$. The inverse domain is $x \le 9$ and the inverse range is $y \ge 2$.)
$$ f^{-1}(x) = 2 + \sqrt{9 - x} $$Domain of $f^{-1}$: $x \le 9$. Range of $f^{-1}$: $y \ge 2$.
🚨 Common Mistakes#
- The sign of $p$: In $y = (x + 4)^2$, students write the turning point as $(4; 0)$. It’s $(-4; 0)$ because $x + 4 = x - (-4)$, so $p = -4$.
- Forgetting the restriction: If an exam asks for $f^{-1}$ to be a function, you must state the domain restriction (e.g., $x \ge 0$).
- Wrong root sign: When the domain is $x \le p$ (left half), the inverse uses the negative root. When $x \ge p$ (right half), use the positive root. But be careful with negative $a$ values — always check by substituting a test point.
- Domain ↔ Range swap: The domain of $f$ becomes the range of $f^{-1}$, and vice versa. Many students forget to state this.
💡 Pro Tip: The “Symmetry” Shortcut#
If you know one x-intercept and the axis of symmetry, you can find the other x-intercept instantly. The axis of symmetry is always the midpoint of the two roots.
If one root is at $x = -1$ and the axis of symmetry is $x = 2$:
$$ \frac{-1 + x_2}{2} = 2 $$$$ x_2 = 5 $$⏮️ Linear Function | 🏠 Back to Functions & Inverses | ⏭️ Hyperbola