Year 10 Mathematics · Number and Algebra · 20m read

Parabolas — Graphing Quadratic Functions

Ever wonder why a basketball follows a curved path when you shoot it, or why satellite dishes have that exact U-shape? That's a parabola at work. You'll learn to sketch these curves and switch between different forms (standard, vertex, and factored) so you can analyse them from any angle—understanding where they peak, where they cross the axes, and what they're really telling you about real-world situations. This skill unlocks tonnes of VCE Maths problems and shows up everywhere from physics to engineering.

Learning objective: Sketch, analyse and interpret parabolas in multiple forms.

By the end of this lesson you can:

I convert between standard, vertex, and intercept forms.
I find vertex, axis of symmetry, and intercepts.
I interpret parabolas in context.

A parabola is the path of a fountain jet. Water rises, slows, peaks (turning point) and falls symmetrically. Every quadratic graph follows the same graceful arc.

Three forms of a quadratic function

A QUADRATIC FUNCTION $y = ax^2 + bx + c$ has a PARABOLA as its graph.

### Three Forms

| Form | Equation | Best for finding |
|------|---------|-----------------|
| Standard | $y = ax^2 + bx + c$ | y-intercept (= $c$); overall shape |
| Vertex (turning point) | $y = a(x-h)^2 + k$ | Vertex $(h, k)$; axis of symmetry ($x = h$) |
| Intercept (factor) | $y = a(x-p)(x-q)$ | x-intercepts ($p$ and $q$) |

### Key Features of a Parabola

| Feature | How to find |
|---------|------------|
| y-intercept | Set $x = 0$ |
| x-intercepts | Set $y = 0$ and solve quadratic |
| Axis of symmetry | $x = -b/(2a)$ (standard form) OR $x = h$ (vertex form) OR midpoint of x-intercepts $x = (p+q)/2$ |
| Vertex | Substitute axis of symmetry into function |
| Opens up/down | $a > 0$ → up (minimum); $a < 0$ → down (maximum) |

### Converting Between Forms

Standard → Vertex (complete the square):
$y = 2x^2 - 8x + 3$
$= 2(x^2 - 4x) + 3$
$= 2(x^2 - 4x + 4 - 4) + 3$
$= 2(x-2)^2 - 8 + 3$
$= 2(x-2)^2 - 5$
Vertex: $(2, -5)$, axis: $x = 2$.

Vertex → Standard (expand):
$y = 3(x+1)^2 - 4 = 3(x^2 + 2x + 1) - 4 = 3x^2 + 6x - 1$

Intercept → Standard (expand and collect):
$y = 2(x-3)(x+1) = 2(x^2 - 2x - 3) = 2x^2 - 4x - 6$

### Transformations from $y = x^2$

| Transformation | Effect |
|----------------|--------|
| $y = ax^2$ ($a > 1$) | Narrower |
| $y = ax^2$ ($0 < a < 1$) | Wider |
| $y = -x^2$ | Reflected about x-axis |
| $y = x^2 + k$ | Shifted up $k$ |
| $y = (x-h)^2$ | Shifted right $h$ |
| $y = a(x-h)^2 + k$ | All of the above combined |

### Sketching Procedure
1. Determine opening direction ($a > 0$ or $a < 0$).
2. Find y-intercept (set $x = 0$).
3. Find x-intercepts (solve $y = 0$).
4. Find axis of symmetry.
5. Find vertex.
6. Plot all key points and draw smooth curve through them.

### Australian context
A ball kicked in AFL follows approximately $h = -0.49x^2 + 3x + 1$, where $x$ is horizontal distance (m) and $h$ is height (m). Finding the maximum height uses the vertex form.

Parabola Formulas

Axis of symmetry: x = -\dfrac{b}{2a}
Vertex from standard form: \text{vertex} = \left(-\frac{b}{2a},\ f\!\left(-\frac{b}{2a}\right)\right)
Vertex form: y = a(x-h)^2 + k (vertex at $(h, k)$)
x-intercepts: y = a(x-p)(x-q) (intercepts at $x=p$ and $x=q$)

Parabola checklist

a > 0 = smile: Opens upward → minimum vertex.
a < 0 = frown: Opens downward → maximum vertex.
Axis midway: Axis of symmetry is midpoint between two x-intercepts.

Worked example (Medium)

Sketch $y = x^2 - 2x - 3$. Find the vertex, axis, intercepts.

y-intercept. $x = 0$: $y = -3$. Point $(0, -3)$.
x-intercepts. Set $y = 0$: $x^2 - 2x - 3 = (x-3)(x+1) = 0$. $x = 3$ or $x = -1$.
Axis. $x = (3 + (-1))/2 = 1$ (midpoint). Or: $x = -(-2)/(2 \times 1) = 1$.
Vertex. $y = 1 - 2 - 3 = -4$. Vertex: $(1, -4)$.
Shape. $a = 1 > 0$ → opens upward.

Answer: Vertex $(1, -4)$, axis $x = 1$, x-intercepts $(-1, 0)$ and $(3, 0)$, y-intercept $(0, -3)$.

Worked example (Hard)

A ball is thrown upward from a 5 m platform. Its height is $h = -5t^2 + 20t + 5$. Find the maximum height and time to land.

Vertex form. $h = -5(t^2 - 4t) + 5 = -5(t^2 - 4t + 4 - 4) + 5 = -5(t-2)^2 + 20 + 5 = -5(t-2)^2 + 25$.
Max height. Vertex at $t = 2$: max height $= 25$ m.
Land (h = 0). $-5t^2 + 20t + 5 = 0 \Rightarrow t^2 - 4t - 1 = 0$. $t = (4 \pm \sqrt{16+4})/2 = (4 \pm \sqrt{20})/2 = 2 \pm \sqrt{5}$.
Positive time. $t = 2 + \sqrt{5} \approx 4.24$ s (reject negative).

Answer: Maximum height 25 m at $t = 2$ s. Lands at $t \approx 4.24$ s.

Common mistakes

Finding the vertex as $(b/(2a), ...)$ instead of $(-b/(2a), ...)$ The axis of symmetry formula is $x = -b/(2a)$ — the NEGATIVE of $b$ divided by $2a$. Fix: Remember the formula as $x_{\text{vertex}} = -\frac{b}{2a}$. The negative sign is critical.
Plotting only 2–3 points and getting the curve shape wrong With too few points, the parabola shape is guessed and often drawn as a V or too flat/steep. Fix: Always plot at least 5 points: two x-intercepts, y-intercept, vertex, and one more symmetric point. The curve must be smooth and symmetric.
Writing the vertex as $(k, h)$ instead of $(h, k)$ in vertex form $y = a(x-h)^2 + k$ In $y = a(x-h)^2 + k$, the vertex is $(h, k)$ where $h$ is the horizontal shift and $k$ is the vertical shift. Fix: Vertex = $(h, k)$. The horizontal shift $h$ appears INSIDE the bracket with the variable; vertical shift $k$ is ADDED outside.

Practice questions

The axis of symmetry of $y = 2x^2 - 12x + 7$ is:
1. $x = 3$
2. $x = -3$
3. $x = 6$
4. $x = -6$
(1 marks)
Write $y = (x-2)(x+4)$ in standard form and find the vertex.
(2 marks)
Convert $y = x^2 + 6x + 4$ to vertex form and state the vertex.
(3 marks)
A parabola has vertex $(1, 4)$ and passes through $(3, -4)$. Find its equation.
(3 marks)

Going Deeper — Discriminant and the Parabola

The discriminant $\Delta = b^2 - 4ac$ tells you about the graph of $y = ax^2 + bx + c$:

$\Delta > 0$: The parabola crosses the x-axis at TWO points.
$\Delta = 0$: The parabola is TANGENT to the x-axis at one point (vertex touches x-axis).
$\Delta < 0$: The parabola does NOT cross the x-axis (entirely above or below).

This connects algebra to geometry beautifully.

Application — finding tangent conditions:
"For what values of $m$ does $y = mx + 2$ not intersect $y = x^2 + 1$?"

Set equal: $x^2 - mx + (1-2) = 0 \Rightarrow x^2 - mx - 1 = 0$.
Wait, let me redo: $mx + 2 = x^2 + 1 \Rightarrow x^2 - mx - 1 = 0$.
$\Delta = m^2 + 4$. Since $m^2 \geq 0$, $\Delta \geq 4 > 0$ always. So the line always intersects the parabola — they always have 2 intersections for any $m$.

Application — "complete" parabola problem:
When does $y = x^2 - kx + 4$ touch (but not cross) the x-axis?
Need $\Delta = 0$: $k^2 - 16 = 0 \Rightarrow k = \pm 4$.

Summary

Standard form: $y = ax^2 + bx + c$. y-intercept $= c$.
Vertex form: $y = a(x-h)^2 + k$. Vertex at $(h, k)$, axis $x = h$.
Intercept form: $y = a(x-p)(x-q)$. x-intercepts at $p$ and $q$.
Axis: $x = -b/(2a)$. Vertex: substitute axis value into function.
$a > 0$: minimum (opens up). $a < 0$: maximum (opens down).