Lesson 3 ~25 min Unit 2 · Non-Linear +85 XP

Dilations: $y = ax^2$

The coefficient $a$ controls width. Bigger $a$ pinches the curve narrower; smaller $a$ spreads it wider. Same vertex, same axis — different stretch.

Today's hook: $y = 2x^2$ and $y = \tfrac{1}{2}x^2$ both look like parabolas. How do they differ — and what does $a$ actually control?

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Think First

warm-up

For $y = x^2$, when $x = 2$, $y = 4$. Now try $y = 2x^2$ at $x = 2$: what do you get? And $y = \tfrac{1}{2}x^2$ at $x = 2$? Sketch all three points on the same set of axes. Which curve climbs fastest, and which climbs slowest?

Record your answer in your workbook.

The Big Idea

+5 XP

The number $a$ in $y = ax^2$ is the dilation factor. It multiplies every $y$-value of $y = x^2$. When $a > 1$ the parabola is narrower (stretched vertically, pinched horizontally). When $0 < a < 1$ it is wider (flattened). The vertex stays at $(0, 0)$ and the axis stays $x = 0$ either way.

The red curve $y = 2x^2$ is steeper than $y = x^2$ — at $x = 1$ it's already at $y = 2$. The purple curve $y = \tfrac{1}{2}x^2$ is shallower — at $x = 1$ it only reaches $y = 0.5$. The gold curve in the middle is $y = x^2$ for reference.

$y = ax^2$ — vertex $(0,0)$, axis $x = 0$, width controlled by $a$

$a > 1$ = narrow

$y = 3x^2$ is pinched. Climbs faster than $y = x^2$.

$0 < a < 1$ = wide

$y = \tfrac{1}{4}x^2$ is spread out. Climbs slower than $y = x^2$.

Vertex unchanged

Still $(0, 0)$. Only the width changes.

What You'll Master

objectives

Know

$y = ax^2$ has the same vertex $(0, 0)$ as $y = x^2$
When $a > 1$ the parabola is narrower
When $0 < a < 1$ the parabola is wider

Understand

Why multiplying by $a$ stretches $y$ rather than $x$
Why the axis of symmetry stays at $x = 0$
Why the curve still opens upward when $a > 0$

Can Do

Sketch $y = 2x^2$, $y = 3x^2$ and $y = \tfrac{1}{2}x^2$ from a table of values
Compare two parabolas and say which is narrower
Find $a$ when given a point on the curve

Words You Need

vocabulary

DilationA stretch or compression of a graph — here, vertical stretching of $y = x^2$ by factor $a$.

Coefficient $a$The number multiplying $x^2$. Controls width (and later, direction).

NarrowerPinched in — the arms rise more steeply. Happens when $a > 1$.

WiderSpread out — the arms rise more gently. Happens when $0 < a < 1$.

Vertical stretchEach $y$-value is multiplied by $a$, while $x$ stays the same.

Reference curve$y = x^2$ — the baseline we compare every other parabola to.

Spot the Trap

heads-up

Wrong: "$y = \tfrac{1}{2}x^2$ is narrower because $\tfrac{1}{2}$ is a fraction." Fractions between 0 and 1 make $y$ smaller, so the curve is WIDER, not narrower.

Right: Bigger $a$ $\Rightarrow$ narrower. Smaller positive $a$ $\Rightarrow$ wider.

Wrong: "$y = 2x^2$ shifts the vertex up by 2." No — multiplying does not shift the curve; it stretches it.

Right: Vertex of $y = ax^2$ is still $(0, 0)$ for any positive $a$.

Compare the Tables

+5 XP

Lining up tables makes the dilation obvious. The $y$-values of $y = 2x^2$ are double those of $y = x^2$; the $y$-values of $y = \tfrac{1}{2}x^2$ are half.

$x$	$-2$	$-1$	$0$	$1$	$2$
$y = x^2$	$4$	$1$	$0$	$1$	$4$
$y = 2x^2$	$8$	$2$	$0$	$2$	$8$
$y = \tfrac{1}{2}x^2$	$2$	$0.5$	$0$	$0.5$	$2$

Every row is $a$ times the $y = x^2$ row.

Times-table trick

Write $y = x^2$ first, then multiply every $y$ by $a$.

$(0, 0)$ doesn't change

$a \times 0 = 0$, so the vertex sits still no matter what $a$ is.

Use $\pm 1, \pm 2$

These give clean numbers and show the symmetry instantly.

Finding $a$ from a Point

+5 XP

If you're told a parabola has form $y = ax^2$ and you know one point off the vertex, you can solve for $a$. Substitute the $x$ and $y$ values and isolate $a$.

Example: a parabola $y = ax^2$ passes through $(2, 12)$. Then $12 = a(2)^2 = 4a$, so $a = 3$. The equation is $y = 3x^2$. Quick check: at $x = -2$, $y = 3(-2)^2 = 12$ ✓.

Method: $y = ax^2 \Rightarrow a = \dfrac{y}{x^2}$ (any non-vertex point)

Don't pick the vertex

$(0,0)$ gives $0 = 0$ — useless. Always pick an off-vertex point.

Square first

$(2)^2 = 4$ before dividing into $y$.

Verify symmetry

$(-x, y)$ should give the same $a$ — a great self-check.

Watch Me Solve It · 3 examples

Watch Me Solve It · Build a table

+15 XP per step

PROBLEM

Build a table of values for $y = 3x^2$ from $x = -2$ to $x = 2$. State whether the curve is narrower or wider than $y = x^2$.

1
Square each $x$
$x^2$: $4, 1, 0, 1, 4$
Square first, then multiply.
2
Multiply by 3
$y$: $12, 3, 0, 3, 12$
Each value is $3 \times x^2$.
3
Compare to $y = x^2$
$3 > 1$, so $y = 3x^2$ is narrower.

AnswerTable: $12, 3, 0, 3, 12$. Narrower than $y = x^2$.

Watch Me Solve It · Find $a$ from a point

+15 XP per step

PROBLEM

A parabola has equation $y = ax^2$ and passes through $(3, 18)$. Find $a$ and state whether the curve is wider or narrower than $y = x^2$.

1
Substitute the point
$18 = a(3)^2 = 9a$
2
Solve for $a$
$a = \dfrac{18}{9} = 2$
3
Interpret
$a = 2 > 1$, so the curve is narrower than $y = x^2$.

Answer$a = 2$. Equation $y = 2x^2$; narrower.

Watch Me Solve It · Compare widths

+15 XP per step

PROBLEM

Rank from widest to narrowest: $y = 4x^2$, $y = \tfrac{1}{3}x^2$, $y = x^2$.

1
List the $a$ values
$a = 4, \tfrac{1}{3}, 1$
2
Smaller $a$ = wider
Order from smallest to largest: $\tfrac{1}{3}, 1, 4$
3
Write the ranking
Widest $\to$ narrowest: $y = \tfrac{1}{3}x^2$, $y = x^2$, $y = 4x^2$.

Answer$y = \tfrac{1}{3}x^2$ widest, then $y = x^2$, then $y = 4x^2$ narrowest.

Common Pitfalls

heads-up

"Fractions = narrower"

A common reflex is that $\tfrac{1}{2}$ "shrinks" the curve and makes it narrower.

Fix: $\tfrac{1}{2}$ shrinks the $y$-values, which flattens the curve. Smaller positive $a$ = WIDER.

Treating $a$ as a shift

Writing the vertex of $y = 2x^2$ as $(0, 2)$ confuses multiplication with translation.

Fix: Multiplication stretches; addition shifts. Vertex of $y = ax^2$ is always $(0,0)$.

Forgetting to square first

Computing $y = 2x^2$ at $x = 3$ as $(2 \times 3)^2 = 36$ instead of $2 \times 9 = 18$.

Fix: Powers before multiplication — square first, then multiply by $a$.

Copy Into Your Books

$y = ax^2$

Dilation of $y = x^2$
Vertex still $(0, 0)$
Axis still $x = 0$

Width

$a > 1$: narrower
$0 < a < 1$: wider
$a = 1$: standard width

Method

Build $x^2$ row
Multiply by $a$
Plot and smooth-curve

Find $a$

Use a non-vertex point
$a = y / x^2$
Verify with symmetry

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Dilations

4 problems

Four quick problems on $y = ax^2$.

1 Find $y$ when $x = -3$ on $y = 2x^2$.
$y = 2 \times (-3)^2 = 2 \times 9$.$y = 18$
2 Is $y = \tfrac{1}{4}x^2$ wider or narrower than $y = x^2$?
$a = \tfrac{1}{4} < 1$, so wider.Wider
3 The point $(2, 20)$ lies on $y = ax^2$. Find $a$.
$20 = 4a \Rightarrow a = 5$.$a = 5$
4 What is the vertex of $y = 5x^2$?
Dilation doesn't move the vertex.$(0, 0)$

Complete in your workbook.

Quick Check · 5 questions

Compared to $y = x^2$, the curve $y = 4x^2$ is:

+10 XP

For $y = 3x^2$, what is $y$ when $x = -2$?

+10 XP

The vertex of $y = 5x^2$ is:

+10 XP

A parabola $y = ax^2$ passes through $(2, 8)$. Find $a$.

+10 XP

Which parabola is the widest?

+10 XP

Show Your Working · 3 questions

Show Your Working

9 marks total

ApplyMedium3 MARKS

Q6. Complete a table of values for $y = 2x^2$ from $x = -3$ to $x = 3$, and sketch the curve. Label the vertex.

Answer in your workbook.

ApplyMedium2 MARKS

Q7. The parabola $y = ax^2$ passes through $(4, 32)$. Find $a$ and write the equation. State whether the curve is wider or narrower than $y = x^2$.

Answer in your workbook.

ReasonHard4 MARKS

Q8. A student says: "$y = \tfrac{1}{2}x^2$ must be narrower than $y = x^2$ because halving makes things smaller." Explain in your own words why this is wrong. Use at least one numerical comparison.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. C — $a = 4 > 1$, narrower.

2. B — $3 \times (-2)^2 = 12$.

3. A — vertex unchanged at $(0,0)$.

4. D — $a = 8/4 = 2$.

5. B — smallest $a = \tfrac{1}{5}$ gives widest curve.

Show Your Working Model Answers

Q6 (3 marks): Table $y: 18, 8, 2, 0, 2, 8, 18$ for $x = -3, \ldots, 3$ [1]. Plotted as a smooth narrow U-curve [1]. Vertex labelled at $(0,0)$ [1].

Q7 (2 marks): $32 = a(4)^2 = 16a$ so $a = 2$ [1]. Equation $y = 2x^2$; since $a = 2 > 1$ the curve is narrower than $y = x^2$ [1].

Q8 (4 marks): The student is wrong [1]. Smaller positive $a$ makes the $y$-values smaller, so the curve grows more slowly — it is WIDER, not narrower [1]. At $x = 2$, $y = x^2$ gives $4$ but $y = \tfrac{1}{2}x^2$ gives only $2$ — the half-curve is below and flatter [1]. So the graph spreads out further along the $x$-axis to climb to the same height as $y = x^2$ [1].

Stretch Challenge · +25 XP, +10 coins

Mystery Coefficient

A parabola of form $y = ax^2$ passes through both $(2, 6)$ and $(-2, 6)$. (a) Find $a$. (b) Is the curve wider or narrower than $y = x^2$? (c) Why do both points give the same value of $a$?

Reveal solution

(a) $6 = a(2)^2 = 4a$ so $a = \tfrac{3}{2}$. (b) $a = 1.5 > 1$, so narrower than $y = x^2$. (c) Squaring $-2$ also gives $4$, so the same equation works — this is exactly the $y$-axis symmetry of every parabola $y = ax^2$.

Quick Review

Equation

$y = ax^2$ — dilation of $y = x^2$

Vertex

$(0, 0)$ — unchanged by $a$

Axis

$x = 0$ — unchanged by $a$

$a > 1$

Narrower (steeper) than $y = x^2$

$0 < a < 1$

Wider (flatter) than $y = x^2$

Find $a$

$a = y / x^2$ from any non-vertex point

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