Lesson 4 ~25 min Unit 2 · Non-Linear +85 XP

Reflections: $y = -x^2$ and $y = -ax^2$

Flip the sign of $a$ and the parabola flips with it. Now the U becomes an upside-down U — opens downward, vertex on top.

Today's hook: A thrown ball goes up, then comes down — its path is a downward parabola. What value of $a$ produces a downward parabola?

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Think First

warm-up

For $y = x^2$ we get $y$ values $\ldots 9, 4, 1, 0, 1, 4, 9 \ldots$. Now compute $y = -x^2$ at the same $x$ values. What changes about every $y$? What stays the same? Where does the new vertex sit on the page — at the bottom of the U, or at the top of an upside-down U?

Record your answer in your workbook.

The Big Idea

+5 XP

If $a$ in $y = ax^2$ is negative, the parabola is reflected in the $x$-axis. The shape is the same, just flipped upside-down. $y = -x^2$ is the mirror image of $y = x^2$. The vertex is still at $(0,0)$ and the axis is still $x = 0$ — but now the vertex is the maximum, not the minimum, and the parabola opens downward.

The gold curve $y = x^2$ opens upward with vertex at the origin. The red curve $y = -x^2$ is its mirror image across the $x$-axis: same vertex, same axis, but now it opens downward. Each point $(x, y)$ on the gold curve becomes $(x, -y)$ on the red one.

$y = -x^2$ — vertex $(0,0)$, axis $x = 0$, opens DOWN

Negative $a$ flips

Sign of $a$ controls direction: $+$ opens up, $-$ opens down.

Vertex = maximum

For $y = -x^2$, $(0,0)$ is the HIGHEST point. $y \le 0$ everywhere.

Magnitude $\to$ width

$|a|$ still controls width. $y = -3x^2$ is narrow AND opens down.

What You'll Master

objectives

Know

When $a < 0$, the parabola opens downward
$y = -x^2$ is the reflection of $y = x^2$ across the $x$-axis
The vertex of $y = -x^2$ is at $(0, 0)$ — the maximum point

Understand

Why a negative coefficient flips the curve vertically
Why the vertex switches role from minimum to maximum
How magnitude $|a|$ and sign of $a$ act independently

Can Do

Sketch $y = -x^2$, $y = -2x^2$ and $y = -\tfrac{1}{2}x^2$
State direction of opening from the sign of $a$
Match a real-world arch (e.g. ball, bridge) to a $-ax^2$ equation

Words You Need

vocabulary

ReflectionA "flip" of the graph — here, flipping $y = x^2$ over the $x$-axis to get $y = -x^2$.

Opens downwardBoth arms head toward $-\infty$. Happens when $a < 0$.

MaximumThe highest point on a downward parabola — the vertex.

MinimumThe lowest point on an upward parabola — the vertex when $a > 0$.

Magnitude $|a|$How large $a$ is, ignoring sign. Controls width.

Sign of $a$Positive $a$ = opens up; negative $a$ = opens down.

Spot the Trap

heads-up

Wrong: "$y = -x^2$ at $x = 3$ gives $9$." Without brackets, $-x^2$ means $-(x^2)$, so $-(3)^2 = -9$.

Right: $y = -x^2$ means "square first, then make negative." So $y = -(3)^2 = -9$ when $x = 3$.

Wrong: "Downward parabolas don't have a vertex." They do — it's just the highest point now, not the lowest.

Right: A downward parabola has a vertex that is its MAXIMUM. For $y = -x^2$ this is $(0, 0)$.

Side-by-Side Tables

+5 XP

Putting tables side by side shows exactly what changes. The negative sign flips every $y$ to its opposite, but the magnitudes stay the same.

$x$	$-2$	$-1$	$0$	$1$	$2$
$y = x^2$	$4$	$1$	$0$	$1$	$4$
$y = -x^2$	$-4$	$-1$	$0$	$-1$	$-4$
$y = -3x^2$	$-12$	$-3$	$0$	$-3$	$-12$

All $y$-values negative (or zero). Larger $|a|$ $\Rightarrow$ steeper drop.

Flip every $y$

$y = -x^2$ takes each $y$ from $y = x^2$ and negates it.

Sign $\ne$ width

$y = -x^2$ has the SAME width as $y = x^2$, just flipped.

Bracket the $x$

Calculator: type $-(2)^2$, never $-2^2$.

Sign and Magnitude Together

+5 XP

The coefficient $a$ does two independent jobs at once:

Sign of $a$: $+$ opens up, $-$ opens down.
Magnitude $|a|$: bigger $\Rightarrow$ narrower, smaller $\Rightarrow$ wider.

So $y = -3x^2$ opens downward (negative sign) and is narrower than $y = -x^2$ (because $|-3| = 3 > 1$). A thrown ball or a bridge arch is a downward, fairly wide parabola.

Direction = sign($a$); width = $|a|$.

Read sign first

Glance at the sign before anything else — up or down?

Then read magnitude

$|a|$ tells you the width — same rule as L03.

Bridge / ball arch

Real-world arches modelled by $y = -ax^2$ (with $a > 0$).

Watch Me Solve It · 3 examples

Watch Me Solve It · Evaluate $y = -x^2$

+15 XP per step

PROBLEM

For $y = -x^2$, find $y$ when $x = -4$.

1
Square first
$(-4)^2 = 16$
Brackets keep the negative inside the square.
2
Apply the minus sign
$y = -(16) = -16$
3
Write the point
$(-4, -16)$ lies on the curve.

Answer$y = -16$, so the point is $(-4, -16)$.

Watch Me Solve It · Direction & width

+15 XP per step

PROBLEM

Describe the graph of $y = -2x^2$: state direction of opening, width compared to $y = x^2$, and vertex.

1
Read the sign
$a = -2 < 0$, so opens DOWN.
2
Read the magnitude
$|a| = 2 > 1$, so NARROWER than $y = x^2$.
3
State the vertex
$(0, 0)$, which is the maximum.

AnswerOpens down, narrower than $y = x^2$, vertex $(0, 0)$.

Watch Me Solve It · Match the curve

+15 XP per step

PROBLEM

A parabola opens downward, has vertex $(0, 0)$, and passes through $(2, -8)$. Find its equation.

1
Set up the form
$y = ax^2$ with $a < 0$ (opens down)
2
Substitute the point
$-8 = a(2)^2 = 4a \Rightarrow a = -2$
3
Write the equation
$y = -2x^2$

Answer$y = -2x^2$.

Common Pitfalls

heads-up

Order of operations on $-x^2$

Students compute $(-x)^2$ and get $x^2$, missing the minus sign.

Fix: $-x^2$ means $-(x^2)$ — square first, then negate. Always negative (or zero).

Confusing reflection with shift

"Negative $a$ pulls the curve down by $a$" mixes reflection with translation.

Fix: Negative $a$ flips the curve through the $x$-axis. Vertex stays at $(0,0)$.

Mixing sign and width effects

Saying "$y = -3x^2$ is wider because $-3$ is smaller than $1$."

Fix: Use $|a|$ for width, sign($a$) for direction. $|-3| = 3 > 1$ — narrower AND downward.

Copy Into Your Books

$y = -x^2$

Reflection of $y = x^2$
Opens DOWN
Vertex $(0,0)$ = maximum

Sign of $a$

$a > 0$: opens up
$a < 0$: opens down
$a = 0$: not a parabola

Magnitude $|a|$

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

Real world

Thrown ball arc
Bridge arch
Both: $y = -ax^2$ shape

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Reflections

4 problems

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

1 Find $y$ for $y = -x^2$ at $x = 5$.
$-(5)^2 = -25$.$y = -25$
2 Which way does $y = -4x^2$ open?
$a = -4 < 0$.Downward
3 Is the vertex of $y = -x^2$ a maximum or minimum?
The curve opens down, so the vertex sits on top.Maximum
4 A parabola opens down with vertex $(0,0)$ through $(1, -5)$. Find $a$.
$-5 = a(1)^2$.$a = -5$

Complete in your workbook.

Quick Check · 5 questions

Which way does $y = -x^2$ open?

+10 XP

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

+10 XP

For $y = -x^2$, the vertex $(0, 0)$ is the:

+10 XP

Which best describes $y = -3x^2$?

+10 XP

A downward parabola $y = ax^2$ passes through $(2, -12)$. Find $a$.

+10 XP

Show Your Working · 3 questions

Show Your Working

9 marks total

ApplyMedium3 MARKS

Q6. Complete a table of values for $y = -x^2$ from $x = -3$ to $x = 3$, then sketch the curve. Label the vertex and state whether it is a maximum or minimum.

Answer in your workbook.

ApplyMedium2 MARKS

Q7. Describe the graph of $y = -\tfrac{1}{2}x^2$: state the direction of opening, the width compared to $y = x^2$, and the vertex.

Answer in your workbook.

ReasonHard4 MARKS

Q8. A thrown ball follows the path $y = -2x^2$ measured in metres (with the launch point at $(0,0)$ and $x$ horizontal). The ball lands when $y = -8$. Find the horizontal distance from launch when this happens, show working, and explain why the negative sign in the equation makes sense for a thrown ball.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. B — negative $a$, opens downward.

2. A — $-(3)^2 = -9$.

3. C — vertex is the maximum on a downward parabola.

4. D — opens down ($a < 0$), narrow ($|a| = 3 > 1$).

5. A — $-12 = 4a$ gives $a = -3$.

Show Your Working Model Answers

Q6 (3 marks): Table $y: -9, -4, -1, 0, -1, -4, -9$ for $x = -3, \ldots, 3$ [1]. Sketched as a smooth downward U through the points [1]. Vertex $(0, 0)$ labelled as a maximum [1].

Q7 (2 marks): $a = -\tfrac{1}{2}$: sign is negative so it opens downward [1]. $|a| = \tfrac{1}{2} < 1$ so wider than $y = x^2$; vertex $(0, 0)$ [1].

Q8 (4 marks): Set $-8 = -2x^2$ [1]. Divide both sides by $-2$: $x^2 = 4$ [1]. So $x = \pm 2$; horizontal distance is $2$ m (taking the positive value) [1]. The negative sign in the equation makes the curve open downward, which matches a thrown ball that rises then falls (its height never goes above the launch point along this model) [1].

Stretch Challenge · +25 XP, +10 coins

Bridge Arch

An engineer models a bridge arch by $y = -\tfrac{1}{4}x^2$, with the top of the arch at the origin and $x, y$ in metres. (a) Find $y$ when $x = 4$. (b) How far is that point below the top of the arch? (c) What value of $|a|$ would make the arch narrower?

Reveal solution

(a) $y = -\tfrac{1}{4}(4)^2 = -\tfrac{1}{4} \times 16 = -4$. (b) $4$ metres below the top. (c) Any $|a| > \tfrac{1}{4}$ — e.g. $|a| = 1$ or $|a| = 2$ would give a narrower arch. (Still negative to keep it opening downward.)

Quick Review

Equation

$y = -ax^2$ — reflected parabola

Vertex

$(0, 0)$ — now a MAXIMUM

Axis

$x = 0$ — still the $y$-axis

Direction

Sign of $a$: $-$ opens down, $+$ opens up

Width

$|a|$ controls width (same as L03)

Range

$y \le 0$ for any $y = -ax^2$ with $a > 0$

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