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

The Basic Parabola $y = x^2$

Plot it, study it, name its parts. Meet the most important curve in Stage 5: the U-shaped parabola.

Today's hook: When $x = -3$ and $x = 3$, both give $y = 9$. Why does squaring always wipe out the negative sign?

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Think First

warm-up

Try this: pick five different $x$-values (some negative, some positive). Square each one. What pattern do you notice about the answers? Now try $x = -5$ and $x = 5$. Why do you think they give the same $y$ value?

Record your answer in your workbook.

The Big Idea

+5 XP

The equation $y = x^2$ creates a perfectly symmetric U-shaped curve called the basic parabola. It is the parent of all parabolas you'll meet in Stage 5. Because $(-a)^2 = a^2$ for any number $a$, each positive and negative $x$ produces the same $y$.

The vertex is the lowest (or highest) point of the parabola. For $y = x^2$ the vertex is at $(0, 0)$. The axis of symmetry is the vertical line that splits the curve into two mirror halves — here, the $y$-axis ($x = 0$). The parabola opens upward because the coefficient of $x^2$ is positive.

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

Same $y$, opposite $x$

$x = 3$ and $x = -3$ both give $y = 9$. Mirror image across $y$-axis.

Vertex = lowest

$(0, 0)$ is the minimum value of $y$. Nothing dips below it.

Opens up

Both arms head up to $+\infty$ as $|x|$ grows.

What You'll Master

objectives

Know

The equation $y = x^2$ produces a parabola
The vertex is at $(0, 0)$ — the minimum point
The axis of symmetry is the line $x = 0$

Understand

Why squaring always gives a non-negative result
Why the curve is symmetric about the $y$-axis
Why both arms point upward

Can Do

Build a table of values for $y = x^2$ from $x = -3$ to $3$
Plot the points and join them with a smooth curve
Identify vertex, axis, and direction of opening

Words You Need

vocabulary

ParabolaThe U-shaped curve produced by an equation of the form $y = ax^2$ (with extras).

VertexThe turning point of the parabola — either its minimum or maximum point.

Axis of symmetryThe vertical line that splits the parabola into two mirror-image halves.

Opens upwardBoth arms of the parabola point up; the vertex is the lowest point.

SymmetricFor every point $(x, y)$ on the curve, the point $(-x, y)$ is also on the curve.

Smooth curveA flowing line with no corners or breaks — what you should draw, not a polyline.

Spot the Trap

heads-up

Wrong: "$(-3)^2 = -9$". The negative is squared along with the 3. $(-3)^2 = (-3) \times (-3) = 9$.

Right: $(-3)^2 = 9$. Brackets matter — $-3^2$ (without brackets) means $-(3^2) = -9$.

Wrong: "$y = x^2$ can be negative for some $x$." No — squaring always gives $\ge 0$.

Right: The basic parabola sits entirely on or above the $x$-axis.

Table of Values

+5 XP

Always build the table before plotting. For $y = x^2$ from $x = -3$ to $3$:

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$y$	$9$	$4$	$1$	$0$	$1$	$4$	$9$

Y-values mirror: $9, 4, 1, 0, 1, 4, 9$

Mirror table

$y$ row reads the same backwards — a sign of symmetry.

Always include $0$

$x = 0$ gives the vertex. Never skip it.

Use brackets

Write $(-2)^2 = 4$ not $-2^2 = -4$ (calculator mistake!).

Key Features at a Glance

+5 XP

Once you've plotted $y = x^2$, here is everything you need to know in five bullets.

Vertex: $(0, 0)$ — the lowest point.
Axis of symmetry: $x = 0$ (the $y$-axis).
$y$-intercept: $(0, 0)$.
$x$-intercept: $(0, 0)$ — the curve just touches the $x$-axis.
Domain: all real $x$. Range: $y \ge 0$.

Five facts; remember them all.

Touch, not cross

The curve just touches the $x$-axis at the vertex.

No negative $y$

$y$ is always $\ge 0$ for $y = x^2$.

Width

Standard width: passes through $(1,1)$ and $(-1,1)$.

Watch Me Solve It · 3 examples

Watch Me Solve It · Find a value

+15 XP per step

PROBLEM

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

1
Substitute
$y = (-4)^2$
Use brackets to keep the negative inside the square.
2
Evaluate
$(-4)^2 = (-4) \times (-4) = 16$
Negative $\times$ negative $=$ positive.
3
Write the point
$(-4, 16)$ lies on the parabola.

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

Watch Me Solve It · Use symmetry

+15 XP per step

PROBLEM

$(5, 25)$ is a point on $y = x^2$. Use symmetry to find another point with the same $y$.

1
Identify the axis
Axis of symmetry is $x = 0$ (the $y$-axis).
2
Reflect across $x = 0$
$x = 5$ reflects to $x = -5$.
$y$ stays the same under reflection.
3
Check
$(-5)^2 = 25$ ✓

Answer$(-5, 25)$.

Watch Me Solve It · Identify features

+15 XP per step

PROBLEM

State the vertex, axis of symmetry, and direction of opening for $y = x^2$.

1
Vertex
$(0, 0)$
Lowest point on the curve.
2
Axis of symmetry
$x = 0$ (the $y$-axis)
3
Direction
Opens upward (coefficient of $x^2$ is positive).

AnswerVertex $(0,0)$; axis $x = 0$; opens up.

Common Pitfalls

heads-up

Squaring without brackets

Writing $-3^2 = -9$ in a calculator is interpreted as $-(3^2)$.

Fix: Always type $(-3)^2$ or hand-write with brackets to get $9$.

Connecting points with straight lines

A polyline through your plotted points is jagged; a parabola is smooth.

Fix: Draw a single flowing curve through the points.

Drawing the V-shape

$y = x^2$ is rounded at the bottom, not pointed. Pointed $=$ $y = |x|$, not a parabola.

Fix: Round off the bottom; show the curve passing through $(1,1)$ and $(-1,1)$.

Copy Into Your Books

$y = x^2$

The basic parabola
Vertex $(0, 0)$
Axis of symmetry $x = 0$

Direction

Opens upward
Vertex is minimum
$y \ge 0$ always

Symmetry

$(a, a^2)$ and $(-a, a^2)$
Mirror in $y$-axis
Same height for $\pm x$

Method

Build table
Plot points
Draw smooth U-curve

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Basic Parabola

4 problems

Four quick problems on the basic parabola.

1 Find $y$ when $x = -6$ on $y = x^2$.
$(-6)^2 = 36$.$y = 36$
2 What is the vertex of $y = x^2$?
The lowest point of the U.$(0, 0)$
3 Write the axis of symmetry of $y = x^2$.
The vertical line through the vertex.$x = 0$
4 Does $(7, 49)$ lie on $y = x^2$?
$7^2 = 49$ ✓Yes

Complete in your workbook.

Quick Check · 5 questions

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

+10 XP

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

+10 XP

The axis of symmetry of $y = x^2$ is:

+10 XP

Which describes the opening of $y = x^2$?

+10 XP

If $(4, 16)$ is on $y = x^2$, which point also lies on it by symmetry?

+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.

Answer in your workbook.

UnderstandEasy2 MARKS

Q7. Without plotting, state the vertex, axis of symmetry, and range of $y = x^2$.

Answer in your workbook.

ReasonHard4 MARKS

Q8. A student claims $(-7, -49)$ lies on $y = x^2$. Is the student correct? Justify with full working and explain the misconception in one sentence.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. A — vertex $(0, 0)$.

2. C — $(-3)^2 = 9$.

3. B — axis $x = 0$.

4. D — opens upward.

5. B — symmetric image is $(-4, 16)$.

Show Your Working Model Answers

Q6 (3 marks): Table $y: 9, 4, 1, 0, 1, 4, 9$ for $x = -3, \ldots, 3$ [1]. Plotted as smooth U-curve [1]. Vertex labelled at $(0,0)$ [1].

Q7 (2 marks): Vertex $(0,0)$ [0.5]. Axis $x = 0$ [0.5]. Range $y \ge 0$ [1].

Q8 (4 marks): Substitute $x = -7$: $y = (-7)^2 = 49$ [1]. So the actual point is $(-7, 49)$, not $(-7, -49)$ [1]. The student is incorrect [1]. Misconception: they forgot brackets around the negative when squaring, treating $(-7)^2$ as $-49$ when it equals $+49$ [1].

Stretch Challenge · +25 XP, +10 coins

Hit the Vertex

A parabola has equation $y = x^2$. Find all $x$ such that $y = 25$. How many solutions are there, and what does this tell you about the symmetry of the curve?

Reveal solution

$x^2 = 25 \Rightarrow x = \pm 5$. Two solutions: $x = 5$ and $x = -5$. They reflect across the $y$-axis, confirming the curve is symmetric about $x = 0$. Every horizontal line $y = c$ (with $c > 0$) hits the parabola at exactly two symmetric points.

Quick Review

Equation

$y = x^2$ — the basic parabola

Vertex

$(0, 0)$ — minimum point

Axis

$x = 0$ — the $y$-axis

Opens

Upward (both arms head up)

Symmetry

$(a, a^2)$ and $(-a, a^2)$

Range

$y \ge 0$ for all real $x$

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