Lesson 7 ~25 min Unit 2 · Non-Linear +85 XP

Horizontal Translations: $y = (x - h)^2$

Replace $x$ with $(x - h)$ and the parabola slides sideways — but the direction can feel backwards. Master the "minus means right" rule.

Today's hook: $y = (x - 3)^2$: the "$-3$" makes you think it goes left, but it goes RIGHT 3. Why does the minus sign fool people?

0/5QUESTS

Printable Worksheets

Print or save as PDF — or build a custom worksheet from any module's questions.

Build Apply Master Build custom

Think First

warm-up

Look at the equation $y = (x - 3)^2$. If you put $x = 3$ into the bracket, you get $0$, and $y = 0$. So the lowest point of the curve is at $(3, 0)$, not at the origin. The graph of $y = x^2$ has been slid to a new home. Where exactly? Why does the "$-3$" inside the bracket move the curve to the RIGHT rather than to the LEFT?

Record your answer in your workbook.

The Big Idea

+5 XP

Replacing $x$ with $(x - h)$ shifts the parabola HORIZONTALLY by $h$ units. The shape and direction stay the same; the curve simply slides sideways. The vertex moves from $(0, 0)$ to $(h, 0)$, and the axis of symmetry becomes $x = h$. The counter-intuitive bit: $y = (x - 3)^2$ shifts the graph 3 units to the RIGHT, while $y = (x + 2)^2$ shifts it 2 units to the LEFT.

The gold reference is $y = x^2$ with vertex $(0, 0)$. The red curve $y = (x - 3)^2$ has slid 3 right — vertex $(3, 0)$. The purple curve $y = (x + 2)^2$ has slid 2 left — vertex $(-2, 0)$. Same shape, new location.

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

Minus means right

$(x - 3)^2$ shifts right by 3.

Plus means left

$(x + 2)^2 = (x - (-2))^2$ shifts left by 2.

Same shape

No stretch, no flip — just a slide.

What You'll Master

objectives

Know

$y = (x - h)^2$ has vertex $(h, 0)$ and axis $x = h$
"Minus inside" shifts right; "plus inside" shifts left
The $y$-intercept of $y = (x - h)^2$ is $h^2$

Understand

Why the sign of $h$ inside flips compared to outside
Why the vertex sits where the bracket equals zero
Why the shape doesn't change — only position does

Can Do

Read the vertex and axis off any $y = (x - h)^2$
Find the $y$-intercept of a horizontally shifted parabola
Sketch $y = (x - 4)^2$ or $y = (x + 1)^2$ using the slide

Words You Need

vocabulary

Horizontal translationA sideways slide of a graph — left or right, never up or down.

$h$The horizontal shift. Vertex sits at $x = h$.

Vertex (on the $x$-axis)$(h, 0)$ for $y = (x - h)^2$. Same height as the origin.

Axis of symmetryThe vertical line $x = h$ — the parabola is a mirror image across it.

$y$-interceptSubstitute $x = 0$: $y = (0 - h)^2 = h^2$.

SlideA rigid motion: every point moves the same distance in the same direction.

Spot the Trap

heads-up

Wrong: $y = (x - 3)^2$ shifts the graph 3 LEFT because of the minus sign.

Right: The vertex is where $(x - 3) = 0$, i.e. $x = 3$. So it shifts 3 RIGHT to $(3, 0)$.

Wrong: $y = (x + 2)^2$ passes through the origin because of the "$+2$".

Right: At $x = 0$, $y = (0 + 2)^2 = 4$. The graph passes through $(0, 4)$, not the origin. Vertex is at $(-2, 0)$.

Why "Minus Means Right"

+5 XP

The vertex of any parabola of this form is wherever the bracket equals zero (because squaring zero gives zero — the lowest possible value).

For $y = (x - h)^2$, solve $x - h = 0 \Rightarrow x = h$. So the vertex sits at $x = h$, which is $+h$ on the $x$-axis. A minus sign INSIDE the bracket means a POSITIVE shift outside.

Quick check: $(x - 5)^2$ → vertex at $x = 5$. $(x + 7)^2 = (x - (-7))^2$ → vertex at $x = -7$.

Vertex $x$-coordinate = value that makes the bracket zero.

Zero the bracket

Find $x$ that makes $x - h = 0$. That's where the vertex lives.

Squaring kills the sign

The output is $\ge 0$ always — minimum is 0 at the vertex.

Flip the sign

"Minus" in the bracket $\to$ "plus" on the axis. And vice versa.

$y$-Intercept of $y = (x - h)^2$

+5 XP

To find where the curve crosses the $y$-axis, substitute $x = 0$.

$y = (0 - h)^2 = h^2$. So the $y$-intercept of $y = (x - h)^2$ is always $(0, h^2)$. It's positive (or zero), since squaring kills the sign of $h$.

$y = (x - 3)^2$: $y$-intercept $(0, 9)$.
$y = (x + 2)^2$: $y$-intercept $(0, 4)$.
$y = (x - 5)^2$: $y$-intercept $(0, 25)$.

$y$-intercept of $y = (x - h)^2$ is $(0, h^2)$.

Substitute $x = 0$

Standard rule for any $y$-intercept — doesn't change here.

$h^2$ is positive

The $y$-intercept is always at or above the $x$-axis.

It's not the vertex

Vertex is at $(h, 0)$; $y$-intercept is at $(0, h^2)$. Different points!

Watch Me Solve It · 3 examples

Watch Me Solve It · Vertex of $y = (x - 4)^2$

+15 XP per step

PROBLEM

State the vertex, axis of symmetry, and $y$-intercept of $y = (x - 4)^2$.

1
Zero the bracket
$x - 4 = 0 \Rightarrow x = 4$. So $h = 4$.
2
Vertex & axis
Vertex $(4, 0)$. Axis of symmetry $x = 4$.
3
$y$-intercept
$y = (0 - 4)^2 = 16$. So $(0, 16)$.
Same shape as $y = x^2$, just slid 4 units right.

AnswerVertex $(4, 0)$, axis $x = 4$, $y$-intercept $(0, 16)$.

Watch Me Solve It · Vertex of $y = (x + 5)^2$

+15 XP per step

PROBLEM

State the vertex and $y$-intercept of $y = (x + 5)^2$. Describe the shift from $y = x^2$.

1
Rewrite in $(x - h)$ form
$(x + 5)^2 = (x - (-5))^2$. So $h = -5$.
2
Vertex
$(h, 0) = (-5, 0)$. Shift: 5 units to the LEFT.
3
$y$-intercept
$y = (0 + 5)^2 = 25$. So $(0, 25)$.
"Plus inside" $\Rightarrow$ shift left. Same shape.

AnswerVertex $(-5, 0)$; $y$-intercept $(0, 25)$; shifted 5 left from $y = x^2$.

Watch Me Solve It · Equation from a shift

+15 XP per step

PROBLEM

$y = x^2$ is shifted 6 units to the right. Write the new equation and state the vertex.

1
Identify $h$
Shift right by 6 $\Rightarrow$ $h = +6$.
2
Replace $x$ with $(x - h)$
$y = (x - 6)^2$. Note the MINUS sign inside.
3
Vertex
$(h, 0) = (6, 0)$.
Shift direction matches sign of $h$: $h = 6 > 0 \Rightarrow$ right.

Answer$y = (x - 6)^2$ with vertex $(6, 0)$.

Common Pitfalls

heads-up

Following the sign of $h$ instead of flipping it

Saying $y = (x - 3)^2$ shifts 3 LEFT because of the minus.

Fix: Zero the bracket: $x = 3$. The vertex sits at the $x$ that makes the inside zero.

Confusing vertex with $y$-intercept

Calling $(0, 9)$ the vertex of $y = (x - 3)^2$ because $y = 9$ when $x = 0$.

Fix: Vertex is the lowest point $(h, 0)$. $y$-intercept is where the curve crosses the $y$-axis. Two different points.

Squaring before subtracting

Reading $(x - 3)^2$ as "$x^2 - 3$" or "$x^2 - 9$".

Fix: Brackets come first. $(x - 3)^2$ means "take $(x - 3)$, then square the WHOLE thing." Not the same as $x^2 - 9$.

Copy Into Your Books

$y = (x - h)^2$

Vertex $(h, 0)$
Axis $x = h$
Same shape as $y = x^2$

Sign rule

Minus inside $\to$ shift right
Plus inside $\to$ shift left
Flip the sign you read

$y$-intercept

Set $x = 0$
$y = h^2$
Always $\ge 0$

Quick checks

$(x - 3)^2$: vertex $(3, 0)$
$(x + 2)^2$: vertex $(-2, 0)$
$(x - 5)^2$: vertex $(5, 0)$

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Horizontal Shifts

4 problems

Four quick problems on $y = (x - h)^2$.

1 State the vertex of $y = (x - 7)^2$.
Bracket zero at $x = 7$.$(7, 0)$
2 State the vertex of $y = (x + 4)^2$.
$(x + 4) = 0 \Rightarrow x = -4$.$(-4, 0)$
3 Find the $y$-intercept of $y = (x - 6)^2$.
$y = (0 - 6)^2 = 36$.$(0, 36)$
4 Write the equation when $y = x^2$ is shifted 8 left.
Left by 8 $\Rightarrow$ $h = -8$.$y = (x + 8)^2$

Complete in your workbook.

Quick Check · 5 questions

The vertex of $y = (x - 3)^2$ is:

+10 XP

$y = (x + 2)^2$ shifts $y = x^2$:

+10 XP

The $y$-intercept of $y = (x - 5)^2$ is:

+10 XP

$y = x^2$ shifted 4 units RIGHT is:

+10 XP

The axis of symmetry of $y = (x + 1)^2$ is:

+10 XP

Show Your Working · 3 questions

Show Your Working

9 marks total

ApplyMedium3 MARKS

Q6. For $y = (x - 2)^2$, state: (a) the vertex; (b) the axis of symmetry; (c) the $y$-intercept. Show how you find (c).

Answer in your workbook.

ApplyMedium3 MARKS

Q7. Sketch $y = (x + 3)^2$ on a labelled set of axes. Mark the vertex, the $y$-intercept, and one symmetric point (say at $x = -1$).

Answer in your workbook.

ReasonHard3 MARKS

Q8. A parabola has the same shape as $y = x^2$ but its vertex is at $(-7, 0)$. (a) Write its equation. (b) Find its $y$-intercept. (c) Explain in one sentence why the sign inside the bracket is opposite to the sign of the $x$-coordinate of the vertex.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. B — $(x - 3) = 0 \Rightarrow x = 3$, vertex $(3, 0)$.

2. A — $(x + 2) = 0 \Rightarrow x = -2$, shift 2 left.

3. D — $(0 - 5)^2 = 25$.

4. C — right by 4 $\Rightarrow$ replace $x$ with $(x - 4)$.

5. B — vertex $(-1, 0)$, axis $x = -1$.

Show Your Working Model Answers

Q6 (3 marks): (a) $(x - 2) = 0 \Rightarrow x = 2$. Vertex $(2, 0)$ [1]. (b) Axis of symmetry $x = 2$ [1]. (c) Sub $x = 0$: $y = (0 - 2)^2 = 4$. $y$-intercept $(0, 4)$ [1].

Q7 (3 marks): Vertex $(-3, 0)$ [1]. $y$-intercept: $(0 + 3)^2 = 9$, so $(0, 9)$ [1]. Symmetric point at $x = -1$: $y = (-1 + 3)^2 = 4$, point $(-1, 4)$. Mirror at $(-5, 4)$. Smooth U through all four labelled points [1].

Q8 (3 marks): (a) $y = (x + 7)^2 = (x - (-7))^2$ [1]. (b) Sub $x = 0$: $y = 49$, intercept $(0, 49)$ [1]. (c) The vertex sits at the $x$ that makes the bracket zero. Inside "$x + 7$" reaches zero at $x = -7$; inside "$x - 7$" reaches zero at $x = +7$. So the inside sign and the vertex $x$-coordinate are opposites [1].

Stretch Challenge · +25 XP, +10 coins

Two Slides in One

$y = x^2$ is first shifted 5 units right, then a second copy of the SAME shape is shifted 3 units left. (a) Write both equations. (b) Where do the two parabolas intersect? (Set them equal and solve.) (c) At that intersection point, what is the value of $y$?

Reveal solution

(a) $y_1 = (x - 5)^2$, $y_2 = (x + 3)^2$. (b) $(x - 5)^2 = (x + 3)^2 \Rightarrow x^2 - 10x + 25 = x^2 + 6x + 9 \Rightarrow -16x = -16 \Rightarrow x = 1$. (c) $y = (1 - 5)^2 = 16$. They meet at $(1, 16)$ — the midpoint of the two vertices $(5, 0)$ and $(-3, 0)$ horizontally, lifted to $y = 16$.

Quick Review

Equation

$y = (x - h)^2$ — sideways slide

Vertex

$(h, 0)$ — on the $x$-axis

Sign rule

Minus inside $\to$ right; plus inside $\to$ left

Axis

$x = h$ — vertical through the vertex

$y$-intercept

$(0, h^2)$ — always non-negative

Shape

Same as $y = x^2$ — no stretch, no flip

Your Badges

0 of 6

First Steps

3-Day Streak

3 in a Row

Lesson Ace

Stretch Seeker

Daily Warrior

Mark lesson as complete

Tick when you've finished Learn, Practice and the Stretch. Earns +85 XP and +25 coins.

Unit overview · Maths subject page