Lesson 5 ~25 min Unit 2 · Non-Linear +85 XP

Vertical Translations: $y = x^2 + c$

Add $c$ at the end of the equation and the whole parabola slides up or down. Shape stays identical — only the vertex moves.

Today's hook: $y = x^2 + 3$ vs $y = x^2$: same shape, just moved. Where is the new vertex? What's changed and what hasn't?

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Think First

warm-up

List the $y$-values of $y = x^2$ for $x = -2, -1, 0, 1, 2$. Now add $3$ to every one of those numbers and call the new list $y = x^2 + 3$. What pattern do you see in the new list? Where would the lowest point sit on a graph now?

Record your answer in your workbook.

The Big Idea

+5 XP

Adding a constant $c$ at the end of $y = x^2$ translates (slides) the whole parabola vertically. If $c > 0$ the graph moves UP by $c$. If $c < 0$ it moves DOWN by $|c|$. The shape stays identical — same width, same direction — but the vertex moves from $(0, 0)$ to $(0, c)$. The axis of symmetry is still $x = 0$.

The gold curve $y = x^2$ has vertex $(0, 0)$. The red curve $y = x^2 + 3$ is the same shape lifted $3$ units up — vertex now $(0, 3)$. The purple curve $y = x^2 - 2$ is shifted $2$ units down — vertex now $(0, -2)$. Every point moves vertically by the same amount.

$y = x^2 + c$ — vertex $(0, c)$, axis still $x = 0$, same width

$c$ shifts $y$

Every $y$-value goes up by $c$ (down if $c$ negative).

Shape doesn't change

Same width, same direction — only position changes.

Axis stays $x = 0$

Vertical shifts don't change the axis of symmetry.

What You'll Master

objectives

Know

$y = x^2 + c$ shifts $y = x^2$ vertically by $c$
Vertex moves from $(0, 0)$ to $(0, c)$
The axis of symmetry stays $x = 0$

Understand

Why adding $c$ moves $y$ but not $x$
Why the shape (width and direction) is unchanged
Why the $y$-intercept of $y = x^2 + c$ is just $c$

Can Do

Sketch $y = x^2 + 3$ and $y = x^2 - 2$ from a table
State the vertex of $y = x^2 + c$ by inspection
Find $x$-intercepts of $y = x^2 + c$ when they exist

Words You Need

vocabulary

TranslationA "slide" of the graph — here, vertical sliding by $c$ units.

Constant $c$The number added at the end of $y = x^2 + c$. Controls vertical position.

$y$-interceptWhere the curve crosses the $y$-axis. For $y = x^2 + c$, the $y$-intercept is $(0, c)$.

$x$-interceptWhere the curve crosses the $x$-axis — solve $x^2 + c = 0$.

Shift upWhole curve moves up by $c$ when $c > 0$.

Shift downWhole curve moves down by $|c|$ when $c < 0$.

Spot the Trap

heads-up

Wrong: "$y = x^2 + 3$ moves the curve $3$ units to the right." Adding $c$ at the end moves the graph vertically, not sideways.

Right: $y = x^2 + 3$ moves the curve $3$ units UP. Vertex jumps to $(0, 3)$.

Wrong: "$y = x^2 + 4$ has $x$-intercepts." Setting $x^2 + 4 = 0$ gives $x^2 = -4$, which has no real solutions.

Right: $y = x^2 + c$ has real $x$-intercepts only when $c \le 0$ (so the vertex sits at or below the $x$-axis).

Shift Tables

+5 XP

Build a table for $y = x^2$ and then just add $c$ to every $y$. The $x$ values stay exactly the same.

$x$	$-2$	$-1$	$0$	$1$	$2$
$y = x^2$	$4$	$1$	$0$	$1$	$4$
$y = x^2 + 3$	$7$	$4$	$3$	$4$	$7$
$y = x^2 - 2$	$2$	$-1$	$-2$	$-1$	$2$

Every row is the $y = x^2$ row $+ c$.

Vertex value

Vertex $y$ = lowest value in the table = $c$.

Range

For $y = x^2 + c$: $y \ge c$ always.

Mirror still works

Rows still read the same backwards — axis $x = 0$ unchanged.

Intercepts of $y = x^2 + c$

+5 XP

Two intercepts to find: the $y$-intercept (always easy) and the $x$-intercepts (sometimes don't exist).

$y$-intercept: sub $x = 0$: $y = 0^2 + c = c$. So always $(0, c)$.
$x$-intercepts: sub $y = 0$: $x^2 + c = 0 \Rightarrow x^2 = -c \Rightarrow x = \pm \sqrt{-c}$ — only real when $c \le 0$.

So $y = x^2 - 4$ has $x$-intercepts at $x = \pm 2$, but $y = x^2 + 4$ has none (the curve sits entirely above the $x$-axis).

$y$-int: $(0, c)$ — $x$-int: $\pm\sqrt{-c}$ (only if $c \le 0$)

$c > 0$: no $x$-int

Vertex sits above the $x$-axis; curve never crosses.

$c = 0$: just $(0,0)$

Vertex on the $x$-axis — the curve touches at the vertex.

$c < 0$: two $x$-int

$x = \pm\sqrt{-c}$. Symmetric across the $y$-axis.

Watch Me Solve It · 3 examples

Watch Me Solve It · Read the vertex

+15 XP per step

PROBLEM

State the vertex, axis of symmetry, and direction of opening for $y = x^2 + 5$.

1
Read $c$
$c = 5$, so the curve shifts up by $5$.
2
Vertex
$(0, c) = (0, 5)$
3
Axis and direction
Axis: $x = 0$. Coefficient of $x^2$ is $+1$, so opens up.

AnswerVertex $(0, 5)$; axis $x = 0$; opens up.

Watch Me Solve It · $x$-intercepts

+15 XP per step

PROBLEM

Find the $x$-intercepts of $y = x^2 - 9$.

1
Set $y = 0$
$0 = x^2 - 9$
2
Rearrange
$x^2 = 9$
3
Take both square roots
$x = \pm 3$
Don't forget the negative root!

Answer$x$-intercepts at $(3, 0)$ and $(-3, 0)$.

Watch Me Solve It · Equation from vertex

+15 XP per step

PROBLEM

A parabola has the same shape as $y = x^2$, opens upward, and has vertex $(0, -4)$. Write its equation.

1
Identify the form
Same shape as $y = x^2$ $\Rightarrow$ $y = x^2 + c$.
2
Read $c$ from vertex
Vertex is $(0, c)$, so $c = -4$.
3
Write the equation
$y = x^2 - 4$

Answer$y = x^2 - 4$.

Common Pitfalls

heads-up

Confusing $+c$ with sideways shift

Some students claim $y = x^2 + 3$ moves "right by 3".

Fix: $+c$ at the END adds to $y$, so the curve moves VERTICALLY. Horizontal shifts use $(x - h)^2$.

Dropping the negative root

Solving $x^2 = 9$ as $x = 3$ only, missing $x = -3$.

Fix: Always write $x = \pm \sqrt{\text{value}}$ — parabolas are symmetric.

"Vertex is $c$"

Writing the vertex of $y = x^2 + 5$ as just "$5$" instead of the point $(0, 5)$.

Fix: Always give vertices as ordered pairs $(x, y)$. The vertex is $(0, c)$.

Copy Into Your Books

$y = x^2 + c$

Vertical translation of $y = x^2$
Vertex $(0, c)$
Axis $x = 0$ still

Shape

Width unchanged
Direction unchanged
Range $y \ge c$

$y$-intercept

Always $(0, c)$
$c$ is the vertex height

$x$-intercepts

$x = \pm \sqrt{-c}$
Real only when $c \le 0$

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Vertical Translations

4 problems

Four quick problems on $y = x^2 + c$.

1 State the vertex of $y = x^2 + 7$.
Read $c = 7$ off the equation.$(0, 7)$
2 Find $y$ when $x = 4$ on $y = x^2 - 5$.
$y = 16 - 5$.$y = 11$
3 $y$-intercept of $y = x^2 - 6$?
Sub $x = 0$.$(0, -6)$
4 $x$-intercepts of $y = x^2 - 16$?
$x^2 = 16$, so $x = \pm 4$.$(4, 0)$ and $(-4, 0)$

Complete in your workbook.

Quick Check · 5 questions

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

+10 XP

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

+10 XP

The $x$-intercepts of $y = x^2 - 25$ are:

+10 XP

How many $x$-intercepts does $y = x^2 + 4$ have?

+10 XP

A parabola has the same shape as $y = x^2$, opens up, with vertex $(0, -3)$. Its equation is:

+10 XP

Show Your Working · 3 questions

Show Your Working

9 marks total

ApplyMedium3 MARKS

Q6. On the same axes, sketch $y = x^2$ and $y = x^2 + 2$ for $x$ from $-3$ to $3$. Label the vertex of each curve.

Answer in your workbook.

ApplyMedium3 MARKS

Q7. For $y = x^2 - 16$, find: (a) the vertex, (b) the $y$-intercept, (c) the $x$-intercepts (show working).

Answer in your workbook.

ReasonHard3 MARKS

Q8. A student writes: "$y = x^2 + 5$ has $x$-intercepts at $x = \pm\sqrt{5}$." Decide if this is correct, justify your answer with working, and state the correct number of $x$-intercepts with a reason.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. A — vertex $(0, c) = (0, 5)$.

2. C — $9 - 2 = 7$.

3. B — $x^2 = 25 \Rightarrow x = \pm 5$.

4. D — no real solutions when $c > 0$.

5. C — $c = -3$ gives $y = x^2 - 3$.

Show Your Working Model Answers

Q6 (3 marks): Table for $y = x^2$: $9, 4, 1, 0, 1, 4, 9$; table for $y = x^2 + 2$: $11, 6, 3, 2, 3, 6, 11$ [1]. Both curves sketched as smooth U-shapes, same width, the second one $2$ units above [1]. Vertices labelled at $(0, 0)$ and $(0, 2)$ [1].

Q7 (3 marks): (a) Vertex $(0, -16)$ [1]. (b) $y$-intercept $(0, -16)$ [1]. (c) $x^2 - 16 = 0 \Rightarrow x^2 = 16 \Rightarrow x = \pm 4$, so $(4, 0)$ and $(-4, 0)$ [1].

Q8 (3 marks): Incorrect [1]. Setting $y = 0$: $0 = x^2 + 5 \Rightarrow x^2 = -5$. No real number squared is negative, so this has no real solutions [1]. The graph has $0$ $x$-intercepts because the vertex $(0, 5)$ sits above the $x$-axis and the curve opens upward, never crossing it [1].

Stretch Challenge · +25 XP, +10 coins

Move the Vertex

The graph of $y = x^2$ is translated so its new vertex is $(0, -7)$. (a) Write the equation of the translated curve. (b) Find any $x$-intercepts. (c) For what value of $x$ is $y = 9$ on the translated curve?

Reveal solution

(a) Vertex $(0, c)$ with $c = -7$, so $y = x^2 - 7$. (b) $x^2 - 7 = 0 \Rightarrow x^2 = 7 \Rightarrow x = \pm \sqrt{7}$. (c) $9 = x^2 - 7 \Rightarrow x^2 = 16 \Rightarrow x = \pm 4$.

Quick Review

Equation

$y = x^2 + c$ — vertical translation

Vertex

$(0, c)$ — moves up/down by $c$

Axis

$x = 0$ — unchanged

Shape

Same width and direction as $y = x^2$

$y$-intercept

$(0, c)$ — same as the vertex

$x$-intercepts

$x = \pm\sqrt{-c}$ (real only if $c \le 0$)

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