Lesson 8 ~25 min Unit 2 · Non-Linear +85 XP

Vertex and Axis of Symmetry

Vertex form $y = a(x - h)^2 + k$ packages everything: vertex $(h, k)$, axis $x = h$, direction from sign of $a$, width from $|a|$. Read it off — no plotting required.

Today's hook: Without plotting any points, can you give the vertex, axis of symmetry, and direction from $y = -3(x + 2)^2 + 5$?

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

L06 gave you $y = ax^2 + c$ — vertex always on the $y$-axis. L07 added a horizontal slide. What if you slide both ways? The equation $y = a(x - h)^2 + k$ packs a horizontal shift $h$ AND a vertical shift $k$ around the standard parabola. The vertex can land ANYWHERE in the plane. Where is the vertex of $y = 2(x - 1)^2 + 3$? Which way does it open?

Record your answer in your workbook.

The Big Idea

+5 XP

The vertex form $y = a(x - h)^2 + k$ shows you every feature of a parabola in one line. Read each parameter:

The red curve is $y = 2(x - 1)^2 + 3$: vertex $(1, 3)$, axis $x = 1$, opens up ($a > 0$), narrower ($|a| = 2 > 1$). The purple curve is $y = -\tfrac{1}{2}(x + 4)^2 - 5$: vertex $(-4, -5)$, axis $x = -4$, opens down ($a < 0$), wider ($|a| = \tfrac{1}{2} < 1$).

$y = a(x - h)^2 + k$ — vertex $(h, k)$, axis $x = h$

Vertex $(h, k)$

$h$ comes from the bracket; $k$ from the constant outside.

Axis $x = h$

Vertical line through the vertex's $x$-coordinate.

$a$ does shape

Sign $\to$ direction; magnitude $\to$ width.

What You'll Master

objectives

Know

Vertex form: $y = a(x - h)^2 + k$ with vertex $(h, k)$ and axis $x = h$
Direction from sign of $a$; width from $|a|$
Vertex is the minimum if $a > 0$, the maximum if $a < 0$

Understand

Why $h$ in $(x - h)$ has its sign flipped to give the vertex $x$-coordinate
Why $k$ outside the bracket keeps its sign
How vertex form combines a horizontal shift, a vertical shift, and a shape change

Can Do

Read vertex, axis, direction, and width from any $y = a(x - h)^2 + k$
Decide whether the vertex is a maximum or a minimum
Write a vertex-form equation given a vertex and an $a$-value

Words You Need

vocabulary

Vertex form$y = a(x - h)^2 + k$ — shape, vertex, and axis all visible at a glance.

Vertex$(h, k)$ — the lowest point (if $a > 0$) or highest point (if $a < 0$).

Axis of symmetryThe vertical line $x = h$ — mirror line of the parabola.

Minimum / maximumSmallest (or largest) value of $y$ on the whole curve. Equals $k$.

$h$ vs $k$$h$ comes from inside the bracket (sign flips). $k$ comes from outside (sign stays).

Narrow / wide$|a| > 1$: narrower than $y = x^2$. $|a| < 1$: wider.

Spot the Trap

heads-up

Wrong: Vertex of $y = 2(x - 1)^2 + 3$ is $(-1, 3)$ — just read the numbers.

Right: The $h$ inside flips sign. $(x - 1) = 0$ at $x = 1$. Vertex $(1, 3)$.

Wrong: Vertex of $y = -(x + 3)^2 - 7$ is $(3, 7)$.

Right: $(x + 3) = 0$ at $x = -3$; the $-7$ stays. Vertex $(-3, -7)$. $h$ flips, $k$ doesn't.

Read in Four Slots

+5 XP

Every vertex-form equation has exactly four bits of information. Read them like reading a form:

Sign of $a$: $+$ opens up (vertex = min), $-$ opens down (vertex = max).
$|a|$: $> 1$ narrower; $< 1$ wider; $= 1$ standard.
$h$: value that makes the bracket zero. Flip the inside sign.
$k$: the constant added at the end. Sign as written.

Vertex $(h, k)$. Axis $x = h$. Min/max value of $y$ is $k$.

$y = a(x - h)^2 + k$ → $(a$ shape$, h$ slot$, k$ slot$)$

Don't plot

For these questions you don't need any $y$-values — just read.

$k$ keeps its sign

$y = 3(x - 2)^2 - 6$: $k = -6$, NOT $+6$.

Watch coefficient

"$-(x + 3)^2$" means $a = -1$ — opens DOWN.

Writing the Equation from Features

+5 XP

Going the other way: given a vertex and an $a$-value, you can build a vertex-form equation.

Vertex $(4, -2)$, $a = 1$: equation $y = (x - 4)^2 - 2$.
Vertex $(-3, 5)$, $a = -2$: equation $y = -2(x + 3)^2 + 5$.
Vertex $(0, 7)$, $a = \tfrac{1}{2}$: equation $y = \tfrac{1}{2}x^2 + 7$ (the $h = 0$ bracket disappears).

Substitute $h$ and $k$ into the template, flipping the sign on $h$ inside the bracket.

Drop $h, k$ into $y = a(x - h)^2 + k$ — remember to flip $h$'s sign in the bracket.

$h$ flip

Vertex $x = 4 \Rightarrow$ bracket $(x - 4)$.

$k$ keeps sign

Vertex $y = -2 \Rightarrow$ "$- 2$" at the end.

$h = 0$? bracket vanishes

Vertex on the $y$-axis → no bracket needed.

Watch Me Solve It · 3 examples

Watch Me Solve It · Read $y = 2(x - 1)^2 + 3$

+15 XP per step

PROBLEM

For $y = 2(x - 1)^2 + 3$, state the vertex, the axis of symmetry, the direction, and the width.

1
Identify $a$, $h$, $k$
$a = 2$, $h = 1$ (bracket zero at $x = 1$), $k = 3$.
2
Vertex & axis
Vertex $(h, k) = (1, 3)$. Axis $x = 1$.
3
Direction & width
$a = 2 > 0$: opens up; $|a| = 2 > 1$: narrower than $y = x^2$. Min value $y = 3$.
All four features pulled straight from the equation — no plotting.

AnswerVertex $(1, 3)$, axis $x = 1$, opens up, narrower than $y = x^2$.

Watch Me Solve It · Read $y = -\tfrac{1}{2}(x + 4)^2 - 5$

+15 XP per step

PROBLEM

State the vertex, axis, direction and width of $y = -\tfrac{1}{2}(x + 4)^2 - 5$. Is the vertex a min or max?

1
Identify $a$, $h$, $k$
$a = -\tfrac{1}{2}$. Bracket $(x + 4) = (x - (-4))$ so $h = -4$. $k = -5$.
2
Vertex & axis
Vertex $(-4, -5)$. Axis $x = -4$.
3
Direction, width, type
$a < 0$: opens down; $|a| = \tfrac{1}{2} < 1$: wider; vertex is a MAXIMUM with value $-5$.
Both $h$ and $k$ are negative this time — watch each sign carefully.

AnswerVertex $(-4, -5)$, axis $x = -4$, opens down, wider; maximum.

Watch Me Solve It · Equation from features

+15 XP per step

PROBLEM

A parabola opens downward, has $|a| = 3$, and vertex $(5, -1)$. Write a vertex-form equation.

1
Sign of $a$
Opens down $\Rightarrow$ $a < 0$. With $|a| = 3$, take $a = -3$.
2
Build the bracket from $h$
Vertex $x = 5 \Rightarrow h = 5 \Rightarrow$ bracket $(x - 5)$.
3
Add $k$
Vertex $y = -1 \Rightarrow k = -1$. Equation: $y = -3(x - 5)^2 - 1$.
Always flip $h$'s sign inside; $k$ keeps its sign outside.

Answer$y = -3(x - 5)^2 - 1$.

Common Pitfalls

heads-up

Reading $h$ with its sign as written

Saying vertex of $y = (x - 6)^2 + 2$ is $(-6, 2)$.

Fix: Zero the bracket. $(x - 6) = 0 \Rightarrow x = 6$. Vertex $(6, 2)$.

Flipping $k$'s sign too

Saying vertex of $y = (x - 1)^2 + 4$ is $(1, -4)$.

Fix: $k$ is OUTSIDE the bracket. Its sign doesn't flip. Vertex is $(1, 4)$.

Missing the negative on $a = -1$

Reading $y = -(x - 2)^2 + 7$ as if it opens up.

Fix: "$-(x - 2)^2$" means $a = -1$. Curve opens DOWN; vertex $(2, 7)$ is a MAXIMUM.

Copy Into Your Books

Vertex form

$y = a(x - h)^2 + k$
Vertex $(h, k)$
Axis $x = h$

$a$ does shape

$a > 0$: opens up, min at $(h, k)$
$a < 0$: opens down, max at $(h, k)$
$|a| > 1$ narrow; $|a| < 1$ wide

Signs

$h$: flip from bracket
$k$: keep as written
Bracket zero $\to$ vertex $x$

Examples

$2(x-1)^2 + 3$: $(1, 3)$ min
$-(x+3)^2 + 1$: $(-3, 1)$ max
$-\tfrac{1}{2}(x+4)^2 - 5$: $(-4, -5)$ max

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Vertex Form

4 problems

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

1 State the vertex of $y = (x - 6)^2 + 2$.
$(x - 6) = 0$ at $x = 6$; $k = 2$.$(6, 2)$
2 State the axis of symmetry of $y = -2(x + 7)^2 - 1$.
Vertex $x = -7$.$x = -7$
3 Does $y = 4(x - 2)^2 - 9$ have a min or max? State its value.
$a = 4 > 0$, so MIN at vertex.Min $y = -9$
4 Write the equation of a parabola with $a = -1$ and vertex $(-2, 3)$.
Flip $h = -2$ into bracket: $(x + 2)$; $k = 3$.$y = -(x + 2)^2 + 3$

Complete in your workbook.

Quick Check · 5 questions

The vertex of $y = 2(x - 1)^2 + 3$ is:

+10 XP

The axis of symmetry of $y = -\tfrac{1}{2}(x + 4)^2 - 5$ is:

+10 XP

For $y = -3(x + 2)^2 + 5$, the vertex is a:

+10 XP

The equation of a parabola with vertex $(3, -7)$ and $a = 2$ is:

+10 XP

How does the graph of $y = 4(x + 1)^2 - 6$ compare to $y = x^2$?

+10 XP

Show Your Working · 3 questions

Show Your Working

9 marks total

ApplyMedium3 MARKS

Q6. For $y = -2(x - 3)^2 + 8$, state: (a) the vertex; (b) the axis of symmetry; (c) the direction the parabola opens. Briefly justify each.

Answer in your workbook.

ApplyMedium3 MARKS

Q7. Write a vertex-form equation for each parabola: (a) vertex $(-1, 4)$, $a = 1$; (b) vertex $(0, -3)$, $a = -2$; (c) vertex $(5, 0)$, $a = \tfrac{1}{2}$.

Answer in your workbook.

ReasonHard3 MARKS

Q8. A parabola has its MAXIMUM value of $y = 10$ at $x = -2$, and is wider than $y = x^2$. (a) State the vertex. (b) Choose a suitable $a$-value and justify why. (c) Write a possible equation in vertex form.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. C — vertex $(1, 3)$.

2. A — axis $x = -4$.

3. B — max at $y = 5$ ($a < 0$).

4. D — $y = 2(x - 3)^2 - 7$.

5. A — $|a| = 4 > 1$ narrower, $a > 0$ opens up.

Show Your Working Model Answers

Q6 (3 marks): (a) $(x - 3) = 0 \Rightarrow x = 3$; $k = 8$. Vertex $(3, 8)$ [1]. (b) Axis of symmetry $x = 3$ — vertical through the vertex [1]. (c) $a = -2 < 0$, so opens DOWN; vertex is a maximum at $y = 8$ [1].

Q7 (3 marks): (a) $y = (x + 1)^2 + 4$ — flip $h = -1$ inside the bracket [1]. (b) $y = -2x^2 - 3$ — $h = 0$ so no bracket needed [1]. (c) $y = \tfrac{1}{2}(x - 5)^2$ — $k = 0$ so no constant at end [1].

Q8 (3 marks): (a) Maximum at $y = 10$ when $x = -2$ $\Rightarrow$ vertex $(-2, 10)$ [1]. (b) Maximum $\Rightarrow a < 0$; wider $\Rightarrow |a| < 1$. Choose e.g. $a = -\tfrac{1}{2}$ [1]. (c) $y = -\tfrac{1}{2}(x + 2)^2 + 10$ [1].

Stretch Challenge · +25 XP, +10 coins

Find $a$ from a Second Point

A parabola in vertex form has vertex $(2, -1)$ and passes through $(4, 7)$. (a) Write the equation with $a$ unknown using the vertex. (b) Substitute the second point and solve for $a$. (c) State the final equation and say whether the vertex is a min or max.

Reveal solution

(a) $y = a(x - 2)^2 - 1$. (b) Sub $(4, 7)$: $7 = a(4 - 2)^2 - 1 = 4a - 1 \Rightarrow 4a = 8 \Rightarrow a = 2$. (c) $y = 2(x - 2)^2 - 1$. Since $a = 2 > 0$, the curve opens up and vertex $(2, -1)$ is a MINIMUM.

Quick Review

Form

$y = a(x - h)^2 + k$

Vertex

$(h, k)$ — flip $h$ from bracket

Axis

$x = h$

Direction

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

Width

$|a| > 1$ narrow, $< 1$ wide

Min/Max

$a > 0$: min $k$. $a < 0$: max $k$.

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