Lesson 20 ~30 min Unit 2 · Synthesis +100 XP

Unit Synthesis, Non-Linear Review

Twenty lessons in one map. Parabolas, circles, hyperbolas, exponentials, match the shape, name the equation, find the features, solve the question.

Today's hook: "What type of curve is this?" Given $y = \tfrac{6}{x}$, $x^2 + y^2 = 25$, $y = 2^x$ and $y = (x - 3)^2 + 1$, can you name the shape and one key feature of each in under a minute?

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Think First

warm-up

Four equations: $y = (x - 3)^2 + 1$, $y = \tfrac{6}{x}$, $x^2 + y^2 = 25$, $y = 2^x$. For each, write (i) the name of the curve, (ii) ONE key feature you'd label on a sketch. Now do the reverse: I show you a U-shape with vertex $(0, -4)$ opening up, what equation am I?

Record your answer in your workbook.

The Big Idea, One Map for Twenty Lessons

+5 XP

Every non-linear relationship in this unit can be identified by its equation form, sketched by its key features, and analysed by the same toolkit: identify shape $\to$ extract features $\to$ sketch or solve.

Four families. Each has its own signature equation and its own pair of key features. The parabola is the headline act, vertex form, $x$-intercepts, axis of symmetry, transformations, and solving by factoring all live here. Circles, hyperbolas and exponentials round out the family.

Shape $\to$ features $\to$ sketch / solve.

Equation shape

$x^2 \to$ parabola, $\tfrac{k}{x} \to$ hyperbola, $a^x \to$ exponential, $x^2 + y^2 \to$ circle.

Label features

Vertex, intercepts, asymptotes, centre, radius, whichever apply.

Reverse it

Given the sketch, name the shape and write the equation.

Master Summary · Four Families

cheat sheet

Parabola (L01-L13, L19)$y = a(x - h)^2 + k$ or $y = ax^2 + bx + c$. Features: vertex $(h, k)$, axis $x = h$, $y$-int (sub $x = 0$), $x$-ints (set $y = 0$, factor). $a > 0$ up, $a < 0$ down; $|a|$ controls width.

Circle (L14-L15)$x^2 + y^2 = r^2$ (centre origin) or $(x - h)^2 + (y - k)^2 = r^2$. Features: centre $(h, k)$, radius $r$. Not a function (fails vertical line test).

Hyperbola (L16-L17)$y = \tfrac{k}{x}$. Two branches; asymptotes are the $x$- and $y$-axes. $k > 0$: branches in Q1 and Q3. $k < 0$: branches in Q2 and Q4.

Exponential (L18)$y = a^x$ with $a > 0$, $a \neq 1$. $y$-intercept $(0, 1)$. Asymptote: $x$-axis ($y = 0$). $a > 1$ growth; $0 < a < 1$ decay.

Vertex-form transformations$y = a(x - h)^2 + k$: shift right $h$, up $k$, stretch by $a$, flip if $a < 0$.

Solving $ax^2 + bx + c = 0$Factor (trinomial or DOTS), apply null factor law. Roots are the $x$-intercepts of $y = ax^2 + bx + c$.

What You'll Master

objectives

Know

The four key non-linear families and their standard equations
The key feature checklist for each shape
The link between solving $y = 0$ and finding $x$-intercepts

Understand

Why the equation form tells you the shape on sight
Why transformations of $y = x^2$ produce every parabola
Why asymptotes mark "not allowed" $x$ or $y$ values

Can Do

Match any of the four equations to its graph
Sketch any vertex-form parabola from scratch
Solve a quadratic by factoring and read the $x$-intercepts off
Identify any non-linear relationship from a description or sketch

Matching: Equation $\leftrightarrow$ Graph

+5 XP

How to identify a curve from its equation in three seconds:

See $x^2$ only (or $(x - h)^2$): parabola. Look for $a$ to know direction.
See $x^2 + y^2$ on its own line $= r^2$: circle, centre origin, radius $r$.
See $x$ in the denominator: hyperbola. Asymptotes on the axes.
See $x$ in the exponent: exponential. $y$-int $(0, 1)$, asymptote $y = 0$.
See $y = mx + b$ (no powers, no fractions): linear, not a non-linear relationship.

Equation form $\to$ curve family in $\leq 3$ seconds.

Look at the powers

Power of $x$ tells you the family. Power in the denominator or exponent matters too.

Then read features

Once you know the shape, read centre / vertex / asymptote off the equation.

Sketch quickly

A rough labelled sketch beats a perfect plot, mark the features that earn marks.

The Parabola Toolkit Revisited

+5 XP

Parabolas are 60% of this unit. Make sure these five sub-skills are automatic:

Vertex-form features: $y = a(x - h)^2 + k$ has vertex $(h, k)$, axis $x = h$, $y$-int sub $x = 0$.
$x$-intercepts: set $y = 0$, isolate $(x - h)^2$, take $\pm$ square root.
Standard form $\to$ factored: $x^2 + bx + c = (x + p)(x + q)$ with $pq = c$, $p + q = b$.
Solving: roots of $ax^2 + bx + c = 0$ are the $x$-intercepts.
Identify (reverse): vertex from sketch + one other point $\to$ solve for $a$ $\to$ equation.

Vertex form $\leftrightarrow$ factored form $\leftrightarrow$ sketch.

Vertex form for sketching

Features pop out instantly.

Factored form for roots

Set product $= 0$, read off $x$-ints.

Move between forms

Expand, complete the square, same parabola, different lens.

Watch Me Solve It · 3 mixed examples

Watch Me Solve It · Identify the curve

+15 XP per step

PROBLEM

For each equation, name the curve and state ONE key feature: (a) $y = -(x - 4)^2 + 9$, (b) $x^2 + y^2 = 36$, (c) $y = \tfrac{-2}{x}$, (d) $y = 3^x$.

1
(a) and (b)
(a) Parabola, vertex $(4, 9)$, opens down ($a = -1$). (b) Circle, centre $(0, 0)$, radius $\sqrt{36} = 6$.
2
(c) and (d)
(c) Hyperbola, $k = -2 < 0$, branches in Q2 and Q4, asymptotes the axes. (d) Exponential growth ($a = 3 > 1$), $y$-int $(0, 1)$, asymptote $y = 0$.
3
Sanity check shapes
Parabola = U; circle = closed loop; hyperbola = two branches; exponential = J-curve.
Match equation form to shape before sketching.

Answer(a) parabola; (b) circle $r = 6$; (c) hyperbola Q2/Q4; (d) exponential growth.

Watch Me Solve It · Sketch & solve a parabola

+15 XP per step

PROBLEM

Given $y = x^2 - 4x - 5$: (a) find the $x$-intercepts by factoring, (b) find the $y$-intercept, (c) find the axis of symmetry and vertex.

1
$x$-intercepts
$x^2 - 4x - 5 = 0$. $pq = -5$, $p + q = -4$: pick $-5, 1$. $(x - 5)(x + 1) = 0 \Rightarrow x = 5$ or $x = -1$.
2
$y$-intercept
Sub $x = 0$: $y = 0 - 0 - 5 = -5$. Point $(0, -5)$.
3
Axis & vertex
Axis: midpoint of $-1$ and $5$ is $2$. So $x = 2$. Vertex $y$: sub $x = 2$ → $y = 4 - 8 - 5 = -9$. Vertex $(2, -9)$.
Vertex is a MIN (because $a = 1 > 0$).

Answer$x$-ints $(-1, 0), (5, 0)$; $y$-int $(0, -5)$; axis $x = 2$; vertex $(2, -9)$.

Watch Me Solve It · Identify from a description

+15 XP per step

PROBLEM

A curve is described as: passing through $(0, 1)$, increasing through Q1, with the $x$-axis as a horizontal asymptote. Name the type and write a possible equation.

1
Spot the asymptote clue
Horizontal asymptote $y = 0$ + $y$-int $(0, 1)$: classic exponential signature.
2
Increasing means growth
$y = a^x$ with $a > 1$ is exponential growth, increasing.
3
Write a possible equation
$y = 2^x$ (or $y = 3^x$, etc.). Any base $> 1$ works.
Many equations match the same broad description, only a second point would pin one down.

AnswerExponential growth, e.g. $y = 2^x$.

Common Pitfalls Across the Unit

Common Pitfalls

heads-up

Sign of $h$ in vertex form

Reading $y = (x + 3)^2 - 2$ as vertex $(3, -2)$. The sign flips: $h = -3$.

Fix: $(x - h)^2$ means subtract $h$. $(x + 3) = (x - (-3))$, so $h = -3$. Vertex $(-3, -2)$.

Confusing circle with parabola

Treating $x^2 + y^2 = 25$ as if it were a parabola.

Fix: If BOTH $x^2$ and $y^2$ appear, it's a circle. If only $x^2$ (or only $y^2$), it's a parabola.

Forgetting null factor law needs zero

Writing $(x - 2)(x + 5) = 6$ then $x - 2 = 6$ and $x + 5 = 6$.

Fix: Expand and rearrange to $= 0$ first. Always.

Exponential is NOT a parabola

Treating $y = 2^x$ as if it had a vertex.

Fix: Exponentials have NO vertex and NO $x$-intercept. They approach $y = 0$ asymptotically.

Copy Into Your Books

Parabola

$y = a(x - h)^2 + k$
Vertex $(h, k)$, axis $x = h$
Solve $= 0$ to find $x$-ints

Circle

$x^2 + y^2 = r^2$ (origin)
$(x - h)^2 + (y - k)^2 = r^2$
Centre $(h, k)$, radius $r$

Hyperbola

$y = \tfrac{k}{x}$
Asymptotes: both axes
$k > 0$: Q1, Q3; $k < 0$: Q2, Q4

Exponential

$y = a^x$, $a > 0$, $a \neq 1$
$y$-int $(0, 1)$, asymp. $y = 0$
$a > 1$: growth; $0 < a < 1$: decay

How are you completing this lesson?

Brain Trainer · 4 mixed problems

Brain Trainer · Identify the Relationship

4 problems

Four quick identifications spanning all four families.

1 Name the curve $(x - 2)^2 + (y + 1)^2 = 16$ and give its centre and radius.
Both $x^2$ and $y^2$, circle.Centre $(2, -1)$, radius $4$
2 Solve $x^2 - 7x + 10 = 0$ and state the $x$-intercepts of $y = x^2 - 7x + 10$.
$(x - 2)(x - 5) = 0$.$x = 2$ or $5$; ints $(2, 0)$ and $(5, 0)$
3 State the vertex and direction of $y = 2(x + 1)^2 - 8$.
$a = 2 > 0$, opens up.Vertex $(-1, -8)$, MIN
4 For $y = \tfrac{4}{x}$, name the shape and which quadrants contain its branches.
$x$ in denominator, hyperbola. $k = 4 > 0$.Branches in Q1 and Q3

Complete in your workbook.

Quick Check · 5 mixed questions

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

+10 XP

$x^2 + y^2 = 49$ describes:

+10 XP

The $x$-intercepts of $y = x^2 + x - 12$ are:

+10 XP

All exponentials $y = a^x$ (with $a > 0$, $a \neq 1$) share which point?

+10 XP

The asymptotes of $y = \tfrac{6}{x}$ are:

+10 XP

Show Your Working · 3 questions

Show Your Working

9 marks total

ApplyMedium3 MARKS

Q6. For each equation, name the curve type and state ONE labelled feature: (a) $y = (x - 1)^2 - 9$, (b) $x^2 + y^2 = 100$, (c) $y = 5^x$.

Answer in your workbook.

ApplyMedium3 MARKS

Q7. Given $y = x^2 - 6x + 8$. (a) Solve $x^2 - 6x + 8 = 0$ by factoring. (b) State the $x$- and $y$-intercepts of the parabola. (c) Find the axis of symmetry and vertex.

Answer in your workbook.

ReasonHard3 MARKS

Q8. A parabola has vertex $(1, -4)$ and one $x$-intercept at $(3, 0)$. (a) Use the symmetry of the parabola to find the other $x$-intercept. (b) Substitute $(3, 0)$ into $y = a(x - 1)^2 - 4$ to find $a$. (c) Hence write the equation in vertex form AND find the equation in expanded form $y = ax^2 + bx + c$.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. C$(x + 2)^2 = (x - (-2))^2$, vertex $(-2, 5)$.

2. B circle, centre origin, $r = 7$.

3. A$(x - 3)(x + 4) = 0 \Rightarrow x = 3$ or $-4$.

4. D$a^0 = 1$, so $(0, 1)$ on all exponentials.

5. B$y = \tfrac{k}{x}$ has the axes as asymptotes.

Show Your Working Model Answers

Q6 (3 marks): (a) Parabola, vertex $(1, -9)$, opens up [1]. (b) Circle, centre $(0, 0)$, radius $10$ [1]. (c) Exponential growth, $y$-intercept $(0, 1)$, asymptote $y = 0$ [1].

Q7 (3 marks): (a) $pq = 8$, $p + q = -6$: $-2, -4$. $(x - 2)(x - 4) = 0 \Rightarrow x = 2$ or $x = 4$ [1]. (b) $x$-ints $(2, 0)$ and $(4, 0)$; $y$-int $(0, 8)$ [1]. (c) Axis: midpoint of $2$ and $4$ is $3$, so $x = 3$. Vertex: sub $x = 3$ → $y = 9 - 18 + 8 = -1$. Vertex $(3, -1)$ [1].

Q8 (3 marks): (a) Axis $x = 1$. $(3, 0)$ is $2$ right of the axis, so the other intercept is $2$ left: $(-1, 0)$ [1]. (b) $0 = a(3 - 1)^2 - 4 = 4a - 4 \Rightarrow a = 1$ [1]. (c) $y = (x - 1)^2 - 4$. Expand: $y = x^2 - 2x + 1 - 4 = x^2 - 2x - 3$ [1].

Stretch Challenge · +30 XP, +15 coins

What Type Of Relationship Is This?

Five clues. For each, name the curve and write a possible equation. (i) Passes through $(0, -16)$, $x$-intercepts at $\pm 4$. (ii) Closed loop, passes through $(0, 5)$ and $(5, 0)$, centred at origin. (iii) Approaches but never touches $y = 0$, passing through $(0, 1)$ and $(1, 4)$. (iv) Two branches in Q2 and Q4, passing through $(2, -3)$. (v) Parabola with axis $x = -2$, $y$-intercept $(0, -3)$, opening down with $a = -1$.

Reveal solution

(i) Parabola. Roots $\pm 4$, $y$-int $-16$: $y = x^2 - 16$. (ii) Circle, centre $(0, 0)$, radius $5$: $x^2 + y^2 = 25$. (iii) Exponential. $(0, 1)$ always works; $(1, 4)$ gives base $4$: $y = 4^x$. (iv) Hyperbola $y = \tfrac{k}{x}$, $k < 0$. Sub $(2, -3)$: $-3 = \tfrac{k}{2} \Rightarrow k = -6$. So $y = \tfrac{-6}{x}$. (v) $y = -(x + 2)^2 + k$. Sub $(0, -3)$: $-3 = -(2)^2 + k = -4 + k \Rightarrow k = 1$. So $y = -(x + 2)^2 + 1$.

Unit Quick Review

Parabola

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

Circle

$x^2 + y^2 = r^2$

Hyperbola

$y = \tfrac{k}{x}$

Exponential

$y = a^x$

Solve

Factor + null factor law

Roots

$=$ $x$-intercepts

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