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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?
0/5QUESTS
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.
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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.
2
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 $y = ax^2 + bx + c$.
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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
4
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.
5
The Parabola Toolkit Revisited
+5 XP

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

  1. Vertex-form features: $y = a(x - h)^2 + k$ has vertex $(h, k)$, axis $x = h$, $y$-int sub $x = 0$.
  2. $x$-intercepts: set $y = 0$, isolate $(x - h)^2$, take $\pm$ square root.
  3. Standard form $\to$ factored: $x^2 + bx + c = (x + p)(x + q)$ with $pq = c$, $p + q = b$.
  4. Solving: roots of $ax^2 + bx + c = 0$ are the $x$-intercepts.
  5. 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 · Identify the curve
+15 XP per step
Q1
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. 1
    (a) and (b)
    (a) Parabola, vertex $(4, 9)$, opens down ($a = -1$). (b) Circle, centre $(0, 0)$, radius $\sqrt{36} = 6$.
  2. 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. 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
Q2
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. 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. 2
    $y$-intercept
    Sub $x = 0$: $y = 0 - 0 - 5 = -5$. Point $(0, -5)$.
  3. 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
Q3
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. 1
    Spot the asymptote clue
    Horizontal asymptote $y = 0$ + $y$-int $(0, 1)$: classic exponential signature.
  2. 2
    Increasing means growth
    $y = a^x$ with $a > 1$ is exponential growth, increasing.
  3. 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. $y = 2^x$.
8
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?

D
Brain Trainer · Identify the Relationship
4 problems

Four quick identifications spanning all four families.

  1. 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. 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. 3 State the vertex and direction of $y = 2(x + 1)^2 - 8$.

    $a = 2 > 0$ — opens up.Vertex $(-1, -8)$, MIN
  4. 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.
1
The vertex of $y = (x + 2)^2 + 5$ is:
+10 XP
2
$x^2 + y^2 = 49$ describes:
+10 XP
3
The $x$-intercepts of $y = x^2 + x - 12$ are:
+10 XP
4
All exponentials $y = a^x$ (with $a > 0$, $a \neq 1$) share which point?
+10 XP
5
The asymptotes of $y = \tfrac{6}{x}$ are:
+10 XP
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 $y = -(x + 2)^2 + 1$.

R
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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