Mathematics • Year 9 • Unit 2 • Lesson 20

Unit Synthesis — Final Mixed Challenge

Six unit-spanning problems: identify, sketch, solve, reverse-engineer. One classic "all $x^2$ are parabolas" misclassification to catch. One open-ended challenge: invent a curve from clues, the way HSC sketch-from-features questions test you.

Master · Mixed Challenge

1. Mixed problems

Pull together every tool from Unit 2: identify, extract features, solve, link equation to graph. 3 marks each

1.1 For each equation, name the curve and state ONE labelled feature: (a) $y = (x - 1)^2 - 9$, (b) $x^2 + y^2 = 100$, (c) $y = 5^x$, (d) $y = \dfrac{-8}{x}$, (e) $y = 3$ (constant line).

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

1.3 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 in expanded $y = ax^2 + bx + c$ form.

1.4 Build a small comparison: at which integer $x$-value(s) in $\{1, 2, 3, 4\}$ does $y = x^2$ equal $y = 2^x$? Tabulate both and identify the matches. Which curve grows faster from $x = 5$ onwards?

1.5 A circle has equation $(x - 3)^2 + (y + 2)^2 = 25$. (a) State the centre and radius. (b) Verify that $(3, 3)$ lies on the circle. (c) Is the point $(8, -2)$ on the circle? Justify.

1.6 Sort each equation into the correct family (parabola / circle / hyperbola / exponential / linear) and state ONE key feature: (i) $y = x^2 - 9$, (ii) $y = 2x + 5$, (iii) $y = \tfrac{6}{x}$, (iv) $x^2 + y^2 = 1$, (v) $y = (\tfrac{1}{3})^x$, (vi) $y = -(x + 2)^2 + 1$.

Stuck on 1.6? Make a 6-row table: equation | family | one feature.

2. Find the mistake

A student tried to classify several equations from across the unit. Exactly two of their answers are wrong. Spot them, explain why, and write the corrected classification. 3 marks

Student's classifications:

A: $y = (x - 2)^2 + 3$ → parabola, vertex $(2, 3)$ ✓

B: $x^2 + y^2 = 16$ → parabola (because it has $x^2$)

C: $y = \dfrac{5}{x}$ → hyperbola, asymptotes the axes ✓

D: $y = 4^x$ → parabola (because it has a power)

E: $y = -2x + 7$ → linear, gradient $-2$ ✓

(a) Which two classifications are wrong?

(b) For each wrong one, explain in one sentence why the student's reasoning is mistaken.

(c) Write out the corrected classification (with family name and one feature) for each wrong one.

Stuck? Lesson § "Common Pitfalls": "Confusing circle with parabola" and "Exponential is NOT a parabola" are both flagged.

3. Open-ended challenge — sketch from clues

This question has many valid answers. 4 marks

3.1 Five descriptions. For each: (i) name the curve, and (ii) write a possible equation that matches.

(a) Passes through $(0, -16)$, with $x$-intercepts at $\pm 4$.
(b) Closed loop, passes through $(0, 5)$ and $(5, 0)$, centred at the origin.
(c) Approaches but never touches $y = 0$, passes through $(0, 1)$ and $(1, 4)$.
(d) Two branches in Q2 and Q4, passes through $(2, -3)$.
(e) Parabola with axis of symmetry $x = -2$, $y$-intercept $(0, -3)$, opens DOWN with $a = -1$.

Stuck? (a) parabola through $\pm 4$ with $y$-int $-16$: $y = x^2 - 16$. (b) circle radius $5$. (c) exponential $y = 4^x$. (d) hyperbola $y = -6/x$. (e) $y = -(x + 2)^2 + k$, find $k$.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — Five identifications

(a) Parabola, vertex $(1, -9)$ MIN.
(b) Circle, centre $(0, 0)$, radius $10$.
(c) Exponential growth, $y$-intercept $(0, 1)$, asymptote $y = 0$.
(d) Hyperbola, $k = -8 < 0 \Rightarrow$ branches in Q2 and Q4, asymptotes the axes.
(e) Linear (horizontal line), $y$-intercept $(0, 3)$, gradient $0$.

1.2 — Parabola $y = x^2 - 6x + 8$

(a) Product $8$, sum $-6$: try $-2, -4$. $(x - 2)(x - 4) = 0 \Rightarrow x = 2$ or $x = 4$.
(b) $x$-intercepts: $(2, 0)$ and $(4, 0)$. $y$-intercept: $y = 0 - 0 + 8 = 8$, so $(0, 8)$.
(c) Axis: midpoint of $2$ and $4$ is $3$, so $x = 3$. Vertex $y$: $9 - 18 + 8 = -1$. Vertex $(3, -1)$.

1.3 — Reverse-engineer a parabola

(a) Axis $x = 1$. $(3, 0)$ is $2$ right of the axis; reflection gives the other intercept at $1 - 2 = -1$. So other $x$-intercept: $(-1, 0)$.
(b) Sub $(3, 0)$ into $y = a(x - 1)^2 - 4$: $0 = a(3 - 1)^2 - 4 = 4a - 4 \Rightarrow a = 1$.
(c) Vertex form: $y = (x - 1)^2 - 4$. Expanded: $y = x^2 - 2x + 1 - 4 = x^2 - 2x - 3$.

1.4 — $y = x^2$ vs $y = 2^x$

$x = 1$: $x^2 = 1$, $2^x = 2$ (different). $x = 2$: $x^2 = 4$, $2^x = 4$ (MATCH ✓). $x = 3$: $x^2 = 9$, $2^x = 8$ (different — parabola is bigger). $x = 4$: $x^2 = 16$, $2^x = 16$ (MATCH ✓). From $x = 5$ onwards, the exponential grows much faster: at $x = 5$, $x^2 = 25$ vs $2^x = 32$; at $x = 10$, $x^2 = 100$ vs $2^x = 1024$. Exponential growth eventually outpaces polynomial growth.

1.5 — Circle $(x - 3)^2 + (y + 2)^2 = 25$

(a) Centre $(3, -2)$ (sign flip on $h$ and $k$). Radius $\sqrt{25} = 5$.
(b) Sub $(3, 3)$: $(3 - 3)^2 + (3 + 2)^2 = 0 + 25 = 25$ ✓. Yes, $(3, 3)$ is on the circle.
(c) Sub $(8, -2)$: $(8 - 3)^2 + (-2 + 2)^2 = 25 + 0 = 25$ ✓. Yes, $(8, -2)$ is also on the circle.

1.6 — Sort and feature

(i) $y = x^2 - 9$: parabola, $x$-intercepts $\pm 3$ (or vertex $(0, -9)$).
(ii) $y = 2x + 5$: linear, gradient $2$, $y$-int $(0, 5)$.
(iii) $y = \tfrac{6}{x}$: hyperbola, branches Q1 and Q3.
(iv) $x^2 + y^2 = 1$: circle, centre $(0, 0)$, radius $1$.
(v) $y = (\tfrac{1}{3})^x$: exponential decay (base $\tfrac{1}{3} < 1$), $y$-int $(0, 1)$.
(vi) $y = -(x + 2)^2 + 1$: parabola, vertex $(-2, 1)$ MAX (opens down).

2 — Find the mistake (two errors: B and D)

(a) The wrong classifications are B and D.
(b) B: "$x^2 + y^2 = 16$ is a parabola because it has $x^2$" — wrong because parabolas have only ONE squared variable. When BOTH $x^2$ AND $y^2$ appear together (with a $+r^2$ on the other side), it's a CIRCLE, not a parabola. D: "$y = 4^x$ is a parabola because it has a power" — wrong because the power is on the BASE ($4$), not on $x$. Parabolas come from $x^2$ (power on $x$); when $x$ is the EXPONENT, the family is EXPONENTIAL.
(c) B: Circle, centre $(0, 0)$, radius $\sqrt{16} = 4$. D: Exponential growth (base $4 > 1$), $y$-intercept $(0, 1)$, horizontal asymptote $y = 0$.

3 — Open-ended challenge (sample solutions)

(a) Parabola. Roots $\pm 4$, $y$-int $-16$: $y = x^2 - 16$. (Check: $y(0) = -16$ ✓; roots $\pm 4$ ✓.)

(b) Circle, centre $(0, 0)$, radius $5$: $x^2 + y^2 = 25$. (Check: $(0)^2 + (5)^2 = 25$ ✓; $(5)^2 + (0)^2 = 25$ ✓.)

(d) Hyperbola $y = \dfrac{k}{x}$ with $k < 0$ (Q2/Q4). Sub $(2, -3)$: $-3 = \dfrac{k}{2} \Rightarrow k = -6$. So $y = \dfrac{-6}{x}$.

(e) Parabola with axis $x = -2$, opens down with $a = -1$: $y = -(x + 2)^2 + k$. Sub $(0, -3)$: $-3 = -(2)^2 + k = -4 + k \Rightarrow k = 1$. So $y = -(x + 2)^2 + 1$.

Marking: 1 mark per pair (name + equation, with substitution check) for any four out of five. Award full marks for all five correctly identified with a valid equation each.