Mathematics • Year 9 • Unit 2 • Lesson 3
Dilations — Mixed Challenge
Pull together every $y = ax^2$ idea from Lesson 3: evaluation, width comparison, finding $a$ from a point, and combining with the basic parabola. Catch a tricky order-of-operations mistake, then take on an open-ended challenge.
1. Mixed problems
Each question uses a different combination of the dilation ideas from Lesson 3 (and the basic parabola from Lesson 2). Show your working. 3 marks each
1.1 For $y = 2x^2$, build a table of values for $x = -3, -2, -1, 0, 1, 2, 3$ and state the vertex.
1.2 Find $a$ for each parabola $y = ax^2$ passing through the given point. State whether each is wider or narrower than $y = x^2$. (a) $(3, 18)$ (b) $(2, 1)$ (c) $(5, 5)$.
1.3 Two parabolas have equations $y = 3x^2$ and $y = \tfrac{1}{3}x^2$. Find both their $y$-values at $x = 6$, and state which curve is steeper at that point.
1.4 Solve $5x^2 = 80$. Show both roots.
1.5 Decide whether each statement about $y = ax^2$ (with $a > 0$) is TRUE or FALSE, with a one-line justification: (a) the vertex moves up as $a$ increases; (b) the axis of symmetry is always $x = 0$; (c) doubling $a$ doubles every $y$-value; (d) when $a = \tfrac{1}{2}$, the curve sits below the $x$-axis.
1.6 The parabola $y = ax^2$ passes through both $(2, 12)$ and $(-2, 12)$. Show that ONLY ONE value of $a$ is needed (because of symmetry), and find it.
2. Find the mistake
Another student has tried to compute $y$ for $y = 3x^2$ at $x = 4$. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why, then re-do the working correctly. 3 marks
Student's working — compute $y$ for $y = 3x^2$ at $x = 4$:
Line 1: $y = 3 \times 4^2$
Line 2: $y = (3 \times 4)^2$
Line 3: $y = 12^2$
Line 4: $y = 144$
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong (refer to order of operations).
(c) Write out the corrected working in full, including the corrected final answer.
Stuck? In $y = 3x^2$ the $x^2$ comes before the multiplication. So $3 \times 4^2 \ne (3 \times 4)^2$.3. Open-ended challenge — design three parabolas
This challenge has many valid answers. 4 marks
3.1 Design three different parabolas of the form $y = ax^2$ (with $a > 0$): one wider than $y = x^2$, one with the same width as $y = x^2$, and one narrower than $y = x^2$.
For each parabola:
(i) State your chosen $a$ and write the equation.
(ii) Build a small table at $x = -2, -1, 0, 1, 2$.
(iii) State whether it's wider, the same, or narrower than $y = x^2$, with a one-line reason.
Bonus: Your three values of $a$ must all be different and the "same width" one cannot literally be $y = x^2$ — you should pick a different equation that happens to have the same width (think carefully about what that means).
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Table for $y = 2x^2$
$x^2$: $9, 4, 1, 0, 1, 4, 9$. $y$: $\mathbf{18, 8, 2, 0, 2, 8, 18}$. Vertex: $\mathbf{(0, 0)}$.
1.2 — Find $a$ from points
(a) $(3, 18)$: $18 = 9a \Rightarrow a = \mathbf{2}$. Narrower than $y = x^2$.
(b) $(2, 1)$: $1 = 4a \Rightarrow a = \mathbf{\tfrac{1}{4}}$. Wider than $y = x^2$.
(c) $(5, 5)$: $5 = 25a \Rightarrow a = \mathbf{\tfrac{1}{5}}$. Wider than $y = x^2$.
1.3 — Compare at $x = 6$
$y = 3x^2$ at $x = 6$: $y = 3 \times 36 = \mathbf{108}$.
$y = \tfrac{1}{3}x^2$ at $x = 6$: $y = \tfrac{1}{3} \times 36 = \mathbf{12}$.
The first curve is steeper at that point (it has climbed higher for the same $x$).
1.4 — Solve $5x^2 = 80$
Divide both sides by $5$: $x^2 = 16$. Take both roots: $x = \pm\sqrt{16} = \mathbf{\pm 4}$. So $x = 4$ or $x = -4$.
1.5 — TRUE / FALSE for $y = ax^2$ ($a > 0$)
(a) FALSE — the vertex is always $(0, 0)$, never moves with $a$.
(b) TRUE — the axis of symmetry is $x = 0$ for every $y = ax^2$.
(c) TRUE — if $y = ax^2$ becomes $y = 2ax^2$, every output is doubled at the same $x$.
(d) FALSE — with $a = \tfrac{1}{2} > 0$, the curve opens upward and sits at or above the $x$-axis ($y \ge 0$). Negative $a$ is what would push it below.
1.6 — Both $(2, 12)$ and $(-2, 12)$
Sub $(2, 12)$: $12 = a(2)^2 = 4a \Rightarrow a = 3$.
Sub $(-2, 12)$: $12 = a(-2)^2 = 4a \Rightarrow a = 3$.
Both give the same equation because $(-2)^2 = (2)^2 = 4$, which is exactly the symmetry of $y = ax^2$ across the $y$-axis. So only one value of $a$ is needed: $a = \mathbf{3}$, giving $y = 3x^2$.
2 — Find the mistake
(a) The mistake is on Line 2.
(b) By order of operations, $3 \times 4^2$ means "square the $4$ first, then multiply by $3$" — i.e. $3 \times 16 = 48$, NOT $(3 \times 4)^2$. The student has incorrectly grouped the $3$ and the $4$ inside the square, which is the rule for $(ax)^2$, not for $ax^2$.
(c) Corrected working:
$y = 3 \times 4^2$
$= 3 \times 16$ (square first)
$= \mathbf{48}$.
This is precisely the trap flagged in the lesson's "Common Pitfalls" card — squaring $ax$ instead of computing $a \times x^2$.
3 — Design three parabolas (sample solution)
Wider than $y = x^2$: $a = \tfrac{1}{4}$, equation $y = \tfrac{1}{4}x^2$. Table at $x = -2, -1, 0, 1, 2$: $y = 1, 0.25, 0, 0.25, 1$. Reason: $a < 1$, so every $y$-value is smaller than for $y = x^2$, making the curve flatter / wider.
Same width as $y = x^2$: This is the trick — with $a > 0$ and $|a| = 1$ we are forced to have $a = 1$, which is the original $y = x^2$. So there is no OTHER $y = ax^2$ with the exact same width. (A student who recognises this earns full marks for the entry.)
Narrower than $y = x^2$: $a = 3$, equation $y = 3x^2$. Table: $y = 12, 3, 0, 3, 12$. Reason: $a > 1$, so every $y$-value is larger, making the curve steeper / narrower.
Marking: 1 mark for the wider example with correct table; 1 mark for spotting the "same width" trick (or accepting $y = x^2$ as the only answer and explaining why no other works); 1 mark for the narrower example with correct table; 1 mark for clear reasoning across all three.