Mathematics • Year 9 • Unit 2 • Lesson 2
The Basic Parabola — Mixed Challenge
Pull together every $y = x^2$ tool from Lesson 2: substitution with brackets, symmetry, key features, and graph reading. Catch one tricky mistake, then take on an open-ended challenge.
1. Mixed problems
Show your working. Brackets matter when negatives are involved. 3 marks each
1.1 For $y = x^2$, find $y$ when (a) $x = -8$, (b) $x = 0.5$, (c) $x = -\tfrac{1}{2}$. Show brackets and final values.
1.2 Use symmetry of $y = x^2$ to fill in: if $(9, 81)$ lies on the curve, then $(\_\_\_\_, \_\_\_\_)$ also lies on it. Then state which point is the vertex.
1.3 Find the two values of $x$ on $y = x^2$ for which $y = 36$. Show your working.
1.4 Build a table of values for $y = x^2$ from $x = -3$ to $x = 3$. Then compute the first differences in $y$ and explain in one sentence what those differences reveal about the steepness of the curve as $x$ moves away from the vertex.
1.5 State the vertex, axis of symmetry, $y$-intercept, $x$-intercept, domain and range of $y = x^2$.
1.6 Decide whether each statement about $y = x^2$ is TRUE or FALSE, with a one-line justification: (a) the curve has only one $x$-intercept; (b) the curve passes through $(10, -100)$; (c) the curve is symmetric about the $x$-axis; (d) for every $y > 0$ on the curve there are exactly two $x$-values.
2. Find the mistake
Another student has tried to build a table of values and state the vertex of $y = x^2$. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why, then re-do that line correctly. 3 marks
Student's working — build a table and state the vertex for $y = x^2$:
Line 1: At $x = -3$: $y = (-3)^2 = 9$.
Line 2: At $x = -2$: $y = -2^2 = -4$.
Line 3: At $x = 0$: $y = 0^2 = 0$.
Line 4: At $x = 2$: $y = 2^2 = 4$.
Line 5: Vertex: $(0, 0)$.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected line in full.
Stuck? Compare Line 2 with Line 1 — one uses brackets around the negative, one doesn't.3. Open-ended challenge — the staircase under the parabola
This challenge has many valid answers. 4 marks
3.1 Consider the parabola $y = x^2$ between $x = 0$ and $x = 4$.
(i) Plot the points on the curve at $x = 0, 1, 2, 3, 4$.
(ii) These five points form a "staircase" of vertical bars. Find the heights of each bar.
(iii) Explain in one sentence why the heights form the sequence $0, 1, 4, 9, 16$ rather than something like $0, 4, 8, 12, 16$.
(iv) The differences between consecutive heights are $1, 3, 5, 7$ — the odd numbers. Find these differences and propose a rule for the $n^{\text{th}}$ difference.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Substitutions
(a) $y = (-8)^2 = (-8) \times (-8) = \mathbf{64}$.
(b) $y = (0.5)^2 = 0.5 \times 0.5 = \mathbf{0.25}$.
(c) $y = (-\tfrac{1}{2})^2 = \tfrac{1}{4} = \mathbf{0.25}$. (Same answer as (b) — symmetric partners.)
1.2 — Symmetric partner
If $(9, 81)$ lies on $y = x^2$, then $(\mathbf{-9}, \mathbf{81})$ also lies on it (reflect across the $y$-axis). The vertex is $\mathbf{(0, 0)}$.
1.3 — Solve $x^2 = 36$
Take both square roots: $x = \pm\sqrt{36} = \pm 6$. So $x = \mathbf{6}$ or $x = \mathbf{-6}$. Don't forget the negative root — the parabola is symmetric.
1.4 — Table and first differences
$y$ values: $9, 4, 1, 0, 1, 4, 9$. First differences: $-5, -3, -1, 1, 3, 5$.
As $x$ moves away from the vertex, the differences grow in size, meaning the curve gets steeper. Near the vertex the differences are small (curve is nearly flat); far from the vertex they're large (curve is steep).
1.5 — Key features of $y = x^2$
Vertex $(0, 0)$. Axis of symmetry $x = 0$. $y$-intercept $(0, 0)$. $x$-intercept $(0, 0)$ (the curve touches the $x$-axis here, doesn't cross). Domain: all real $x$. Range: $y \ge 0$.
1.6 — TRUE / FALSE
(a) TRUE — the curve only touches the $x$-axis at the vertex $(0, 0)$.
(b) FALSE — $10^2 = 100$ (positive), not $-100$. The basic parabola has no negative $y$-values.
(c) FALSE — the curve is symmetric about the $y$-axis ($x = 0$), not the $x$-axis. Symmetry about $x$-axis would put points below it, which never happens.
(d) TRUE — for any $y > 0$, $x^2 = y$ has two solutions $x = \pm\sqrt{y}$.
2 — Find the mistake
(a) The mistake is on Line 2.
(b) The student wrote $-2^2 = -4$, which (by order of operations) means $-(2^2) = -4$. But substituting $x = -2$ into $y = x^2$ should give $y = (-2)^2 = 4$, with brackets keeping the negative inside the square. Squaring always gives a non-negative result.
(c) Corrected line: at $x = -2$: $y = (-2)^2 = (-2) \times (-2) = +4$.
This is precisely the trap flagged in the lesson's "Spot the Trap" card.
3 — Staircase challenge (sample solution)
(i) Points: $(0, 0), (1, 1), (2, 4), (3, 9), (4, 16)$.
(ii) Heights: $0, 1, 4, 9, 16$ — the square numbers.
(iii) The heights are $n^2$ values, not multiples of a fixed number, because area grows with the SQUARE of the side — not in a straight line. Equal steps in $x$ produce unequal jumps in $y$, which is exactly what makes $y = x^2$ non-linear.
(iv) Differences between consecutive heights: $1 - 0 = 1$, $4 - 1 = 3$, $9 - 4 = 5$, $16 - 9 = 7$ — the odd numbers. A rule for the $n^{\text{th}}$ difference is $\mathbf{2n - 1}$ (where $n = 1, 2, 3, 4, \ldots$). Check: $n = 1$ gives $1$ ✓, $n = 2$ gives $3$ ✓, $n = 3$ gives $5$ ✓, $n = 4$ gives $7$ ✓.
Marking: 1 mark for correct heights; 1 mark for the non-linear explanation; 1 mark for spotting the odd-number pattern; 1 mark for the $2n - 1$ rule (or equivalent).