Mathematics • Year 9 • Unit 2 • Lesson 2
The Basic Parabola $y = x^2$
Build the table-then-plot-then-name routine for $y = x^2$. Work through one fully-worked example with brackets shown, then a faded guided example, then eight independent practice problems graduated from foundation to extension.
1. I do — fully worked example
Read every line. The brackets matter — watch where they go.
Problem. For $y = x^2$, find $y$ when $x = -5$, then state another point with the same $y$.
Step 1 — Substitute with brackets.
$y = (-5)^2$
Reason: brackets keep the negative INSIDE the square. Without them, $-5^2$ would mean $-(5^2) = -25$.
Step 2 — Evaluate.
$(-5)^2 = (-5) \times (-5) = 25$
Reason: negative $\times$ negative $=$ positive. The result is $+25$, not $-25$.
Step 3 — Write the point.
$(-5, 25)$ lies on the parabola.
Reason: always give a point as an ordered pair $(x, y)$.
Step 4 — Use symmetry to find another point.
Reflect across the axis $x = 0$: $x = -5$ becomes $x = +5$. $y$ stays the same.
Reason: the basic parabola is symmetric about the $y$-axis, so $(a, a^2)$ and $(-a, a^2)$ are both on the curve.
Step 5 — Write the symmetric partner.
$(5, 25)$ also lies on the parabola.
Reason: quick check $5^2 = 25$ ✓.
Answer: $y = \mathbf{25}$. Symmetric partner: $\mathbf{(5, 25)}$.
2. We do — fill in the missing steps
Same structure as Section 1, but with the working faded. Fill in each blank. 4 marks
Problem. For $y = x^2$, find $y$ when $x = -6$ and give the symmetric partner.
Step 1 — Substitute with brackets:
$y = (\_\_\_\_\,)^2$
Step 2 — Evaluate:
$y = \_\_\_\_ \times \_\_\_\_ = \_\_\_\_$
Step 3 — Write the point:
$(\_\_\_\_, \_\_\_\_)$ lies on the parabola.
Step 4 — Use symmetry: reflect across $x = 0$. The symmetric partner is
$(\_\_\_\_, \_\_\_\_)$.
3. You do — independent practice
Show your working in the space under each problem. The first four are foundation (one substitution). The middle two are standard (symmetry or feature recall). The last two are extension (table or proof).
Foundation — substitute into $y = x^2$
3.1 For $y = x^2$, find $y$ when $x = 4$. 1 mark
3.2 For $y = x^2$, find $y$ when $x = -7$. 1 mark
3.3 Does $(8, 64)$ lie on $y = x^2$? Show your check. 1 mark
3.4 State the vertex and the axis of symmetry of $y = x^2$. 1 mark
Standard — symmetry and key features
3.5 $(6, 36)$ lies on $y = x^2$. Use symmetry to find another point with the same $y$-value. 2 marks
3.6 Without plotting, state the vertex, axis of symmetry, and range of $y = x^2$, and explain in one sentence why the range cannot include negative numbers. 2 marks
Extension — table and reasoning
3.7 Complete a table of values for $y = x^2$ using $x = -3, -2, -1, 0, 1, 2, 3$, then sketch the curve in the space below. Label the vertex and mark the symmetry of the table. 3 marks
3.8 A student claims that $(-7, -49)$ lies on $y = x^2$. Test the claim with full working and explain in one sentence what mistake the student likely made. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (faded $x = -6$)
Step 1: $y = (\mathbf{-6})^2$.
Step 2: $y = \mathbf{(-6)} \times \mathbf{(-6)} = \mathbf{36}$.
Step 3: $(\mathbf{-6}, \mathbf{36})$ lies on the parabola.
Step 4: symmetric partner $(\mathbf{6}, \mathbf{36})$.
3.1 — $x = 4$
$y = 4^2 = \mathbf{16}$.
3.2 — $x = -7$
$y = (-7)^2 = (-7) \times (-7) = \mathbf{49}$.
3.3 — Does $(8, 64)$ lie on $y = x^2$?
Sub $x = 8$: $y = 8^2 = 64$. Yes, $64 = 64$, so the point does lie on the curve.
3.4 — Vertex and axis
Vertex $(0, 0)$ (the lowest point). Axis of symmetry $x = 0$ (the $y$-axis).
3.5 — Symmetric partner of $(6, 36)$
The axis of symmetry is $x = 0$, so reflect: $x = 6$ becomes $x = -6$, $y$ unchanged. Partner point: $\mathbf{(-6, 36)}$. Check: $(-6)^2 = 36$ ✓.
3.6 — Features without plotting
Vertex $(0, 0)$. Axis of symmetry $x = 0$. Range $y \ge 0$. The range cannot include negatives because squaring any real number gives a non-negative result — the curve sits entirely on or above the $x$-axis.
3.7 — Table and sketch
Table: $(-3, 9), (-2, 4), (-1, 1), (0, 0), (1, 1), (2, 4), (3, 9)$. The $y$-row reads $9, 4, 1, 0, 1, 4, 9$ — identical forwards and backwards (mirror), confirming symmetry across the $y$-axis.
Sketch: a smooth U passing through $(0, 0)$ at the bottom and through $(\pm 1, 1)$, $(\pm 2, 4)$, $(\pm 3, 9)$. Label the vertex $(0, 0)$ at the bottom.
3.8 — Test $(-7, -49)$
Sub $x = -7$: $y = (-7)^2 = (-7) \times (-7) = 49$, NOT $-49$. So $(-7, -49)$ does not lie on $y = x^2$. The student likely wrote $-7^2 = -49$ (missing the brackets), which would interpret the expression as $-(7^2)$. The correct calculation is $(-7)^2 = +49$.