Mathematics • Year 9 • Unit 2 • Lesson 3
Dilations: $y = ax^2$
Build fluency with the dilation factor $a$ in $y = ax^2$: square first, then multiply by $a$. Work through one fully-worked example, then a faded guided example, then eight graduated independent problems.
1. I do — fully worked example
Read every line. Order of operations matters: square first, then multiply by $a$.
Problem. Build a table of values for $y = 3x^2$ from $x = -2$ to $x = 2$, and state whether the curve is narrower or wider than $y = x^2$.
Step 1 — Square each $x$ first.
$x = -2, -1, 0, 1, 2$ → $x^2 = 4, 1, 0, 1, 4$
Reason: BEDMAS / order of operations — the exponent applies before the multiplication.
Step 2 — Multiply each $x^2$ value by $a = 3$.
$y = 3 \times 4, 3 \times 1, 3 \times 0, 3 \times 1, 3 \times 4 = 12, 3, 0, 3, 12$
Reason: the formula $y = ax^2$ means "$y$ is $a$ times $x$-squared", not "$ax$ all squared".
Step 3 — Check the vertex.
At $x = 0$: $y = 3 \times 0 = 0$. Vertex still $(0, 0)$.
Reason: $a \times 0 = 0$ for any $a$, so dilation NEVER moves the vertex.
Step 4 — Compare width to $y = x^2$.
$a = 3 > 1$, so $y = 3x^2$ is NARROWER than $y = x^2$.
Reason: every $y$-value is multiplied by $3$, so the curve climbs three times as fast for the same $x$ — pinched in horizontally.
Answer: Table $y = \mathbf{12, 3, 0, 3, 12}$. The curve is narrower.
2. We do — fill in the missing steps
Same structure as Section 1, but with the working faded. 4 marks
Problem. Build a table for $y = \tfrac{1}{2}x^2$ from $x = -2$ to $x = 2$, and state whether the curve is narrower or wider than $y = x^2$.
Step 1 — Square each $x$:
$x^2$: $\_\_\_\_, \_\_\_\_, \_\_\_\_, \_\_\_\_, \_\_\_\_$
Step 2 — Multiply each $x^2$ by $a = \tfrac{1}{2}$:
$y$: $\_\_\_\_, \_\_\_\_, \_\_\_\_, \_\_\_\_, \_\_\_\_$
Step 3 — Vertex check: at $x = 0$, $y = \tfrac{1}{2} \times 0 = \_\_\_$. Vertex: $(\_\_\_, \_\_\_)$.
Step 4 — Width: $a = \tfrac{1}{2}$ is less than $1$, so the curve is __________________ than $y = x^2$.
3. You do — independent practice
Show your working. Foundation problems are single substitutions. Standard problems combine evaluation with width comparison. Extension problems find $a$ from given information.
Foundation — single substitutions
3.1 Find $y$ when $x = 3$ on $y = 2x^2$. 1 mark
3.2 Find $y$ when $x = -3$ on $y = 2x^2$. 1 mark
3.3 Find $y$ when $x = 4$ on $y = \tfrac{1}{4}x^2$. 1 mark
3.4 State the vertex of $y = 7x^2$. 1 mark
Standard — evaluate then compare width
3.5 For $y = 5x^2$, build a small table for $x = -2, -1, 0, 1, 2$, and state whether the curve is wider or narrower than $y = x^2$. 2 marks
3.6 Rank from widest to narrowest: $y = 4x^2$, $y = \tfrac{1}{3}x^2$, $y = x^2$, $y = 2x^2$. Justify in one sentence. 2 marks
Extension — find $a$ from a point
3.7 A parabola $y = ax^2$ passes through $(2, 20)$. Find $a$, write the equation, and state whether the curve is wider or narrower than $y = x^2$. 3 marks
3.8 A parabola $y = ax^2$ passes through $(4, 2)$. Find $a$ and write the equation. Use a non-vertex point to verify by symmetry. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (faded $y = \tfrac{1}{2}x^2$)
Step 1: $x^2$: $\mathbf{4, 1, 0, 1, 4}$.
Step 2: $y$: $\mathbf{2, 0.5, 0, 0.5, 2}$.
Step 3: at $x = 0$, $y = \tfrac{1}{2} \times 0 = \mathbf{0}$. Vertex: $\mathbf{(0, 0)}$.
Step 4: $a = \tfrac{1}{2} < 1$, so the curve is wider than $y = x^2$.
3.1 — $y = 2x^2$ at $x = 3$
$y = 2 \times (3)^2 = 2 \times 9 = \mathbf{18}$.
3.2 — $y = 2x^2$ at $x = -3$
$y = 2 \times (-3)^2 = 2 \times 9 = \mathbf{18}$. (Same as 3.1 — symmetric across $y$-axis.)
3.3 — $y = \tfrac{1}{4}x^2$ at $x = 4$
$y = \tfrac{1}{4} \times (4)^2 = \tfrac{1}{4} \times 16 = \mathbf{4}$.
3.4 — Vertex of $y = 7x^2$
Vertex of $y = ax^2$ is always $\mathbf{(0, 0)}$ for any $a$ — dilation does not shift the vertex.
3.5 — Table for $y = 5x^2$
$x = -2, -1, 0, 1, 2$ → $x^2 = 4, 1, 0, 1, 4$ → $y = \mathbf{20, 5, 0, 5, 20}$.
$a = 5 > 1$, so the curve is narrower than $y = x^2$.
3.6 — Rank from widest to narrowest
$a$ values: $4, \tfrac{1}{3}, 1, 2$. Smaller positive $a$ $\Rightarrow$ wider. Ordering smallest to largest: $\tfrac{1}{3} < 1 < 2 < 4$.
Widest to narrowest: $\mathbf{y = \tfrac{1}{3}x^2, \; y = x^2, \; y = 2x^2, \; y = 4x^2}$.
3.7 — Find $a$ from $(2, 20)$
Sub the point: $20 = a(2)^2 = 4a$. Solve: $a = \tfrac{20}{4} = \mathbf{5}$. Equation: $y = 5x^2$. Since $a = 5 > 1$, the curve is narrower than $y = x^2$.
3.8 — Find $a$ from $(4, 2)$
Sub the point: $2 = a(4)^2 = 16a$. Solve: $a = \tfrac{2}{16} = \mathbf{\tfrac{1}{8}}$. Equation: $y = \tfrac{1}{8}x^2$.
Symmetry check: at $x = -4$, $y = \tfrac{1}{8} \times (-4)^2 = \tfrac{1}{8} \times 16 = 2$. So $(-4, 2)$ also lies on the curve ✓.