Mathematics • Year 9 • Unit 2 • Lesson 4
Reflections: $y = -x^2$ and $y = -ax^2$
Build the two-step "read the sign, then read the magnitude" routine for $y = -ax^2$. Work through one fully-worked example, then a faded guided one, then eight independent problems graduated from foundation to extension.
1. I do — fully worked example
Read every line. The sign decides direction; the magnitude decides width.
Problem. Describe the graph of $y = -2x^2$: direction of opening, width compared to $y = x^2$, vertex, and find $y$ when $x = -3$.
Step 1 — Read the sign of $a$.
$a = -2 < 0 \Rightarrow$ opens DOWN.
Reason: negative coefficient flips the basic parabola through the $x$-axis.
Step 2 — Read the magnitude $|a|$.
$|a| = 2 > 1 \Rightarrow$ NARROWER than $y = x^2$.
Reason: width depends only on $|a|$. Bigger magnitude = narrower curve, same rule as Lesson 3.
Step 3 — State the vertex.
Vertex still at $(0, 0)$ — but now it's the MAXIMUM.
Reason: $-2 \times 0^2 = 0$, so the vertex sits on the axis. Because the curve opens down, that point is the highest, not the lowest.
Step 4 — Find $y$ when $x = -3$ (square first, multiply by $a$).
$y = -2 \times (-3)^2 = -2 \times 9 = -18$
Reason: brackets keep the negative inside the square; $(-3)^2 = +9$, then $-2 \times 9 = -18$.
Answer: Opens down, narrower than $y = x^2$, vertex $\mathbf{(0, 0)}$ (a maximum). At $x = -3$: $y = \mathbf{-18}$.
2. We do — fill in the missing steps
Same template as Section 1, with the working faded. 4 marks
Problem. Describe $y = -\tfrac{1}{2}x^2$: direction, width vs $y = x^2$, vertex, and find $y$ at $x = -4$.
Step 1 — Sign of $a$: $a = -\tfrac{1}{2}$ is __________________ , so opens __________________ .
Step 2 — Magnitude $|a|$: $|-\tfrac{1}{2}| = \_\_\_\_$ , which is __________________ than $1$, so the curve is __________________ than $y = x^2$.
Step 3 — Vertex: $(0, 0)$, which is a __________________ (max or min)?
Step 4 — Find $y$ at $x = -4$:
$y = -\tfrac{1}{2} \times (\_\_\_\_)^2 = -\tfrac{1}{2} \times \_\_\_\_ = \_\_\_\_$
3. You do — independent practice
Show your working. Brackets when substituting negatives. Foundation = one job. Standard = both sign and width. Extension = find $a$ given direction.
Foundation — single evaluations
3.1 Find $y$ for $y = -x^2$ at $x = 5$. 1 mark
3.2 Find $y$ for $y = -x^2$ at $x = -5$. 1 mark
3.3 Which way does $y = -4x^2$ open? 1 mark
3.4 Is the vertex of $y = -x^2$ a maximum or a minimum? 1 mark
Standard — sign and width together
3.5 Describe $y = -3x^2$: direction of opening, width compared to $y = x^2$, and vertex. 2 marks
3.6 Build a small table for $y = -2x^2$ at $x = -2, -1, 0, 1, 2$. 2 marks
Extension — find $a$ for a downward parabola
3.7 A parabola opens downward, has vertex $(0, 0)$, and passes through $(1, -5)$. Find $a$ and write the equation. 3 marks
3.8 A parabola has equation $y = ax^2$ and passes through $(2, -12)$. Find $a$, write the equation, and state direction and width vs $y = x^2$. 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: $a$ is negative, so opens downward.
Step 2: $|a| = \mathbf{\tfrac{1}{2}}$, which is less than $1$, so the curve is wider than $y = x^2$.
Step 3: vertex $(0, 0)$ — a maximum.
Step 4: $y = -\tfrac{1}{2} \times (\mathbf{-4})^2 = -\tfrac{1}{2} \times \mathbf{16} = \mathbf{-8}$.
3.1 — $y = -x^2$ at $x = 5$
$y = -(5)^2 = -(25) = \mathbf{-25}$.
3.2 — $y = -x^2$ at $x = -5$
$y = -(-5)^2 = -(+25) = \mathbf{-25}$. Same answer as 3.1 — the curve is symmetric across the $y$-axis.
3.3 — Opening of $y = -4x^2$
$a = -4 < 0$, so the curve opens downward.
3.4 — Vertex of $y = -x^2$
The curve opens downward, so the vertex sits on top — it is a maximum.
3.5 — Describe $y = -3x^2$
Sign: $a = -3 < 0$, opens downward. Magnitude: $|-3| = 3 > 1$, so narrower than $y = x^2$. Vertex: $(0, 0)$, a maximum.
3.6 — Table for $y = -2x^2$
$x^2$: $4, 1, 0, 1, 4$. Multiply by $-2$: $y = \mathbf{-8, -2, 0, -2, -8}$. All values are $\le 0$, consistent with the curve opening downward and the vertex being a maximum.
3.7 — Downward parabola through $(1, -5)$
Sub the point: $-5 = a(1)^2 = a$, so $a = \mathbf{-5}$. Equation: $y = -5x^2$. (Sign is negative as required.)
3.8 — Parabola through $(2, -12)$
Sub the point: $-12 = a(2)^2 = 4a \Rightarrow a = \mathbf{-3}$. Equation: $y = -3x^2$. Direction: opens downward ($a < 0$). Width: $|-3| = 3 > 1$, so narrower than $y = x^2$.