Mathematics • Year 9 • Unit 2 • Lesson 4
Reflections — Mixed Challenge
Pull together every $y = -ax^2$ idea from Lessons 2-4: sign decides direction, magnitude decides width, vertex sits at $(0, 0)$. Catch a tricky order-of-operations mistake, then take on an open-ended challenge.
1. Mixed problems
Each question pulls on a different combination of Lessons 2-4. Show your working. 3 marks each
1.1 For $y = -x^2$, build a table for $x = -3, -2, -1, 0, 1, 2, 3$ and state the vertex.
1.2 Describe each curve in two words (direction + width): (a) $y = -x^2$ (b) $y = -\tfrac{1}{4}x^2$ (c) $y = 5x^2$ (d) $y = -10x^2$.
1.3 Find $a$ for each parabola $y = ax^2$ passing through the given point, then state direction: (a) $(2, 8)$ (b) $(2, -8)$ (c) $(4, -2)$.
1.4 Solve $-2x^2 = -18$ (find both roots).
1.5 Decide TRUE or FALSE with a one-line justification: (a) the vertex of $y = -ax^2$ is at $(0, 0)$ for any $a$; (b) $y = -x^2$ has the same width as $y = x^2$; (c) increasing $|a|$ on a downward parabola makes the curve wider; (d) at $x = 4$, $y = -3x^2$ gives $y = -48$.
1.6 Sketch (in the margin or in your book) all three on the same axes: $y = x^2$, $y = -x^2$, and $y = -2x^2$. Label the vertex and mark which arms point up vs down.
2. Find the mistake
Another student has tried to compute $y$ for $y = -x^2$ 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 $y = -x^2$ at $x = 4$:
Line 1: $y = -(4)^2$
Line 2: $y = (-4)^2$
Line 3: $y = 16$
Line 4: So $(4, 16)$ lies on $y = -x^2$.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the corrected final answer.
Stuck? $-(4)^2$ and $(-4)^2$ are NOT the same. The first squares the $4$ and then negates; the second squares a negative.3. Open-ended challenge — same vertex, different shapes
This challenge has many valid answers. 4 marks
3.1 Design four different parabolas of the form $y = ax^2$, all sharing the same vertex $(0, 0)$, one of each type:
(i) opens up, narrow,
(ii) opens up, wide,
(iii) opens down, narrow,
(iv) opens down, wide.
For each:
• State your chosen $a$ and write the equation.
• Compute the $y$-value at $x = 2$ to show the shape.
• Confirm in one sentence that the sign and magnitude do the jobs they should.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Table for $y = -x^2$
$x^2$: $9, 4, 1, 0, 1, 4, 9$. Negate: $y = \mathbf{-9, -4, -1, 0, -1, -4, -9}$. Vertex: $\mathbf{(0, 0)}$ — a maximum.
1.2 — Describe each curve
(a) Down, standard width. (b) Down, wide. (c) Up, narrow. (d) Down, narrow.
1.3 — Find $a$ and state direction
(a) $(2, 8)$: $8 = 4a \Rightarrow a = \mathbf{2}$. Opens up.
(b) $(2, -8)$: $-8 = 4a \Rightarrow a = \mathbf{-2}$. Opens down.
(c) $(4, -2)$: $-2 = 16a \Rightarrow a = \mathbf{-\tfrac{1}{8}}$. Opens down.
1.4 — Solve $-2x^2 = -18$
Divide both sides by $-2$: $x^2 = 9$. Take both roots: $x = \mathbf{\pm 3}$. (Don't forget the negative root.)
1.5 — TRUE / FALSE
(a) TRUE — $a \times 0^2 = 0$ for any $a$, so the vertex is always $(0, 0)$.
(b) TRUE — same $|a| = 1$ means same width; the negative just flips the curve.
(c) FALSE — increasing $|a|$ makes the curve NARROWER (regardless of sign). Smaller $|a|$ makes it wider.
(d) TRUE — $y = -3 \times (4)^2 = -3 \times 16 = -48$ ✓.
1.6 — Three on the same axes
$y = x^2$: upward U through $(\pm 1, 1), (\pm 2, 4)$, vertex $(0, 0)$.
$y = -x^2$: downward U (mirror of $y = x^2$ through $x$-axis) through $(\pm 1, -1), (\pm 2, -4)$, vertex $(0, 0)$.
$y = -2x^2$: narrower downward U through $(\pm 1, -2), (\pm 2, -8)$, vertex $(0, 0)$.
All three share the vertex $(0, 0)$ and the axis $x = 0$. $y = x^2$ has arms pointing up; the other two have arms pointing down.
2 — Find the mistake
(a) The mistake is on Line 2.
(b) The student has rewritten $-(4)^2$ as $(-4)^2$, but those are not equal. $-(4)^2$ means "square the $4$, then negate" $= -(16) = -16$. $(-4)^2$ means "square a negative four" $= (-4)(-4) = +16$. Swapping the brackets changes whether the negative is inside or outside the square.
(c) Corrected working:
$y = -(4)^2$
$= -(16)$ (square first)
$= \mathbf{-16}$.
So the point on the curve is $(4, -16)$, not $(4, 16)$. The corrected statement: $(4, -16)$ lies on $y = -x^2$.
This is exactly the trap from the lesson's "Common Pitfalls" card.
3 — Four parabolas (sample solution)
(i) Up, narrow: $a = 4$, equation $y = 4x^2$. At $x = 2$: $y = 16$. Sign positive (opens up); $|a| = 4 > 1$ (narrow). ✓
(ii) Up, wide: $a = \tfrac{1}{4}$, equation $y = \tfrac{1}{4}x^2$. At $x = 2$: $y = 1$. Sign positive (opens up); $|a| = 0.25 < 1$ (wide). ✓
(iii) Down, narrow: $a = -3$, equation $y = -3x^2$. At $x = 2$: $y = -12$. Sign negative (opens down); $|a| = 3 > 1$ (narrow). ✓
(iv) Down, wide: $a = -\tfrac{1}{5}$, equation $y = -\tfrac{1}{5}x^2$. At $x = 2$: $y = -\tfrac{4}{5} = -0.8$. Sign negative (opens down); $|a| = 0.2 < 1$ (wide). ✓
Marking: 1 mark per correctly designed parabola with valid $a$, correct $y(2)$ value, and a one-line confirmation of sign + magnitude jobs.