Mathematics • Year 9 • Unit 2 • Lesson 6
$y = ax^2 + c$ — Mixed Challenge
Mixed practice on every move from Lesson 6: read the equation, sketch the curve, find intercepts, predict crossings, and reverse-engineer an equation from two conditions. One tricky mistake to catch, then an open-ended invention task.
1. Mixed problems — read, sketch, intercept, reverse
Each problem mixes the four core skills from the lesson. Show your working. 3 marks each
1.1 For each equation, state the vertex, direction (up / down) and width (narrow / standard / wide): (a) $y = 4x^2 + 1$ (b) $y = -\tfrac{1}{3}x^2 - 2$ (c) $y = -x^2$ (d) $y = \tfrac{1}{2}x^2 + 6$.
1.2 Sketch $y = -2x^2 + 8$ using the 4-step method. Label the vertex, the anchor pair at $x = \pm 1$, and both $x$-intercepts.
1.3 Without solving algebraically, decide how many $x$-intercepts each parabola has. Justify each from the signs of $a$ and $c$. (a) $y = x^2 + 4$ (b) $y = -x^2 + 7$ (c) $y = 3x^2 - 1$ (d) $y = -\tfrac{1}{2}x^2 - 5$.
1.4 Find the $x$-intercepts of $y = 3x^2 - 12$. Then state the $y$-intercept and the vertex. How many of these three labelled points are different from each other?
1.5 A parabola has vertex $(0, -4)$ and passes through $(2, 8)$. (a) Use the vertex to state $c$. (b) Substitute $(2, 8)$ into $y = ax^2 + c$ to find $a$. (c) Write the equation, and state the direction and width.
1.6 Two parabolas: P: $y = x^2 - 4$ and Q: $y = -x^2 + 4$. (a) Find the $x$-intercepts of each. (b) Find the points where the two parabolas intersect each other (set their $y$-values equal). (c) On the same axes, sketch both and circle the two intersection points.
2. Find the mistake
A classmate has tried to sketch $y = -3x^2 + 12$ in five steps. Their working is shown below. Exactly two of their steps are wrong. Spot them, explain why, and fix them. 3 marks
Student's working for $y = -3x^2 + 12$:
Step 1: $a = -3$, so opens UP (negative means up). ✗
Step 2: $|a| = 3 > 1$, so NARROWER than $y = x^2$. ✓
Step 3: Vertex $(0, 12)$. ✓
Step 4: Anchor at $x = 1$: $y = -3(1)^2 + 12 = 9$, so $(1, 9)$ and $(-1, 9)$. ✓
Step 5: $x$-intercepts: $0 = -3x^2 + 12 \Rightarrow x^2 = 4 \Rightarrow x = 4$ only. ✗
(a) Which two steps are wrong?
(b) For each wrong step, explain in one sentence why the student's reasoning is mistaken.
(c) Write the corrected version of each wrong step.
Stuck? Compare each step to the lesson rules: sign of $a$ → direction; $\pm$ when taking square roots.3. Open-ended challenge — design four parabolas
This question has many valid answers. Be creative but precise. 4 marks
3.1 Invent FOUR equations of the form $y = ax^2 + c$, one matching each of these descriptions:
A. Opens UP, narrower than $y = x^2$, vertex BELOW the $x$-axis. (Should have two $x$-intercepts.)
B. Opens DOWN, wider than $y = x^2$, vertex ABOVE the $x$-axis. (Should have two $x$-intercepts.)
C. Opens UP, vertex ON the $x$-axis. (Exactly one $x$-intercept.)
D. Opens DOWN, vertex BELOW the $x$-axis. (No real $x$-intercepts.)
For each equation:
(i) Write your chosen equation.
(ii) State the vertex and the number of $x$-intercepts.
(iii) Show working for the $x$-intercepts (or explain why there are none).
Bonus: Use a different value of $a$ for each of the four equations.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Read four equations
(a) Vertex $(0, 1)$, opens up, narrow ($|a| = 4 > 1$).
(b) Vertex $(0, -2)$, opens down, wide ($|a| = \tfrac{1}{3} < 1$).
(c) Vertex $(0, 0)$, opens down, standard ($|a| = 1$).
(d) Vertex $(0, 6)$, opens up, wide ($|a| = \tfrac{1}{2} < 1$).
1.2 — Sketch $y = -2x^2 + 8$
Direction: down ($a = -2 < 0$). Width: narrow ($|a| = 2 > 1$). Vertex $(0, 8)$ (a max). Anchor at $x = 1$: $y = -2 + 8 = 6$, so $(\pm 1, 6)$. $x$-intercepts: $-2x^2 + 8 = 0 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$, so $(2, 0)$ and $(-2, 0)$. Smooth narrow inverted U through the five labelled points.
1.3 — Count $x$-intercepts without solving
(a) $a = 1 > 0$, $c = 4 > 0$, SAME sign $\Rightarrow$ 0 $x$-intercepts (vertex above, opens up).
(b) $a = -1$, $c = 7$, OPPOSITE signs $\Rightarrow$ 2 $x$-intercepts.
(c) $a = 3$, $c = -1$, OPPOSITE signs $\Rightarrow$ 2 $x$-intercepts.
(d) $a = -\tfrac{1}{2}$, $c = -5$, SAME sign $\Rightarrow$ 0 $x$-intercepts.
1.4 — $y = 3x^2 - 12$
$x$-intercepts: $3x^2 - 12 = 0 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$, so $(2, 0)$ and $(-2, 0)$. $y$-intercept: sub $x = 0$, $y = -12$, so $(0, -12)$. Vertex: $(0, c) = (0, -12)$ (same as the $y$-intercept). Three labelled points listed but only TWO distinct points: $(\pm 2, 0)$ and the single point $(0, -12)$ that doubles as both vertex and $y$-intercept.
1.5 — Vertex $(0, -4)$ through $(2, 8)$
(a) Vertex $(0, c) = (0, -4) \Rightarrow c = -4$. (b) Sub $(2, 8)$: $8 = a(2)^2 + (-4) = 4a - 4$, so $4a = 12 \Rightarrow a = 3$. (c) Equation: $y = 3x^2 - 4$. Direction: up ($a > 0$). Width: narrow ($|a| = 3 > 1$).
1.6 — Two parabolas P and Q
(a) P: $x^2 - 4 = 0 \Rightarrow x = \pm 2$. Q: $-x^2 + 4 = 0 \Rightarrow x = \pm 2$. Both share $x$-intercepts $(\pm 2, 0)$. (b) Set equal: $x^2 - 4 = -x^2 + 4 \Rightarrow 2x^2 = 8 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$. At $x = 2$: $y = 4 - 4 = 0$. At $x = -2$: same. Intersection points $(2, 0)$ and $(-2, 0)$ — the two shared $x$-intercepts. (c) Sketch: P is an upward U with vertex $(0, -4)$; Q is an inverted U with vertex $(0, 4)$; they meet exactly at the two $x$-axis points, forming a leaf shape.
2 — Find the mistake
(a) The two wrong steps are Step 1 and Step 5.
(b) Step 1: "Negative $a$ opens UP" is wrong — the sign of $a$ tells you direction, and a NEGATIVE $a$ means the parabola opens DOWN. Step 5: "$x = 4$ only" is wrong — when you square-root both sides of $x^2 = 4$ you must include $\pm$, otherwise you lose half the solutions. The parabola is symmetric, so both roots must be there.
(c) Step 1 corrected: $a = -3 < 0$, so opens DOWN. Step 5 corrected: $0 = -3x^2 + 12 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$. $x$-intercepts: $(2, 0)$ and $(-2, 0)$.
Both mistakes appear in the lesson's "Common Pitfalls" — sign-of-$a$ confusion and forgetting the negative root.
3 — Open-ended challenge (sample solutions)
A. (opens up, narrow, vertex below): $y = 2x^2 - 8$. Vertex $(0, -8)$. $2x^2 - 8 = 0 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$. Two $x$-intercepts.
B. (opens down, wide, vertex above): $y = -\tfrac{1}{2}x^2 + 8$. Vertex $(0, 8)$. $-\tfrac{1}{2}x^2 + 8 = 0 \Rightarrow x^2 = 16 \Rightarrow x = \pm 4$. Two $x$-intercepts.
C. (opens up, vertex on $x$-axis): $y = 3x^2$. Vertex $(0, 0)$. $3x^2 = 0 \Rightarrow x = 0$. Exactly one $x$-intercept (repeated).
D. (opens down, vertex below $x$-axis): $y = -4x^2 - 1$. Vertex $(0, -1)$. $-4x^2 - 1 = 0 \Rightarrow x^2 = -\tfrac{1}{4}$, no real solutions. Zero $x$-intercepts.
Marking: 1 mark per part with valid equation, correct vertex, and correct intercept count + working. Full marks for any four equations that satisfy the brief and use four different $a$-values.