Mathematics • Year 9 • Unit 2 • Lesson 6
Combining $y = ax^2 + c$
Build the 4-step sketch habit for parabolas with both a stretch ($a$) and a vertical shift ($c$): direction, width, vertex, anchor pair. One fully-worked example, one guided fill-in, then eight independent problems from foundation to extension.
1. I do — fully worked example
Read every line. Each step explains why we make the call, not just what the answer is.
Problem. Sketch $y = 2x^2 - 3$ using direction, width, vertex, and one anchor pair. State the vertex and the $x$-intercepts (if any).
Step 1 — Direction.
Read the sign of $a$: here $a = 2 > 0$, so the parabola opens UP.
Reason: positive $a$ means $y$ grows large for $|x|$ large, so the U sits the right way up.
Step 2 — Width.
$|a| = 2 > 1$, so the curve is NARROWER than $y = x^2$.
Reason: a bigger coefficient pushes $y$ up faster for the same $x$, squeezing the U inward.
Step 3 — Vertex.
Vertex at $(0, c) = (0, -3)$. This is the minimum.
Reason: $c$ alone sets the vertex height for $y = ax^2 + c$ (axis is $x = 0$).
Step 4 — Anchor pair at $x = \pm 1$.
$y = 2(1)^2 - 3 = 2 - 3 = -1$. So $(1, -1)$ and by symmetry $(-1, -1)$.
Reason: three points + symmetry = enough to draw a clean U; no need for a full table.
Step 5 — $x$-intercepts (bonus).
Set $y = 0$: $2x^2 - 3 = 0 \Rightarrow x^2 = \tfrac{3}{2} \Rightarrow x = \pm\sqrt{1.5} \approx \pm 1.22$.
Reason: vertex below $x$-axis ($c < 0$) and opens up ($a > 0$) — two crossings guaranteed.
Answer: Narrow upward parabola; vertex $(0, -3)$; anchors $(\pm 1, -1)$; $x$-ints $\approx \pm 1.22$.
2. We do — fill in the missing steps
Same 4-step structure as Section 1, but with the working faded. Fill each blank. 4 marks
Problem. Sketch $y = -x^2 + 4$ using the 4-step method, and find the $x$-intercepts.
Step 1 — Direction: $a = $ __________ , so the parabola opens __________________ .
Step 2 — Width: $|a| = $ __________ , so the curve is __________________ (narrower / wider / standard) compared to $y = x^2$.
Step 3 — Vertex: $c = $ __________ , so vertex $(0, $__________ $)$. This is a (max / min) __________________ .
Step 4 — Anchor pair: at $x = 1$, $y = -(1)^2 + 4 = $ __________ . Anchor points $($__________ , __________$)$ and $($__________ , __________$)$.
$x$-intercepts: $0 = -x^2 + 4 \Rightarrow x^2 = $ __________ $\Rightarrow x = \pm$ __________ . Points: $($__________ , $0)$ and $($__________ , $0)$.
3. You do — independent practice
Show your working in the space under each problem. The first four are foundation (read a single feature). The middle two are standard (vertex + anchor or intercepts). The last two are extension (build the full sketch or reverse-engineer the equation).
Foundation — read one feature off the equation
3.1 State the vertex of $y = 3x^2 - 5$. 1 mark
3.2 Find $y$ at $x = 2$ on the parabola $y = -2x^2 + 1$. Show working. 1 mark
3.3 For $y = -\tfrac{1}{2}x^2 + 3$: (a) does it open up or down? (b) is it narrower or wider than $y = x^2$? (c) where is the vertex? 1 mark
3.4 Find the $x$-intercepts of $y = x^2 - 9$. 1 mark
Standard — combine vertex and anchor or intercepts
3.5 Sketch $y = x^2 - 4$ using the 4-step method. Label the vertex and the two $x$-intercepts. 2 marks
3.6 For $y = -x^2 + 9$, find: (a) the vertex; (b) the $y$-intercept; (c) the $x$-intercepts. Show working for (c). 2 marks
Extension — full sketch and reverse engineering
3.7 Sketch $y = 2x^2 - 8$ using the 4-step method. Find the vertex, the anchor pair at $x = \pm 1$, and both $x$-intercepts. State how many $x$-intercepts it has and why. 3 marks
3.8 A parabola opens upward, is narrower than $y = x^2$, and has its vertex at $(0, -4)$. (a) Write one possible equation. (b) Find the $x$-intercepts of your equation. (c) Briefly justify why your equation meets each of the three given features. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (faded $y = -x^2 + 4$)
Step 1: $a = \mathbf{-1}$, opens DOWN.
Step 2: $|a| = \mathbf{1}$, standard width (same as $y = x^2$).
Step 3: $c = \mathbf{4}$, vertex $(0, \mathbf{4})$. This is a MAX.
Step 4: $y = -(1)^2 + 4 = \mathbf{3}$. Anchors $(\mathbf{1}, \mathbf{3})$ and $(\mathbf{-1}, \mathbf{3})$.
$x$-ints: $x^2 = \mathbf{4} \Rightarrow x = \pm\mathbf{2}$. Points: $(\mathbf{2}, 0)$ and $(\mathbf{-2}, 0)$.
3.1 — Vertex of $y = 3x^2 - 5$
Vertex is $(0, c) = \mathbf{(0, -5)}$.
3.2 — $y$ at $x = 2$ on $y = -2x^2 + 1$
$y = -2(2)^2 + 1 = -2 \times 4 + 1 = -8 + 1 = \mathbf{-7}$. (Square BEFORE multiplying.)
3.3 — Read $y = -\tfrac{1}{2}x^2 + 3$
(a) $a = -\tfrac{1}{2} < 0$, opens down. (b) $|a| = \tfrac{1}{2} < 1$, wider than $y = x^2$. (c) Vertex $(0, 3)$.
3.4 — $x$-intercepts of $y = x^2 - 9$
$0 = x^2 - 9 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3$. Points: $(3, 0)$ and $(-3, 0)$.
3.5 — Sketch $y = x^2 - 4$
Direction: opens up ($a = 1 > 0$). Width: standard ($|a| = 1$). Vertex $(0, -4)$. Anchor at $x = 1$: $y = 1 - 4 = -3$, so $(\pm 1, -3)$. $x$-intercepts: $x^2 = 4 \Rightarrow x = \pm 2$, so $(2, 0)$ and $(-2, 0)$. Smooth U through five labelled points.
3.6 — $y = -x^2 + 9$
(a) Vertex $(0, 9)$. (b) $y$-intercept $(0, 9)$ (same point as the vertex). (c) Set $y = 0$: $-x^2 + 9 = 0 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3$. $x$-intercepts $(3, 0)$ and $(-3, 0)$.
3.7 — Sketch $y = 2x^2 - 8$
Direction: up ($a = 2 > 0$). Width: narrower ($|a| = 2 > 1$). Vertex $(0, -8)$. 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 $(-2, 0)$. Two $x$-intercepts because the vertex is below the $x$-axis ($c < 0$) and the curve opens up ($a > 0$) — opposite signs of $a$ and $c$ guarantee two crossings.
3.8 — Build a parabola
(a) Example: $y = 3x^2 - 4$ (any $a > 1$ with $c = -4$ works). (b) Set $y = 0$: $3x^2 = 4 \Rightarrow x^2 = \tfrac{4}{3} \Rightarrow x = \pm \tfrac{2}{\sqrt{3}} \approx \pm 1.15$. (c) Justify: $a = 3 > 0$ gives upward opening; $|a| = 3 > 1$ gives narrower than $y = x^2$; $c = -4$ puts the vertex at $(0, -4)$.