Mathematics • Year 9 • Unit 2 • Lesson 11
Vertex Form — Mixed Challenge
Pull together every vertex-form skill: read $a$, $h$, $k$ from any equation; expand; compare two parabolas; spot a sign-flip mistake; then take an open-ended challenge that asks you to design your own parabolas to spec.
1. Mixed problems
Each question targets a different vertex-form skill. Show working. 3 marks each
1.1 For each equation, state the vertex, axis, direction, and whether the vertex is a max or min: (a) $y = 3(x - 2)^2 + 4$ (b) $y = -(x + 1)^2 - 3$ (c) $y = \tfrac{1}{2}(x - 5)^2$.
1.2 Expand $y = 3(x - 2)^2 + 1$ into the form $y = ax^2 + bx + c$. Show every step.
1.3 Parabola P: $y = 2(x - 1)^2 + 3$. Parabola Q: $y = -3(x + 2)^2 + 3$. (a) Which is wider? Justify using $|a|$. (b) Compare the vertex heights. (c) Which opens downward?
1.4 Write the vertex form of a parabola that opens UP, has its vertex at $(-3, 4)$, and is narrower than $y = x^2$. (Many correct answers exist; choose any $|a| > 1$.) Then state $a$, $h$ and $k$ for your equation.
1.5 A parabola has $a = -2$ and vertex at $(4, 9)$. (a) Write the vertex form. (b) Find $y$ when $x = 2$. (c) Find $y$ when $x = 6$. (d) Why are your answers to (b) and (c) the same?
1.6 Convert from general form back to vertex form by completing the square: $y = x^2 - 6x + 11$. Show your steps. Then state the vertex and axis.
2. Find the mistake
Another student has tried to convert $y = -(x - 4)^2 + 3$ into general form. Their working is below. There is one mistake somewhere. Find it, fix it, and write the correct expanded form. 3 marks
Student's working:
Step 1: $(x - 4)^2 = x^2 - 8x + 16$ ✓
Step 2: Multiply by $a = -1$: $-x^2 - 8x - 16$
Step 3: Add $k = 3$: $-x^2 - 8x - 16 + 3 = -x^2 - 8x - 13$
Final: $y = -x^2 - 8x - 13$
(a) Which step contains the mistake?
(b) Explain in one sentence what the student did wrong.
(c) Write the corrected expanded form.
Stuck? When you multiply $-1$ through $x^2 - 8x + 16$, EVERY term changes sign. The $-8x$ becomes $+8x$.3. Open-ended challenge — design three parabolas
This question has many valid answers. Be creative. 4 marks
3.1 Design three different parabolas in vertex form that ALL share the same vertex at $(2, -1)$ but have different shapes. Your three parabolas must include:
(i) ONE that opens up and is narrower than $y = x^2$ (so $a > 1$).
(ii) ONE that opens up and is wider than $y = x^2$ (so $0 < a < 1$).
(iii) ONE that opens down (so $a < 0$).
For each one: (a) write the equation, (b) state $a$, $h$ and $k$, (c) state whether the vertex is a MAX or MIN, (d) compute $y$ at $x = 4$ to confirm the three parabolas give DIFFERENT values there.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Mixed read
(a) $a = 3$, vertex $(2, 4)$, axis $x = 2$, opens UP, vertex is MIN.
(b) $a = -1$, vertex $(-1, -3)$, axis $x = -1$, opens DOWN, vertex is MAX.
(c) $a = \tfrac{1}{2}$, vertex $(5, 0)$, axis $x = 5$, opens UP, vertex is MIN.
1.2 — Expand $y = 3(x - 2)^2 + 1$
$(x - 2)^2 = x^2 - 4x + 4$. Multiply by $3$: $3x^2 - 12x + 12$. Add $1$: $\mathbf{y = 3x^2 - 12x + 13}$.
1.3 — P vs Q
(a) P: $|a| = 2$; Q: $|a| = 3$. P is WIDER (smaller $|a|$).
(b) Vertex of P is $(1, 3)$; vertex of Q is $(-2, 3)$. Both have height $k = 3$ — SAME vertex height.
(c) Q opens down (since $a = -3 < 0$); P opens up.
1.4 — Design a parabola
Many valid answers. Example: $\mathbf{y = 3(x + 3)^2 + 4}$. Here $a = 3$ (UP and narrower than $y = x^2$ since $|a| > 1$), $h = -3$, $k = 4$. Vertex $(-3, 4)$ as required.
1.5 — Build and test
(a) $\mathbf{y = -2(x - 4)^2 + 9}$.
(b) At $x = 2$: $y = -2(2 - 4)^2 + 9 = -2(4) + 9 = -8 + 9 = 1$.
(c) At $x = 6$: $y = -2(6 - 4)^2 + 9 = -2(4) + 9 = 1$.
(d) Both $x = 2$ and $x = 6$ are exactly $2$ units away from the axis of symmetry $x = 4$. Because parabolas are mirror-symmetric about their axis, they give the same $y$-value.
1.6 — Completing the square
Half of $-6$ is $-3$; squared is $9$.
$y = x^2 - 6x + 11 = x^2 - 6x + 9 - 9 + 11 = (x - 3)^2 + 2$.
Vertex $(3, 2)$, axis $x = 3$. (Check by expanding $(x - 3)^2 + 2 = x^2 - 6x + 9 + 2 = x^2 - 6x + 11$. Matches.)
2 — Find the mistake
(a) The mistake is in Step 2.
(b) When multiplying $-1$ through every term of $x^2 - 8x + 16$, the middle term $-8x$ should become $+8x$, not $-8x$. The student forgot to flip ALL the signs.
(c) Correct: $-1 \times (x^2 - 8x + 16) = -x^2 + 8x - 16$. Add $k = 3$: $\mathbf{y = -x^2 + 8x - 13}$.
3 — Open-ended challenge (sample solutions)
(i) $y = 4(x - 2)^2 - 1$. $a = 4$, $h = 2$, $k = -1$. Vertex is MIN (opens up). At $x = 4$: $y = 4(4) - 1 = 15$.
(ii) $y = \tfrac{1}{2}(x - 2)^2 - 1$. $a = \tfrac{1}{2}$, $h = 2$, $k = -1$. Vertex is MIN (opens up). At $x = 4$: $y = \tfrac{1}{2}(4) - 1 = 1$.
(iii) $y = -3(x - 2)^2 - 1$. $a = -3$, $h = 2$, $k = -1$. Vertex is MAX (opens down). At $x = 4$: $y = -3(4) - 1 = -13$.
The three values at $x = 4$ ($15$, $1$, $-13$) are all different, confirming three genuinely different parabolas through the same vertex.
Marking: 1 mark per valid parabola (correct $a$ direction/width and vertex preserved), plus 1 mark for the comparison at $x = 4$.