Mathematics • Year 9 • Unit 2 • Lesson 11
The General Form $y = a(x - h)^2 + k$
Build the read-the-three-numbers habit for vertex form. Watch one worked example, fill in a guided one, then run eight independent drills graduated from quick reads to expansion.
1. I do — fully worked example
Read every line. The goal is to extract the parabola's full fingerprint — $a$, vertex, axis, direction, width — without graphing.
Problem. For $y = 2(x - 3)^2 - 5$, state $a$, the vertex, the axis of symmetry, the direction the parabola opens and whether the vertex is a maximum or minimum.
Step 1 — Identify $a$.
The coefficient out the front of the bracket is $a = 2$.
Reason: $a$ does two jobs — sign sets direction, size sets width. $a = 2 > 0$ means UP; $|a| = 2 > 1$ means narrower than $y = x^2$.
Step 2 — Identify $h$ (flip the sign).
Inside the bracket we see $(x - 3)$, so $h = 3$.
Reason: vertex form is $y = a(x - h)^2 + k$. A minus inside means $h$ is positive; a plus inside would mean $h$ is negative.
Step 3 — Identify $k$ (keep the sign).
The constant on the end is $-5$, so $k = -5$.
Reason: no sign flip on $k$ — what you see is what you get.
Step 4 — Put it all together.
Vertex $(h, k) = (3, -5)$. Axis of symmetry $x = h = 3$. $a = 2 > 0$ so opens UP — vertex is a MINIMUM.
Reason: in vertex form everything sits in one tidy line.
Answer: $a = 2$ (up, narrow), vertex $(3, -5)$, axis $x = 3$, vertex is a MIN.
2. We do — fill in the missing steps
Same structure as Section 1, but with the working faded. Fill in each blank line. 4 marks
Problem. For $y = -\tfrac{1}{2}(x + 4)^2 + 3$, state $a$, the vertex, the axis, the direction and whether the vertex is a max or min.
Step 1 — Identify $a$: $a = $ __________________ . Sign $\Rightarrow$ opens __________________ . Size $\Rightarrow$ __________________ than $y = x^2$.
Step 2 — Identify $h$: rewrite $(x + 4)$ as $(x - $ __________________ $)$. So $h = $ __________________ .
Step 3 — Identify $k$: constant on the end is $+3$, so $k = $ __________________ .
Step 4 — Put it together: vertex $($ __________________ $,$ __________________ $)$, axis $x = $ __________________ , vertex is a __________________ (max / min).
3. You do — independent practice
Show your working in the space under each problem. Drills 3.1–3.4 are foundation (read vertex + axis). 3.5–3.6 are standard (full feature set). 3.7–3.8 are extension (expand to general form).
Foundation — read vertex and axis
3.1 State the vertex and axis of symmetry of $y = 4(x - 1)^2 + 7$. 1 mark
3.2 State the vertex and axis of $y = (x + 5)^2 - 2$. 1 mark
3.3 State the vertex and direction (up/down) of $y = -(x + 2)^2 - 6$. 1 mark
3.4 State the vertex of $y = 3(x - 6)^2$. (Hint: the constant on the end is $0$.) 1 mark
Standard — full feature set
3.5 For $y = -2(x - 4)^2 + 9$, state $a$, the vertex, the axis, the direction the parabola opens, and whether the vertex is a maximum or minimum. 2 marks
3.6 For $y = \tfrac{1}{3}(x + 7)^2 - 1$, state $a$, the vertex, the axis, and describe the width (narrower or wider than $y = x^2$). 2 marks
Extension — expand to general form
3.7 Expand $y = 2(x - 1)^2 + 5$ into the form $y = ax^2 + bx + c$. Show every step (square the bracket, multiply by $a$, add $k$). 2 marks
3.8 Expand $y = -(x - 5)^2 + 2$ to general form. Then check your answer is consistent: the coefficient of $x^2$ should equal $a$ from the vertex form. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (faded $y = -\tfrac{1}{2}(x + 4)^2 + 3$)
Step 1: $a = \mathbf{-\tfrac{1}{2}}$. Opens DOWN; WIDER than $y = x^2$ (because $|a| = \tfrac{1}{2} < 1$).
Step 2: $(x + 4) = (x - (\mathbf{-4}))$, so $h = \mathbf{-4}$.
Step 3: $k = \mathbf{3}$.
Step 4: vertex $(\mathbf{-4}, \mathbf{3})$, axis $x = \mathbf{-4}$, vertex is a MAX (since $a < 0$).
3.1 — $y = 4(x - 1)^2 + 7$
$h = 1$, $k = 7$. Vertex $(1, 7)$, axis $x = 1$.
3.2 — $y = (x + 5)^2 - 2$
$(x + 5) \Rightarrow h = -5$. $k = -2$. Vertex $(-5, -2)$, axis $x = -5$.
3.3 — $y = -(x + 2)^2 - 6$
$h = -2$, $k = -6$, $a = -1$. Vertex $(-2, -6)$. Opens DOWN (since $a < 0$).
3.4 — $y = 3(x - 6)^2$
$h = 6$, $k = 0$. Vertex $(6, 0)$. (Axis $x = 6$ if asked.)
3.5 — $y = -2(x - 4)^2 + 9$
$a = -2$; vertex $(4, 9)$; axis $x = 4$; opens DOWN; vertex is a MAX.
3.6 — $y = \tfrac{1}{3}(x + 7)^2 - 1$
$a = \tfrac{1}{3}$; vertex $(-7, -1)$; axis $x = -7$. $|a| = \tfrac{1}{3} < 1$, so the parabola is WIDER than $y = x^2$.
3.7 — Expand $y = 2(x - 1)^2 + 5$
$(x - 1)^2 = x^2 - 2x + 1$. Multiply by $2$: $2x^2 - 4x + 2$. Add $k = 5$: $\mathbf{y = 2x^2 - 4x + 7}$.
3.8 — Expand $y = -(x - 5)^2 + 2$
$(x - 5)^2 = x^2 - 10x + 25$. Multiply by $a = -1$: $-x^2 + 10x - 25$. Add $k = 2$: $\mathbf{y = -x^2 + 10x - 23}$.
Check: coefficient of $x^2$ is $-1$, which matches $a = -1$ from vertex form. Consistent.