Mathematics • Year 9 • Unit 2 • Lesson 8
Vertex Form $y = a(x - h)^2 + k$
Build the "read in four slots" habit: $a$ for direction and width, $h$ for the horizontal shift (sign flips!), $k$ for the vertical shift (sign stays). One worked example, one guided fill-in, then eight independent problems.
1. I do — fully worked example
Read every line. Each step explains why we make the call, not just what the answer is.
Problem. For $y = 2(x - 1)^2 + 3$, state the vertex, the axis of symmetry, the direction, the width, and whether the vertex is a maximum or a minimum.
Step 1 — Identify $a$, $h$, $k$.
$a = 2$, bracket zero at $x = 1 \Rightarrow h = 1$, constant after the bracket: $k = 3$.
Reason: vertex form $y = a(x - h)^2 + k$ has four slots — read them all before doing anything else.
Step 2 — Vertex and axis.
Vertex $(h, k) = (1, 3)$. Axis of symmetry $x = 1$.
Reason: $h$ sets the $x$-coordinate of the vertex; $k$ sets the $y$-coordinate. The axis is the vertical line through the vertex.
Step 3 — Direction.
$a = 2 > 0$, so opens UP.
Reason: positive $a$ means $y$ grows large for $|x - h|$ large, so the U sits the right way up.
Step 4 — Width.
$|a| = 2 > 1$, so NARROWER than $y = x^2$.
Reason: a bigger coefficient pushes $y$ up faster, squeezing the U inward.
Step 5 — Min or max.
$a > 0$ $\Rightarrow$ opens up $\Rightarrow$ vertex is a MINIMUM with $y_{min} = k = 3$.
Reason: an upward-opening parabola has its lowest point at the vertex.
Answer: Vertex $(1, 3)$, axis $x = 1$, opens up, narrower than $y = x^2$, MIN value $3$.
2. We do — fill in the missing steps
Same structure as Section 1, but with the working faded. Fill each blank. 4 marks
Problem. For $y = -\tfrac{1}{2}(x + 4)^2 - 5$, state the vertex, axis, direction, width, and whether it has a min or a max.
Step 1 — Identify $a$, $h$, $k$: $a = $ __________ ; $(x + 4) = 0$ at $x = $__________ so $h = $__________ ; $k = $ __________ .
Step 2 — Vertex and axis: Vertex $($__________ , __________$)$. Axis $x = $ __________ .
Step 3 — Direction: sign of $a$ is __________ , so opens __________________ (UP / DOWN).
Step 4 — Width: $|a| = $ __________ , so the curve is __________________ (narrower / wider) than $y = x^2$.
Step 5 — Min or max: Because $a$ is __________ (positive / negative), the vertex is a __________________ (MAX / MIN) with value $y = $ __________ .
3. You do — independent practice
Foundation: read one feature off the equation. Standard: read multiple features together. Extension: build the equation from features.
Foundation — read one feature
3.1 State the vertex of $y = (x - 6)^2 + 2$. 1 mark
3.2 State the axis of symmetry of $y = -2(x + 7)^2 - 1$. 1 mark
3.3 Does $y = 4(x - 2)^2 - 9$ have a maximum or a minimum? State the value. 1 mark
3.4 Write the equation of a parabola with $a = -1$ and vertex $(-2, 3)$. 1 mark
Standard — read multiple features
3.5 For $y = -2(x - 3)^2 + 8$, state: (a) the vertex; (b) the axis of symmetry; (c) the direction; (d) whether it has a max or min. Briefly justify each. 2 marks
3.6 Compare $y = 4(x + 1)^2 - 6$ with $y = x^2$. (a) Is it narrower or wider? (b) Does it open up or down? (c) Where is its vertex? 2 marks
Extension — write the equation
3.7 Write a vertex-form equation for each parabola: (a) vertex $(-1, 4)$, $a = 1$; (b) vertex $(0, -3)$, $a = -2$; (c) vertex $(5, 0)$, $a = \tfrac{1}{2}$. 3 marks
3.8 A parabola has its MAXIMUM value of $y = 10$ at $x = -2$, and is wider than $y = x^2$. (a) State the vertex. (b) Choose a suitable $a$-value and justify your choice. (c) Write a possible equation in vertex form. 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (faded $y = -\tfrac{1}{2}(x + 4)^2 - 5$)
Step 1: $a = \mathbf{-\tfrac{1}{2}}$; $(x + 4) = 0$ at $x = \mathbf{-4}$, so $h = \mathbf{-4}$; $k = \mathbf{-5}$.
Step 2: Vertex $(\mathbf{-4}, \mathbf{-5})$. Axis $x = \mathbf{-4}$.
Step 3: Sign of $a$ is negative, so opens DOWN.
Step 4: $|a| = \mathbf{\tfrac{1}{2}}$, so wider than $y = x^2$.
Step 5: $a$ is negative, so vertex is a MAX with value $y = \mathbf{-5}$.
3.1 — Vertex of $y = (x - 6)^2 + 2$
$(x - 6) = 0$ at $x = 6$; $k = 2$. Vertex $(6, 2)$.
3.2 — Axis of $y = -2(x + 7)^2 - 1$
$(x + 7) = 0$ at $x = -7$. Axis $x = -7$.
3.3 — Min or max of $y = 4(x - 2)^2 - 9$
$a = 4 > 0$, so opens up. Vertex is a MINIMUM at $y = k = -9$.
3.4 — Equation with $a = -1$, vertex $(-2, 3)$
Flip $h = -2$ into bracket as $(x + 2)$; $k = 3$. Equation: $y = -(x + 2)^2 + 3$.
3.5 — Read $y = -2(x - 3)^2 + 8$
(a) $h = 3$ (bracket zero), $k = 8$. Vertex $(3, 8)$. (b) Axis $x = 3$. (c) $a = -2 < 0$, opens DOWN. (d) Opens down $\Rightarrow$ vertex is a MAXIMUM with value $y = 8$.
3.6 — Compare $y = 4(x + 1)^2 - 6$ with $y = x^2$
(a) $|a| = 4 > 1$, so NARROWER. (b) $a = 4 > 0$, so opens UP. (c) $(x + 1) = 0$ at $x = -1$, $k = -6$. Vertex $(-1, -6)$.
3.7 — Write the equation
(a) Vertex $(-1, 4)$, $a = 1$: $y = (x + 1)^2 + 4$.
(b) Vertex $(0, -3)$, $a = -2$: $y = -2(x - 0)^2 + (-3) = -2x^2 - 3$.
(c) Vertex $(5, 0)$, $a = \tfrac{1}{2}$: $y = \tfrac{1}{2}(x - 5)^2 + 0 = \tfrac{1}{2}(x - 5)^2$.
3.8 — Max value 10 at $x = -2$, wider than $y = x^2$
(a) Vertex $(-2, 10)$ (since the max happens there). (b) For "max" we need $a < 0$. For "wider" we need $|a| < 1$. So pick e.g. $a = -\tfrac{1}{2}$ (any value with $-1 < a < 0$ works). (c) Equation: $y = -\tfrac{1}{2}(x + 2)^2 + 10$. Justification: negative $a$ flips the parabola to open down (so the vertex is a max), and a small $|a|$ stretches it sideways making it wider.