Mathematics • Year 9 • Unit 2 • Lesson 8

Vertex Form, Mixed Challenge

Use every move from vertex form: read four slots, write the equation from features, sub a second point to find $a$, and decide min vs max. Catch one common mistake, then design your own family of parabolas.

Master · Mixed Challenge

1. Mixed problems

Each problem mixes the four-slot read with another skill. Show working. 3 marks each

1.1 For each equation, state the vertex, the axis of symmetry, and the direction (up / down): (a) $y = 3(x - 2)^2 - 5$ (b) $y = -(x + 1)^2 + 4$ (c) $y = \tfrac{1}{4}(x - 6)^2$ (d) $y = -2(x + 3)^2 - 7$.

1.2 Write the vertex-form equation for each parabola: (a) vertex $(3, -7)$ with $a = 2$; (b) vertex $(-4, 1)$ with $a = -1$; (c) vertex $(0, 0)$ with $a = -\tfrac{1}{3}$.

1.3 For $y = 4(x - 1)^2 - 16$, find: (a) the vertex; (b) the $y$-intercept (sub $x = 0$); (c) whether the curve has a min or max, and its value.

1.4 A parabola has vertex $(2, -1)$ and passes through $(4, 7)$. (a) Write the equation with $a$ as an unknown using the vertex. (b) Substitute $(4, 7)$ to find $a$. (c) State the final equation and whether the vertex is a min or max.

1.5 For each statement, decide TRUE or FALSE. If false, write the corrected statement. (a) "The vertex of $y = -(x - 2)^2 + 7$ is a maximum at $y = 7$." (b) "The axis of symmetry of $y = 2(x + 5)^2 + 1$ is $x = 5$." (c) "Both $y = (x - 3)^2 + 4$ and $y = (x + 3)^2 + 4$ have the same vertex." (d) "If $a = 1$, the parabola has the same shape as $y = x^2$."

1.6 Two parabolas share a vertex at $(2, 3)$. Parabola P opens UP with $a = 2$. Parabola Q opens DOWN with $a = -2$. (a) Write both equations. (b) Find the $y$-intercept of each. (c) Sketch them on the same axes (just describe in words if no paper handy) and describe how they relate.

Stuck on 1.6(c)? Two parabolas with the same vertex but opposite signs of $a$ are mirror images of each other across the horizontal line $y = k$.

2. Find the mistake

A classmate has tried to read features off five vertex-form parabolas. Exactly two answers are wrong. Spot them, explain why, and fix them. 3 marks

Student's answers:

A: Vertex of $y = (x - 1)^2 + 4$ is $(1, -4)$. (Flip both signs.)

B: Vertex of $y = -2(x + 3)^2 - 1$ is $(-3, -1)$. ✓

C: $y = -(x - 2)^2 + 7$ opens UP because the bracket is squared, and squares are always positive.

D: Vertex of $y = 4(x - 5)^2 + 2$ is a MINIMUM at $y = 2$ (because $a = 4 > 0$). ✓

E: Axis of symmetry of $y = (x + 6)^2$ is $x = -6$. ✓

(a) Which two are wrong?

(b) For each wrong one, explain in one sentence why the student's reasoning is mistaken.

(c) Write the correct version of each wrong answer.

Stuck on C? "$-(x - 2)^2$" means $a = -1$, even though $(x - 2)^2$ on its own is non-negative.

3. Open-ended challenge, invent a family

Many valid answers. Be creative but precise. 4 marks

3.1 Invent FOUR vertex-form equations $y = a(x - h)^2 + k$ that ALL share the same vertex $(2, -3)$, but differ in $a$:

• Parabola 1: opens UP, narrow.
• Parabola 2: opens UP, wide.
• Parabola 3: opens DOWN, standard width ($|a| = 1$).
• Parabola 4: opens DOWN, narrow.

For each:
(i) Write the equation.
(ii) State whether the vertex is a min or max and the value.
(iii) Find the $y$-intercept (sub $x = 0$).

Bonus: Which two of your four parabolas have the same $y$-intercept? Explain why.

Stuck? Pick four $a$-values: e.g. $3$, $\tfrac{1}{2}$, $-1$, $-2$. Then for each, write $y = a(x - 2)^2 - 3$.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers, Do not peek before attempting

1.1, Four reads

(a) Vertex $(2, -5)$, axis $x = 2$, opens UP.
(b) Vertex $(-1, 4)$, axis $x = -1$, opens DOWN.
(c) Vertex $(6, 0)$, axis $x = 6$, opens UP.
(d) Vertex $(-3, -7)$, axis $x = -3$, opens DOWN.

1.2, Write the equation

(a) $y = 2(x - 3)^2 - 7$.
(b) $y = -(x + 4)^2 + 1$.
(c) $y = -\tfrac{1}{3}x^2$ (or $y = -\tfrac{1}{3}(x - 0)^2 + 0$).

1.3, $y = 4(x - 1)^2 - 16$

(a) Vertex $(1, -16)$. (b) Sub $x = 0$: $y = 4(-1)^2 - 16 = 4 - 16 = -12$. $y$-intercept $(0, -12)$. (c) $a = 4 > 0$, opens up, so vertex is a MIN with value $y = -16$.

1.4, Vertex $(2, -1)$ through $(4, 7)$

(a) $y = a(x - 2)^2 - 1$. (b) Sub $(4, 7)$: $7 = a(4 - 2)^2 - 1 = 4a - 1 \Rightarrow 4a = 8 \Rightarrow a = 2$. (c) Equation: $y = 2(x - 2)^2 - 1$. Since $a = 2 > 0$, vertex is a MIN.

1.5, True / False

(a) TRUE. $a = -1 < 0$ opens down, so the vertex $(2, 7)$ is a max with $y = 7$.
(b) FALSE. $(x + 5) = 0$ at $x = -5$, so axis is $x = -5$.
(c) FALSE. First has vertex $(3, 4)$; second has vertex $(-3, 4)$. Different $x$-coordinates.
(d) TRUE. $a = 1$ means same direction (up) and same width, same shape as $y = x^2$, just possibly shifted.

1.6, Two parabolas sharing vertex $(2, 3)$

(a) P: $y = 2(x - 2)^2 + 3$. Q: $y = -2(x - 2)^2 + 3$. (b) P at $x = 0$: $y = 2(4) + 3 = 11$, so $(0, 11)$. Q at $x = 0$: $y = -2(4) + 3 = -5$, so $(0, -5)$. (c) They share the same vertex $(2, 3)$ but P opens up (vertex = min) and Q opens down (vertex = max). They are mirror images of each other across the horizontal line $y = 3$, flip one over and you get the other.

2, Find the mistake

(a) The two wrong answers are A and C.
(b) A: The student flipped BOTH signs. Only $h$ flips (it comes from inside the bracket). $k$ keeps its sign (it sits OUTSIDE the bracket and is added directly). C: The "$-$" in "$-(x - 2)^2$" applies to the WHOLE squared term, making $a = -1$. So the parabola opens DOWN, even though $(x - 2)^2$ on its own is non-negative.
(c) A corrected: Vertex of $y = (x - 1)^2 + 4$ is $(1, 4)$, $h$ flips to $+1$, $k$ stays as $+4$. C corrected: $a = -1 < 0$, so $y = -(x - 2)^2 + 7$ opens DOWN. Vertex $(2, 7)$ is a MAXIMUM.
Both pitfalls are flagged in the lesson: "flipping $k$'s sign too" and "missing the negative on $a = -1$".

3, Open-ended challenge (sample solutions)

Parabola 1 (UP, narrow): $y = 3(x - 2)^2 - 3$. Vertex $(2, -3)$: MIN, value $-3$. $y$-intercept: $y = 3(4) - 3 = 9$, so $(0, 9)$.

Parabola 2 (UP, wide): $y = \tfrac{1}{2}(x - 2)^2 - 3$. MIN, value $-3$. $y$-int: $y = \tfrac{1}{2}(4) - 3 = -1$, so $(0, -1)$.

Parabola 3 (DOWN, standard): $y = -(x - 2)^2 - 3$. MAX, value $-3$. $y$-int: $y = -(4) - 3 = -7$, so $(0, -7)$.

Parabola 4 (DOWN, narrow): $y = -2(x - 2)^2 - 3$. MAX, value $-3$. $y$-int: $y = -2(4) - 3 = -11$, so $(0, -11)$.

Bonus: No two of these four share a $y$-intercept exactly (they're all different here). But Parabolas 1 and 4 would share their $y$-intercept's magnitude from the vertex difference if we chose $a$ values with opposite signs and equal $|a|$ (e.g. $a = 3$ and $a = -3$ would give $y$-intercepts of $9$ and $-15$ symmetric about $-3$). (Accept any valid pair with reasoning; if the student picks $a = 3$ and $a = -3$, then $y$-intercepts $9$ and $-15$ are symmetric about $k = -3$.)

Marking: 1 mark per parabola with correct equation, vertex type, and $y$-intercept calculation. Bonus mark for thoughtful pattern explanation.