MYP Inquiry Framework
Key Concept Relationships
Related Concepts Form, Models, Space
Global Context Orientation in space and time (Trajectories, ballistic curves & structural arches)
Statement of Inquiry Representing parabolic trajectories through equivalent algebraic forms reveals optimal vertices and critical boundary conditions in physical space.
3.1

The Three Forms of a Quadratic Function

  • 1. Standard (General) Form: $$f(x) = ax^2 + bx + c \quad (a \ne 0)$$ Reveals the $y$-intercept at $(0, c)$. The sign of $a$ dictates concavity: $a > 0$ opens upwards (concave up, minimum vertex), while $a < 0$ opens downwards (concave down, maximum vertex).
  • 2. Vertex (Turning Point) Form: $$f(x) = a(x - h)^2 + k$$ Directly reveals the coordinates of the turning point (vertex) at $\mathbf{(h, k)}$ and the vertical axis of symmetry equation $\mathbf{x = h}$.
  • 3. Factorised (Intercept) Form: $$f(x) = a(x - p)(x - q)$$ Directly reveals the $x$-intercepts (roots/zeros) at $(p, 0)$ and $(q, 0)$. By symmetry, the axis of symmetry is the midpoint: $h = \frac{p + q}{2}$.
x y O Axis of Symmetry: x = h Vertex (h, k) (p, 0) (q, 0) y-int (0, c)
3.2

Completing the Square & Deriving the Quadratic Formula

  • Completing the Square (Monic Case $x^2 + bx + c$): Add and subtract $\left(\frac{b}{2}\right)^2$: $$x^2 + bx + c = \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c$$
  • Completing the Square (Non-Monic $ax^2 + bx + c$): Factor out $a$ from the first two terms before completing the square: $$ax^2 + bx + c = a\left[x^2 + \frac{b}{a}x\right] + c = a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c$$ This confirms the vertex coordinates are $h = -\frac{b}{2a}$ and $k = c - \frac{b^2}{4a}$.
  • The Quadratic Formula: Solving $a\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a}$ directly yields: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
3.3

The Discriminant: $\Delta = b^2 - 4ac$

  • Definition: The discriminant $\Delta = b^2 - 4ac$ is the quantity beneath the radical in the quadratic formula. It classifies the nature and number of real roots:
    • $\Delta > 0$: Two distinct real roots ($2$ unique $x$-intercepts). If $\Delta$ is also a perfect square and coefficients are rational, roots are rational; otherwise irrational conjugates.
    • $\Delta = 0$: Exactly one repeated real root ($1$ $x$-intercept; the vertex lies on the $x$-axis and the line $y = 0$ is tangent to the parabola).
    • $\Delta < 0$: No real roots ($0$ $x$-intercepts; the parabola lies entirely above or below the $x$-axis).
  • Geometric Tangency Condition: A line $y = mx + c$ is tangent to a parabola $y = ax^2 + bx + d$ if and only if the discriminant of their difference equation $ax^2 + (b-m)x + (d-c) = 0$ equals zero ($\Delta = 0$).
3.4

Real-World Quadratic Modeling & Optimization

  • Projectile Motion: Height $h(t)$ under gravity $g$ (where $g \approx 9.8\text{ m/s}^2$): $$h(t) = -\frac{1}{2}gt^2 + v_0 t + h_0$$ Maximum height is achieved at time $t = -\frac{v_0}{2(-4.9)} = \frac{v_0}{g}$, and flight duration is found by solving $h(t) = 0$.
  • Revenue & Profit Optimization: Revenue $R(x) = (\text{price}) \times (\text{quantity})$. If price $p(x) = A - Bx$, then $R(x) = Ax - Bx^2$, which has a maximum vertex at $x = \frac{A}{2B}$.

Worked Examples (Criterion A & C)

Example 1 Converting to Vertex Form & Parabola Sketching

Problem: Write $f(x) = 2x^2 - 12x + 11$ in vertex form $f(x) = a(x - h)^2 + k$. State the coordinates of the vertex, equation of the axis of symmetry, and find all intercepts.

Step 1: Factor out $a = 2$ from variable terms $$f(x) = 2(x^2 - 6x) + 11$$
Step 2: Complete the square inside the bracket

Take half of $-6$, which is $-3$, and square it: $(-3)^2 = 9$.

$$f(x) = 2(x^2 - 6x + 9 - 9) + 11 = 2(x - 3)^2 - 2(9) + 11$$ $$f(x) = 2(x - 3)^2 - 18 + 11 = \mathbf{2(x - 3)^2 - 7}$$
Step 3: Extract vertex and axis of symmetry

Vertex is $\mathbf{(3, -7)}$. Since $a = 2 > 0$, this is a minimum turning point.

Axis of symmetry: $\mathbf{x = 3}$.

Step 4: Determine intercepts

$y$-intercept ($x = 0$): $f(0) = 11$, giving $\mathbf{(0, 11)}$.

$x$-intercepts ($f(x) = 0$): $2(x - 3)^2 - 7 = 0 \implies (x - 3)^2 = \frac{7}{2} \implies x = 3 \pm \sqrt{3.5} \approx 3 \pm 1.871$.

Criterion A: Complete mathematical accuracy across all key curve features
Example 2 Discriminant Tangency Problem

Problem: Find the values of $k$ for which the straight line $y = 2x + k$ is tangent to the parabola $y = x^2 - 4x + 7$.

Step 1: Set equations equal to find intersection points $$x^2 - 4x + 7 = 2x + k \implies x^2 - 6x + (7 - k) = 0$$
Step 2: Apply the tangency condition

A straight line is tangent to a parabola if and only if they intersect at exactly one repeated point, meaning $\Delta = 0$.

Here $a = 1$, $b = -6$, $c = (7 - k)$.

$$\Delta = b^2 - 4ac = (-6)^2 - 4(1)(7 - k) = 36 - 28 + 4k = 8 + 4k$$
Step 3: Solve for $k$ $$8 + 4k = 0 \implies 4k = -8 \implies \mathbf{k = -2}$$ Criterion C: Coherent algebraic reasoning connecting geometric tangency to zero discriminant

MYP 4–5 Practice Exercises: Quadratic Functions

Tier 1 (Core skills), Tier 2 (Reasoning), and Tier 3 (Extension and Criterion D modelling).

Question 1 Tier 1: Standard 3 Marks

Solve the quadratic equation by factoring: $x^2 + 7x - 18 = 0$.

View Full Method Markscheme
[M1] Identify factor pairs of $-18$ that sum to $+7$: $+9$ and $-2$.
[M1] Write in factorised form: $(x + 9)(x - 2) = 0$.
[A1] Apply Null Factor Law: $\mathbf{x = -9 \quad \text{or} \quad x = 2}$.
Question 2 Tier 1: Standard 3 Marks

Use the quadratic formula to solve $3x^2 - 5x - 4 = 0$. Give answers to 3 significant figures.

View Full Method Markscheme
[M1] Identify $a = 3, b = -5, c = -4$. Substitute into quadratic formula: $$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(-4)}}{2(3)}$$
[M1] Evaluate discriminant: $\Delta = 25 + 48 = 73$. Thus $x = \frac{5 \pm \sqrt{73}}{6}$.
[A1] Approximate to 3 s.f.: $\mathbf{x \approx 2.26 \quad \text{or} \quad x \approx -0.591}$.
Question 3 Tier 1: Standard 3 Marks

Calculate the discriminant for $2x^2 - 6x + 5 = 0$ and describe the number and nature of its roots.

View Full Method Markscheme
[M1] Identify $a = 2, b = -6, c = 5$. Compute $\Delta = b^2 - 4ac = (-6)^2 - 4(2)(5)$.
[A1] Calculate value: $\Delta = 36 - 40 = \mathbf{-4}$.
[R1] Since $\Delta < 0$, there are $\mathbf{\text{no real roots}}$ (and $0$ real $x$-intercepts).
Question 4 Tier 2: Intermediate 4 Marks

Convert $y = x^2 + 8x - 5$ to vertex form by completing the square, and state its minimum value.

View Full Method Markscheme
[M1] Half of $8$ is $4$, and $4^2 = 16$.
[M1] Rewrite: $y = (x^2 + 8x + 16) - 16 - 5$.
[A1] Vertex form: $\mathbf{y = (x + 4)^2 - 21}$.
[A1] Since $(x+4)^2 \ge 0$, the minimum value is $\mathbf{y = -21}$ occurring at $x = -4$.
Question 5 Tier 2: Intermediate 4 Marks

A parabola has its vertex at $(2, -5)$ and passes through $(4, 7)$. Find its equation in standard form $y = ax^2 + bx + c$.

View Full Method Markscheme
[M1] Write in vertex form: $y = a(x - 2)^2 - 5$.
[M1] Substitute point $(4, 7)$: $7 = a(4 - 2)^2 - 5 \implies 7 = 4a - 5$.
[A1] Solve for $a$: $4a = 12 \implies a = 3$.
[A1] Expand to standard form: $y = 3(x^2 - 4x + 4) - 5 = \mathbf{3x^2 - 12x + 7}$.
Question 6 Tier 2: Intermediate 4 Marks

Find the values of $p$ for which the quadratic equation $px^2 + 8x + p = 0$ has two distinct real solutions.

View Full Method Markscheme
[M1] Two distinct real solutions require $\Delta > 0$ and $p \ne 0$ (for it to remain quadratic).
[M1] Compute discriminant: $\Delta = 8^2 - 4(p)(p) = 64 - 4p^2$.
[M1] Solve inequality: $64 - 4p^2 > 0 \implies 4p^2 < 64 \implies p^2 < 16$.
[A1] Taking square root gives $-4 < p < 4$. Excluding $p = 0$: $\mathbf{-4 < p < 4, \; p \ne 0}$ or $\mathbf{(-4, 0) \cup (0, 4)}$.
Question 7 Tier 2: Intermediate 4 Marks

Find all real solutions to the equation: $(2x - 3)^2 = 49$.

View Full Method Markscheme
[M1] Take the square root of both sides, keeping both positive and negative branches: $2x - 3 = \pm 7$.
[M1] Case 1: $2x - 3 = 7 \implies 2x = 10 \implies x = 5$.
[M1] Case 2: $2x - 3 = -7 \implies 2x = -4 \implies x = -2$.
[A1] Final roots: $\mathbf{x = 5 \quad \text{or} \quad x = -2}$.
Question 8 Tier 3: Extended 5 Marks

Criterion D Projectile Investigation: A model rocket is launched upwards from a platform. Its height $h(t)$ in metres at time $t$ seconds is modeled by: $$h(t) = -5t^2 + 30t + 35$$ (a) Find the initial launch height of the rocket.
(b) Calculate the maximum height reached and the time at which this occurs.
(c) Find the total flight time until the rocket crashes onto the ground.

View Full Method Markscheme
[M1] Part (a): Initial height is at $t = 0$: $h(0) = -5(0)^2 + 30(0) + 35 = \mathbf{35\text{ metres}}$.
[M1] Part (b): Time to maximum height is at the axis of symmetry: $t = -\frac{b}{2a} = -\frac{30}{2(-5)} = \mathbf{3\text{ seconds}}$.
[A1] Maximum height: $h(3) = -5(3)^2 + 30(3) + 35 = -45 + 90 + 35 = \mathbf{80\text{ metres}}$.
[M1] Part (c): Ground impact occurs when $h(t) = 0$: $$-5t^2 + 30t + 35 = 0 \implies t^2 - 6t - 7 = 0 \implies (t - 7)(t + 1) = 0$$
[A1] Since time must be non-negative ($t \ge 0$), discard $t = -1$. Total flight time is $\mathbf{7\text{ seconds}}$.
Question 9 Tier 3: Extended 5 Marks

A farmer has $120\text{ metres}$ of fencing to enclose a rectangular pasture partitioned into two equal pens by an internal fence parallel to one of the sides.
(a) Show that the total area $A(x)$ as a function of the pen width $x$ is $A(x) = 60x - 1.5x^2$.
(b) Find the dimensions that maximise the total enclosed area and state the maximum area.

View Full Method Markscheme
[M1] Let width be $x$ and length be $y$. Fencing comprises three widths and two lengths: $3x + 2y = 120$.
[M1] Express $y$ in terms of $x$: $2y = 120 - 3x \implies y = 60 - 1.5x$.
[A1] Area: $A(x) = x \cdot y = x(60 - 1.5x) = \mathbf{60x - 1.5x^2}$. [Proved]
[M1] This is a downward parabola ($a = -1.5 < 0$). Maximum occurs at vertex: $$x = -\frac{b}{2a} = -\frac{60}{2(-1.5)} = \frac{60}{3} = \mathbf{20\text{ metres}}$$
[A1] Then $y = 60 - 1.5(20) = 30\text{ metres}$. Maximum area: $A(20) = 20 \times 30 = \mathbf{600\text{ m}^2}$.
Question 10 Tier 3: Extended 5 Marks

Find the set of values of $m$ for which the parabola $y = x^2 + 4$ and the line $y = mx$ do not intersect.

View Full Method Markscheme
[M1] Equate expressions: $x^2 + 4 = mx \implies x^2 - mx + 4 = 0$.
[M1] No intersection means the resulting quadratic has no real roots, so $\Delta < 0$.
[M1] Calculate discriminant: $\Delta = (-m)^2 - 4(1)(4) = m^2 - 16$.
[M1] Set $\Delta < 0$: $m^2 - 16 < 0 \implies (m - 4)(m + 4) < 0$.
[A1] The solution is $\mathbf{-4 < m < 4}$ or in interval notation $\mathbf{(-4, 4)}$.

Quadratic Functions Formulas

  • Standard Form: $y = ax^2 + bx + c$
  • Vertex Form: $y = a(x - h)^2 + k \quad \implies \text{Vertex at } (h, k)$
  • Factorised Form: $y = a(x - p)(x - q) \quad \implies x\text{-intercepts at } (p, 0), (q, 0)$
  • Axis of Symmetry: $x = -\dfrac{b}{2a}$
  • Quadratic Formula: $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
  • The Discriminant: $\Delta = b^2 - 4ac$
    • $\Delta > 0 \implies 2$ distinct real roots
    • $\Delta = 0 \implies 1$ repeated real root
    • $\Delta < 0 \implies 0$ real roots

Top Common MYP Pitfalls

1. Sign Error in $(-b)^2$

When $b = -5$, evaluating $b^2$ as $-25$ instead of $+25$ in calculator syntax. Always use brackets: $(-5)^2 = 25$.

2. Division Bar in the Quadratic Formula

Writing $-b \pm \frac{\sqrt{\Delta}}{2a}$. The entire numerator $(-b \pm \sqrt{\Delta})$ is divided by $2a$!

3. Sign of $h$ in Vertex Form

In $y = (x + 3)^2 - 5$, the horizontal coordinate of the vertex is $h = -3$, NOT $+3$.

4. Forgetting the Negative Branch of Square Roots

Writing $x^2 = 25 \implies x = 5$ while omitting $x = -5$.

Interactive Mini Test: Quadratic Functions

Answer the questions to see your score and feedback.

Question 1 of 5 Parabola Vertex Coordinates

What are the vertex coordinates of the parabola $y = -2(x - 4)^2 + 9$?

Question 2 of 5 Discriminant Interpretation

If the discriminant $\Delta = b^2 - 4ac = 0$, what does this indicate about the graph of $y = ax^2 + bx + c$?

Question 3 of 5 Axis of Symmetry Calculation

What is the equation of the axis of symmetry for $f(x) = 3x^2 - 18x + 5$?

Question 4 of 5 Completing the Square

What number must be added to $x^2 - 14x$ to complete the square?

Question 5 of 5 Quadratic Factoring from Roots

A parabola has $x$-intercepts at $(-4, 0)$ and $(1, 0)$ and passes through $(0, -8)$. What is its equation in standard form?