Essential Core Foundation

Vertex Form, Roots & Parabolic Trajectories

Quadratic functions bridge algebra with geometric curves. Master completing the square, discriminant analysis ($\Delta = b^2 - 4ac$), sum and product of roots, quadratic inequalities, and real-world trajectory optimization.

1. Core Mathematical Theorems & Formulas

Vertex (Completed Square) Form

$f(x) = a(x - h)^2 + k$
  • • If $a > 0$, minimum point $(h, k)$, min value $= k$.
  • • If $a < 0$, maximum point $(h, k)$, max value $= k$.
  • • Axis of symmetry is line $x = h$.

Discriminant ($\Delta = b^2 - 4ac$)

$\Delta = b^2 - 4ac$
  • • $\Delta > 0$: Two distinct real roots (intersects $x$-axis twice).
  • • $\Delta = 0$: Two equal real roots (tangent to $x$-axis).
  • • $\Delta < 0$: No real roots (does not touch $x$-axis).
  • • $\Delta \ge 0$: Has real roots.

Sum & Product of Roots

$x^2 - (\text{SOR})x + (\text{POR}) = 0$
  • • Roots $\alpha, \beta$: $\text{SOR} = \alpha + \beta = -\frac{b}{a}$.
  • • $\text{POR} = \alpha\beta = \frac{c}{a}$.
  • • $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$.

2. SPM Examiner Pitfalls & Frequent Blunders

Trap 1: Factoring Out $a$ when Completing the Square:

When $f(x) = -2x^2 + 8x - 5$, you MUST factor out $-2$ from the quadratic and linear terms first: $-2[x^2 - 4x] - 5$. Adding and subtracting $(b/2)^2$ inside the bracket must be properly multiplied back by $-2$. Forgetting this produces an inverted vertex!

Trap 2: "Always Positive" / "Always Negative" Conditions:

Students often write $b^2 - 4ac > 0$ for "always positive". That is dead wrong! If a curve is always positive ($f(x) > 0$ for all $x$), it NEVER intersects the $x$-axis, which strictly requires $b^2 - 4ac < 0$ and $a > 0$!

Trap 3: Quadratic Inequalities Sign Flipping:

Never divide or multiply an inequality by an algebraic variable (e.g. dividing by $x$). Always bring all terms to one side, factorize $(x - p)(x - q) \le 0$, and sketch the parabola to determine the interval $p \le x \le q$.

3. Progressive Worked Examples with SPM Marking Rubrics

Example 1 • Completing the Square & Vertex [4 Marks]
Express $f(x) = -3x^2 + 12x - 5$ in the form $a(x - h)^2 + k$. Hence, state:

(a) The coordinates of the maximum turning point. [1 mark]

(b) The equation of the axis of symmetry. [1 mark]

$f(x) = -3(x^2 - 4x) - 5$ [K1: Factor out $-3$ correctly]
$= -3\left[\left(x - \frac{4}{2}\right)^2 - 2^2\right] - 5 = -3[(x - 2)^2 - 4] - 5 = -3(x - 2)^2 + 12 - 5$ [K1: Complete square inside bracket]
$= -3(x - 2)^2 + 7$ [N1: Form $a(x-h)^2+k$]
(a) Maximum point = $(2, 7)$ [P1: Correct coordinates]
(b) Axis of symmetry: $x = 2$ [P1: Stated line $x = 2$]
Example 2 • Forming New Equation with Roots $\alpha, \beta$ [5 Marks]
The quadratic equation $2x^2 - 6x + 3 = 0$ has roots $\alpha$ and $\beta$. Form a quadratic equation with integer coefficients which has roots $(2\alpha + 1)$ and $(2\beta + 1)$.
From $2x^2 - 6x + 3 = 0$: $\alpha + \beta = -\left(-\frac{6}{2}\right) = 3$ and $\alpha\beta = \frac{3}{2}$. [K1: Correct $\text{SOR}$ and $\text{POR}$ of original eq]
New $\text{SOR} = (2\alpha + 1) + (2\beta + 1) = 2(\alpha + \beta) + 2 = 2(3) + 2 = 8$. [K1: Calculate new sum of roots]
New $\text{POR} = (2\alpha + 1)(2\beta + 1) = 4\alpha\beta + 2(\alpha + \beta) + 1 = 4\left(\frac{3}{2}\right) + 2(3) + 1 = 6 + 6 + 1 = 13$. [K1: Calculate new product of roots]
New equation: $x^2 - (\text{new SOR})x + (\text{new POR}) = 0 \implies x^2 - 8x + 13 = 0$. [N1: Correct quadratic equation]
Example 3 • KBAT Parabolic Arch Bridge Modeling [6 Marks]
An architectural concrete bridge arch has a parabolic profile given by $y = -0.05x^2 + 2x$, where $x$ is the horizontal distance (in meters) from the left riverbank origin $(0,0)$, and $y$ is the height of the arch above water level.

(a) Find the total span (width) of the bridge across the river. [2 marks]

(b) Determine the maximum clearance height of the bridge arch. [2 marks]

(c) A cargo barge of height $15\text{ m}$ above water level requires a minimum horizontal clearance of $12\text{ m}$. Determine whether the barge can safely navigate underneath the arch. [2 marks]

(a) Riverbank level is $y = 0 \implies -0.05x^2 + 2x = 0 \implies x(-0.05x + 2) = 0$. [K1: Set $y = 0$]
$x = 0$ or $x = \frac{2}{0.05} = 40\text{ m}$. Total span $= 40\text{ meters}$. [N1: Correct span]
(b) Max height occurs at midpoint axis of symmetry: $x = \frac{0 + 40}{2} = 20\text{ m}$. [K1: Use symmetry $x = 20$ or complete square]
$y_{\max} = -0.05(20)^2 + 2(20) = -0.05(400) + 40 = -20 + 40 = 20\text{ meters}$. [N1: Max height $= 20\text{ m}$]
(c) At height $y = 15$: $15 = -0.05x^2 + 2x \implies 0.05x^2 - 2x + 15 = 0 \implies x^2 - 40x + 300 = 0$. [K1: Substitute $y = 15$]
$(x - 10)(x - 30) = 0 \implies x_1 = 10\text{ m}, x_2 = 30\text{ m}$. [K1: Solve for $x$ boundaries]
Horizontal width at $15\text{ m}$ is $30 - 10 = 20\text{ m}$. Since $20\text{ m} > 12\text{ m}$, the cargo barge can safely pass through. [N1: Valid justification & conclusion]

Casio fx-570EX / 991CW: Finding Vertex & Roots Instantly

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