Three Canonical Forms of a Quadratic Function
- Standard (Expanded) Form: $f(x) = ax^2 + bx + c$ ($a \ne 0$).
- The coefficient $a$ dictates orientation: $a > 0 \implies$ opens upwards ($\cup$), minimum vertex; $a < 0 \implies$ opens downwards ($\cap$), maximum vertex.
- The constant $c$ gives the $y$-intercept: $(0, c)$.
- The axis of symmetry is $x = -\frac{b}{2a}$ (Formula Booklet).
- Factorized (Root) Form: $f(x) = a(x - p)(x - q)$.
- Explicitly reveals the $x$-intercepts (roots): $(p, 0)$ and $(q, 0)$.
- The axis of symmetry lies midway between roots: $x = \frac{p + q}{2}$.
- Vertex (Completed Square) Form: $f(x) = a(x - h)^2 + k$.
- The vertex coordinates are $(h, k)$.
- If $a > 0$, the minimum value is $k$, so the range is $\{y \in \mathbb{R} \mid y \ge k\}$.
- If $a < 0$, the maximum value is $k$, so the range is $\{y \in \mathbb{R} \mid y \le k\}$.
(a) Express $f(x)$ in factorized form and write down the $x$-intercepts.
(b) Express $f(x)$ in vertex form and state the coordinates of the minimum point.
Part (a): Factorized Form
$$f(x) = 2(x^2 - 6x + 5) = 2(x - 1)(x - 5)$$The $x$-intercepts are $(1, 0)$ and $(5, 0)$.
Part (b): Vertex Form via Completing the Square
$$f(x) = 2(x^2 - 6x) + 10 = 2\left[(x - 3)^2 - 9\right] + 10$$ $$f(x) = 2(x - 3)^2 - 18 + 10 = 2(x - 3)^2 - 8$$The vertex coordinates are $(3, -8)$. Since $a = 2 > 0$, this is a minimum point.
Range: $\{y \in \mathbb{R} \mid y \ge -8\}$.
The Discriminant $\Delta = b^2 - 4ac$
- Geometric & Algebraic Classification: For the quadratic equation $ax^2 + bx + c = 0$:
$$\Delta = b^2 - 4ac$$
- $\Delta > 0$: Two distinct real roots. The parabola crosses the $x$-axis at two distinct points.
- $\Delta = 0$: One repeated real root (two equal real roots). The parabola is tangent to the $x$-axis at its vertex.
- $\Delta < 0$: No real roots (two complex conjugate roots). The parabola does not touch or intersect the $x$-axis.
- "Real Roots" Condition: When an IB examination question states that an equation has real roots, you must use: $$\Delta \ge 0 \quad (\text{includes both distinct and repeated cases})$$
- Strictly Positive / Negative Quadratics:
- $ax^2 + bx + c > 0$ for all $x \in \mathbb{R} \iff a > 0$ and $\Delta < 0$.
- $ax^2 + bx + c < 0$ for all $x \in \mathbb{R} \iff a < 0$ and $\Delta < 0$.
Step 1: Identify coefficients $a, b, c$.
$$a = 1, \quad b = k - 2, \quad c = k + 1$$Step 2: Set discriminant $\Delta = 0$ for equal real roots.
$$\Delta = b^2 - 4ac = (k - 2)^2 - 4(1)(k + 1) = 0$$Step 3: Expand and collect terms into a quadratic in $k$.
$$(k^2 - 4k + 4) - (4k + 4) = 0$$ $$k^2 - 8k = 0 \implies k(k - 8) = 0$$Step 4: Solve for $k$.
$$k = 0 \quad \text{or} \quad k = 8$$Answer: $k = 0$ or $k = 8$.
Solving Quadratic Inequalities
- The Critical Roots Method:
- Rearrange all terms to one side so the other side is zero: $ax^2 + bx + c < 0$ or $\ge 0$.
- Find the critical values by solving the boundary equation $ax^2 + bx + c = 0 \implies x = p, q$ (with $p < q$).
- Sketch the parabola:
- If $a > 0$ (opens upwards $\cup$): $$ax^2 + bx + c < 0 \implies p < x < q \quad (\text{between roots})$$ $$ax^2 + bx + c > 0 \implies x < p \text{ or } x > q \quad (\text{outside roots})$$
- Include boundary endpoints for non-strict inequalities ($\le$ or $\ge$).
Step 1: Factor the quadratic to locate critical values.
$$3x^2 - 5x - 2 = 0 \implies (3x + 1)(x - 2) = 0$$ $$\text{Critical values: } x = -\frac{1}{3} \quad \text{and} \quad x = 2$$Step 2: Inspect the sign of the parabola.
Since the leading coefficient $a = 3 > 0$, the parabola opens upwards. The curve lies on or below the $x$-axis ($\le 0$) between and including the two roots.
Solution: $-\frac{1}{3} \le x \le 2$ or in interval notation: $\left[-\frac{1}{3}, 2\right]$.
Line-Parabola Tangency & Linear Systems
- Forming the Characteristic Quadratic: To find the intersections of line $y = mx + d$ and parabola $y = ax^2 + bx + c$, equate the two expressions: $$ax^2 + bx + c = mx + d \implies ax^2 + (b - m)x + (c - d) = 0$$
- Classification via Discriminant:
- $\Delta > 0$: Secant line (intersects at 2 distinct points).
- $\Delta = 0$: Tangent line (touches at exactly 1 point of contact).
- $\Delta < 0$: Non-intersecting line (passes without touching).
Step 1: Equate line and parabola equations.
$$x^2 + 2x + 7 = 4x + c \implies x^2 - 2x + (7 - c) = 0$$Step 2: Set discriminant equal to zero for tangency.
$$A = 1, \quad B = -2, \quad C = 7 - c$$ $$\Delta = B^2 - 4AC = (-2)^2 - 4(1)(7 - c) = 0$$ $$4 - 28 + 4c = 0 \implies -24 + 4c = 0 \implies 4c = 24 \implies c = 6$$Point of contact: $x = -\frac{B}{2A} = -\frac{-2}{2(1)} = 1$. Then $y = 4(1) + 6 = 10 \implies (1, 10)$.
Answer: $c = 6$ (point of tangency is $(1, 10)$).
Line-Parabola Tangency & Discriminant Constraints
(a) Find the values of $k$ for which the line $L$ is a tangent to the curve $C$. [4 marks]
(b) For the value of $k$ found in (a), find the exact coordinates of the point of tangency. [3 marks]
Part (a): Set up the intersection equation:
$$x^2 - 4x + 7 = 2x + k \implies x^2 - 6x + (7 - k) = 0 \quad [M1]$$Here $a = 1$, $b = -6$, and $c = 7 - k$. For tangency, the line touches the curve at exactly one point, requiring $\Delta = 0$ [M1]:
$$\Delta = b^2 - 4ac = (-6)^2 - 4(1)(7 - k) = 0 \quad [M1]$$ $$36 - 28 + 4k = 0 \implies 8 + 4k = 0 \implies 4k = -8 \implies k = -2 \quad [A1]$$Part (b): Find coordinates of point of tangency:
Substitute $k = -2$ back into the quadratic equation:
$$x^2 - 6x + (7 - (-2)) = 0 \implies x^2 - 6x + 9 = 0 \quad [M1]$$ $$(x - 3)^2 = 0 \implies x = 3 \quad [A1]$$Substitute $x = 3$ into the line equation to find $y$:
$$y = 2(3) + (-2) = 6 - 2 = 4 \quad [A1]$$Answers: (a) $k = -2$; (b) Point of tangency is $(3, 4)$
10 Progressive Practice Exercises
Tier 1 tests vertex forms and discriminant evaluation; Tier 2 introduces parameters and inequalities; Tier 3 challenges strictly positive quadratics and real-world optimization models.
Tier 1 Core Fluency (Problems 1–3)
Convert $f(x) = -x^2 + 6x - 5$ into vertex form $a(x - h)^2 + k$. State the coordinates of the vertex and the maximum value of $f$.
View Full Worked Solution
Step 1: Complete the square:
$$f(x) = -(x^2 - 6x) - 5 = -\left[(x - 3)^2 - 9\right] - 5 = -(x - 3)^2 + 9 - 5 = -(x - 3)^2 + 4$$Step 2: State vertex and maximum:
The vertex is $(3, 4)$. Since $a = -1 < 0$, the maximum value is $4$.
Answer: $f(x) = -(x - 3)^2 + 4$; vertex $(3, 4)$; maximum value is $4$.
Evaluate the discriminant and state the number and nature of roots for each equation:
(a) $4x^2 - 12x + 9 = 0$
(b) $2x^2 - 3x + 5 = 0$
View Full Worked Solution
Part (a):
$$\Delta = (-12)^2 - 4(4)(9) = 144 - 144 = 0$$Since $\Delta = 0$, there is one repeated real root (or two equal real roots).
Part (b):
$$\Delta = (-3)^2 - 4(2)(5) = 9 - 40 = -31$$Since $\Delta = -31 < 0$, there are no real roots.
Solve the quadratic inequality $x^2 - 7x + 12 < 0$.
View Full Worked Solution
Step 1: Find critical values:
$$x^2 - 7x + 12 = 0 \implies (x - 3)(x - 4) = 0 \implies x = 3 \quad \text{or} \quad x = 4$$Step 2: Sign check:
Since $a = 1 > 0$, the parabola opens upwards. It lies strictly below zero between the roots.
Answer: $3 < x < 4$ (or $(3, 4)$).
Tier 2 Exam Modeling (Problems 4–7)
Find the exact values of the non-zero constant $k$ for which the equation $(k + 1)x^2 - 6x + 2 = 0$ has equal real roots.
View Full Worked Solution
Step 1: Set discriminant equal to zero:
$$a = k + 1, \quad b = -6, \quad c = 2$$ $$\Delta = b^2 - 4ac = (-6)^2 - 4(k + 1)(2) = 0$$Step 2: Solve for $k$:
$$36 - 8(k + 1) = 0 \implies 36 - 8k - 8 = 0 \implies 28 - 8k = 0$$ $$8k = 28 \implies k = \frac{28}{8} = \frac{7}{2} = 3.5$$Answer: $k = \frac{7}{2}$.
Find the range of values of $m$ for which the equation $x^2 + 4mx + (4m + 3) = 0$ has two distinct real roots.
View Full Worked Solution
Step 1: Condition for two distinct real roots is $\Delta > 0$:
$$\Delta = (4m)^2 - 4(1)(4m + 3) > 0$$ $$16m^2 - 16m - 12 > 0$$Step 2: Divide by 4:
$$4m^2 - 4m - 3 > 0$$Step 3: Factor the quadratic in $m$:
$$(2m - 3)(2m + 1) > 0$$Critical values: $m = \frac{3}{2}$ and $m = -\frac{1}{2}$.
Since the parabola in $m$ opens upwards, the expression is strictly positive outside the roots:
Answer: $m < -\frac{1}{2}$ or $m > \frac{3}{2}$.
A quadratic function $f(x)$ has its vertex at $(3, -8)$ and passes through $(1, 0)$.
(a) Find the equation of $f(x)$ in standard form $ax^2 + bx + c$.
(b) Solve the inequality $f(x) \ge 0$.
View Full Worked Solution
Part (a): Vertex form
$$f(x) = a(x - 3)^2 - 8$$Substitute $(1, 0)$:
$$0 = a(1 - 3)^2 - 8 \implies 0 = 4a - 8 \implies 4a = 8 \implies a = 2$$ $$f(x) = 2(x - 3)^2 - 8 = 2(x^2 - 6x + 9) - 8 = 2x^2 - 12x + 18 - 8 = 2x^2 - 12x + 10$$Standard form: $f(x) = 2x^2 - 12x + 10$.
Part (b): Solve $f(x) \ge 0$
$$2(x - 1)(x - 5) \ge 0$$Roots are $x = 1$ and $x = 5$. Since $a = 2 > 0$, the parabola is non-negative outside the roots:
Answer: $x \le 1$ or $x \ge 5$ (or $(-\infty, 1] \cup [5, \infty)$).
Find the value of $k$ for which the straight line $y = 2x + k$ is tangent to the curve $y = x^2 - 4x + 14$, and find the coordinates of the point of tangency.
View Full Worked Solution
Step 1: Form the intersection equation:
$$x^2 - 4x + 14 = 2x + k \implies x^2 - 6x + (14 - k) = 0$$Step 2: Set discriminant equal to zero:
$$\Delta = (-6)^2 - 4(1)(14 - k) = 0 \implies 36 - 56 + 4k = 0$$ $$-20 + 4k = 0 \implies 4k = 20 \implies k = 5$$Step 3: Point of tangency:
$$x = -\frac{-6}{2(1)} = 3$$ $$y = 2(3) + 5 = 11 \implies (3, 11)$$Answer: $k = 5$; point of tangency is $(3, 11)$.
Tier 3 Complex Applications & Synthesis (Problems 8–10)
Find the exact range of values of $k$ such that the quadratic expression $kx^2 + 4x + k > 0$ for all real numbers $x \in \mathbb{R}$.
View Full Worked Solution
Condition 1: Leading coefficient must be strictly positive:
$$k > 0$$Condition 2: Parabola must have no real roots ($\Delta < 0$):
$$\Delta = 4^2 - 4(k)(k) < 0 \implies 16 - 4k^2 < 0 \implies 4 - k^2 < 0$$ $$k^2 > 4 \implies k > 2 \quad \text{or} \quad k < -2$$Combine Condition 1 and Condition 2:
Since $k > 0$, we reject $k < -2$. Therefore $k > 2$.
Answer: $k > 2$ (or $(2, \infty)$).
The height $h(t)$ in meters of a flare fired into the air after $t$ seconds is modeled by:
$$h(t) = -5t^2 + 30t + 35, \quad t \ge 0$$
(a) Find the initial height from which the flare was fired.
(b) Express $h(t)$ in vertex form and state the maximum height attained and the time at which it occurs.
(c) Find the total time the flare remains in flight until it strikes the ground.
View Full Worked Solution
Part (a): Initial height
$$h(0) = -5(0)^2 + 30(0) + 35 = 35\text{ meters}$$Part (b): Vertex form
$$h(t) = -5(t^2 - 6t) + 35 = -5\left[(t - 3)^2 - 9\right] + 35 = -5(t - 3)^2 + 45 + 35 = -5(t - 3)^2 + 80$$The maximum height is $80\text{ meters}$, attained after $3\text{ seconds}$.
Part (c): Total flight time (set $h(t) = 0$)
$$-5t^2 + 30t + 35 = 0 \implies -5(t^2 - 6t - 7) = 0 \implies -5(t - 7)(t + 1) = 0$$Since $t \ge 0$, we reject $t = -1$. Therefore $t = 7\text{ seconds}$.
Answer: (a) $35\text{ m}$; (b) $80\text{ m}$ at $t = 3\text{ s}$; (c) $7\text{ seconds}$.
A rectangular paddock is enclosed using an existing stone wall as one boundary. Fencing of total length $60\text{ meters}$ is used for the remaining three sides.
(a) If the two sides perpendicular to the wall have length $x$ meters, show that the area $A(x) = 60x - 2x^2$.
(b) Find the dimensions that maximize the enclosed area, and state this maximum area.
(c) Find the range of possible values for $x$ if the area must be at least $400\text{ m}^2$.
View Full Worked Solution
Part (a): Area expression
Let width $= x$. The length parallel to the wall is $60 - 2x$.
$$\text{Area } A(x) = x(60 - 2x) = 60x - 2x^2$$Part (b): Maximum area via vertex
$$A(x) = -2(x^2 - 30x) = -2\left[(x - 15)^2 - 225\right] = -2(x - 15)^2 + 450$$Maximum area occurs at $x = 15\text{ m}$. Length $= 60 - 2(15) = 30\text{ m}$.
Maximum area $= 450\text{ m}^2$. Dimensions: $15\text{ m} \times 30\text{ m}$.
Part (c): Area at least $400\text{ m}^2$
$$60x - 2x^2 \ge 400 \implies 2x^2 - 60x + 400 \le 0 \implies x^2 - 30x + 200 \le 0$$ $$(x - 10)(x - 20) \le 0 \implies 10 \le x \le 20$$Answer: (a) $A(x) = 60x - 2x^2$; (b) $15\text{ m} \times 30\text{ m}$ with area $450\text{ m}^2$; (c) $10\text{ m} \le x \le 20\text{ m}$.
Quadratic Formulas
- Axis of Symmetry: $$x = -\frac{b}{2a}$$
- Quadratic Equation Roots: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
- Discriminant: $$\Delta = b^2 - 4ac$$
Forms, Discriminants & Inequalities
- Vertex Form: $$y = a(x - h)^2 + k \implies \text{Vertex } (h, k)$$
- Factorized Form: $$y = a(x - p)(x - q) \implies \text{Roots } p, q$$
- Discriminant Classification: $$\Delta > 0 \implies 2 \text{ distinct real roots}$$ $$\Delta = 0 \implies 1 \text{ repeated real root (tangent)}$$ $$\Delta < 0 \implies \text{no real roots (floating)}$$
- Strictly Positive Condition: $$ax^2 + bx + c > 0 \; \forall x \iff a > 0 \text{ and } \Delta < 0$$
Critical IB Examination Pitfalls
- Overlooking the "Real Roots" Inclusion ($\Delta \ge 0$): When a question asks for conditions so that an equation has "real roots", writing only $\Delta > 0$ loses method marks. You must use $\Delta \ge 0$.
- Sign Error in Axis of Symmetry: Writing $x = \frac{b}{2a}$ instead of $x = -\frac{b}{2a}$.
- Failing to Flip the Inequality Sign: When dividing or multiplying an inequality by a negative number, the direction of the inequality must be inverted.
- Incorrect Formatting of Disjoint Inequalities: For $x^2 - 5x + 6 > 0$, the solution is $x < 2$ or $x > 3$. Writing $2 > x > 3$ is mathematically invalid and penalized by examiners.
IB AA SL Topic Check: Quadratic Functions & Discriminant
Section A has diagnostic questions with feedback. Section B has Paper 1 and Paper 2 exam-style problems with markschemes.
Section A: Diagnostic Concept Check
Score: 0 / 4Question 1: What is the value of the discriminant for $3x^2 - 2x + 5 = 0$?
Question 2: What are the vertex coordinates of the parabola $y = 2(x + 4)^2 - 9$?
Question 3: What value of $k$ makes the equation $x^2 - 8x + k = 0$ have a single repeated real root?
Question 4: What is the solution set of the inequality $x^2 - 9 \le 0$?
Section B: Exam-Style Free Response Questions
Question 5: Vertex Form & Exact Roots
Consider the quadratic function $f(x) = 2x^2 - 8x + 3$.
(a) Write $f(x)$ in the form $a(x - h)^2 + k$. [3 marks]
(b) State the minimum value of $f(x)$ and its range. [1 mark]
(c) Find the solutions of $f(x) = 0$ in the exact radical form $\frac{p \pm \sqrt{q}}{r}$. [2 marks]
View Official Markscheme & Model Solution
Part (a) [3 Marks]:
$$f(x) = 2(x^2 - 4x) + 3 \quad \text{\bf (M1)}$$ $$f(x) = 2\left[(x - 2)^2 - 4\right] + 3 = 2(x - 2)^2 - 8 + 3 \quad \text{\bf (M1)}$$ $$f(x) = 2(x - 2)^2 - 5 \quad \text{\bf (A1)}$$Part (b) [1 Mark]:
Minimum value is $-5$; Range is $\{y \in \mathbb{R} \mid y \ge -5\}$. (A1)
Part (c) [2 Marks]:
$$2(x - 2)^2 - 5 = 0 \implies (x - 2)^2 = \frac{5}{2} \quad \text{\bf (M1)}$$ $$x - 2 = \pm\sqrt{\frac{5}{2}} = \pm\frac{\sqrt{10}}{2} \implies x = \frac{4 \pm \sqrt{10}}{2} \quad \text{\bf (A1)}$$Question 6: Line-Parabola Tangency Condition
The line $y = 2x + k$ intersects the parabola $y = x^2 - 4x + 10$.
(a) Find an expression for the discriminant of the resulting quadratic equation in terms of $k$. [2 marks]
(b) Find the value of $k$ for which the line is tangent to the parabola. [2 marks]
(c) For this value of $k$, find the coordinates of the point of contact. [3 marks]
View Official Markscheme & Model Solution
Part (a) [2 Marks]:
$$x^2 - 4x + 10 = 2x + k \implies x^2 - 6x + (10 - k) = 0 \quad \text{\bf (M1)}$$ $$\Delta = (-6)^2 - 4(1)(10 - k) = 36 - 40 + 4k = 4k - 4 \quad \text{\bf (A1)}$$Part (b) [2 Marks]:
$$\Delta = 0 \implies 4k - 4 = 0 \quad \text{\bf (M1)}$$ $$4k = 4 \implies k = 1 \quad \text{\bf (A1)}$$Part (c) [3 Marks]:
When $k = 1$:
$$x^2 - 6x + 9 = 0 \implies (x - 3)^2 = 0 \implies x = 3 \quad \text{\bf (M1)(A1)}$$ $$y = 2(3) + 1 = 7 \quad \text{\bf (A1)}$$Point of contact: $(3, 7)$.
Topic 2 Curriculum
All Functions Study Guides
About this guide
Independent SJMaths resource; not an official IB publication. Follows the AA syllabus first assessed in 2021 (final assessment: November 2028). Use the separate 2029 guide for the new course.
Check the IB Mathematics programme page and the official AA subject guide for the authoritative syllabus and examination-session requirements.