Gradient & Forms of Linear Equations
- Gradient (Slope) Definition: The gradient $m$ measures the rate of vertical rise per horizontal unit run through two distinct points $(x_1, y_1)$ and $(x_2, y_2)$: $$m = \frac{y_2 - y_1}{x_2 - x_1} \quad (x_2 \ne x_1)$$ A horizontal line has $m = 0$ ($y = c$). A vertical line has an undefined gradient ($x = k$).
- Three Standard Forms of a Line:
- Gradient-Intercept Form: $y = mx + c$, where $m$ is the gradient and $c$ is the $y$-intercept $(0, c)$.
- Point-Gradient Form: $y - y_1 = m(x - x_1)$, ideal when given a known point $(x_1, y_1)$ and slope $m$.
- General Form: $ax + by + d = 0$, where $a, b, d \in \mathbb{Z}$ and $a > 0$ by standard IB convention.
- Axis Intercepts:
- $x$-intercept: set $y = 0$ and solve for $x$ $\implies \left(-\frac{d}{a}, 0\right)$.
- $y$-intercept: set $x = 0$ and solve for $y$ $\implies \left(0, -\frac{d}{b}\right)$.
Step 1: Calculate the gradient $m$.
$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - (-2)}{-1 - 3} = \frac{6}{-4} = -\frac{3}{2}$$Step 2: Use point-gradient form with point $A(3, -2)$.
$$y - (-2) = -\frac{3}{2}(x - 3) \implies y + 2 = -\frac{3}{2}(x - 3)$$Step 3: Clear fractions by multiplying both sides by 2.
$$2(y + 2) = -3(x - 3) \implies 2y + 4 = -3x + 9$$Step 4: Rearrange into general form $ax + by + d = 0$ with $a > 0$.
$$3x + 2y + 4 - 9 = 0 \implies 3x + 2y - 5 = 0$$Answer: $3x + 2y - 5 = 0$, where $a = 3$, $b = 2$, and $d = -5$.
Parallel & Perpendicular Lines; Perpendicular Bisectors
- Parallel Lines: Two non-vertical lines $L_1$ and $L_2$ are parallel if and only if their gradients are equal: $$L_1 \parallel L_2 \iff m_1 = m_2$$
- Perpendicular Lines: Two non-vertical lines $L_1$ and $L_2$ are perpendicular if and only if the product of their gradients is $-1$: $$L_1 \perp L_2 \iff m_1 \cdot m_2 = -1 \iff m_2 = -\frac{1}{m_1}$$ Rule: The perpendicular gradient is the negative reciprocal of the original gradient.
- Perpendicular Bisector of a Segment $[AB]$:
A line that cuts the line segment $[AB]$ into two equal halves at a $90^\circ$ right angle.
- Find the midpoint $M = \left(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}\right)$.
- Find the gradient of $[AB]$: $m_{AB} = \frac{y_B - y_A}{x_B - x_A}$.
- Calculate the perpendicular gradient: $m_\perp = -\frac{1}{m_{AB}}$.
- Write the line equation passing through midpoint $M$ with gradient $m_\perp$.
Step 1: Find the coordinates of the midpoint $M$.
$$M = \left(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}\right) = \left(\frac{-2 + 4}{2}, \frac{5 + 1}{2}\right) = (1, 3)$$Step 2: Find the gradient of line segment $[AB]$.
$$m_{AB} = \frac{1 - 5}{4 - (-2)} = \frac{-4}{6} = -\frac{2}{3}$$Step 3: Determine the perpendicular gradient $m_\perp$.
$$m_\perp = -\frac{1}{m_{AB}} = -\frac{1}{-2/3} = \frac{3}{2}$$Step 4: Form the equation passing through $M(1, 3)$ with gradient $\frac{3}{2}$.
$$y - 3 = \frac{3}{2}(x - 1) \implies y - 3 = \frac{3}{2}x - \frac{3}{2}$$ $$y = \frac{3}{2}x - \frac{3}{2} + 3 = \frac{3}{2}x + \frac{3}{2}$$Answer: $y = \frac{3}{2}x + \frac{3}{2}$ (or $y = 1.5x + 1.5$).
Concept of a Function & Function Notation
- Mathematical Definition of a Function: A relation $f$ from a set $X$ to a set $Y$ is a function if each element $x \in X$ is mapped to exactly one element $y \in Y$.
- Vertical Line Test: A curve in the Cartesian plane represents a function if and only if no vertical line intersects the curve more than once.
- Function Notation: Written as $y = f(x)$ or $f: x \mapsto \dots$
- $x$ is the independent variable (input / argument).
- $f(x)$ represents the value of the function (output / image) at $x$.
- Evaluating Algebraic Expressions: To evaluate $f(g(x))$ or $f(x + h)$, replace every single occurrence of the variable $x$ in $f(x)$ with the entire bracketed expression.
(a) Evaluate $f(-3)$.
(b) Simplify the expression $\frac{f(x+h) - f(x)}{h}$ where $h \ne 0$.
Part (a): Evaluate $f(-3)$.
$$f(-3) = 2(-3)^2 - 3(-3) + 1 = 2(9) + 9 + 1 = 18 + 9 + 1 = 28$$Part (b): Expand $f(x+h)$.
$$f(x+h) = 2(x+h)^2 - 3(x+h) + 1 = 2(x^2 + 2xh + h^2) - 3x - 3h + 1 = 2x^2 + 4xh + 2h^2 - 3x - 3h + 1$$Subtract $f(x)$:
$$f(x+h) - f(x) = (2x^2 + 4xh + 2h^2 - 3x - 3h + 1) - (2x^2 - 3x + 1) = 4xh + 2h^2 - 3h$$Divide by $h$:
$$\frac{f(x+h) - f(x)}{h} = \frac{h(4x + 2h - 3)}{h} = 4x + 2h - 3 \quad (h \ne 0)$$Answer: (a) $28$; (b) $4x + 2h - 3$.
Determining Domain & Range
- Domain ($\text{dom } f$): The set of all permitted real input values $x$ for which the function produces a real number output.
- Division by Zero: Denominators must never be zero: $\frac{A(x)}{B(x)} \implies B(x) \ne 0$.
- Even Roots (Square Roots): Radicands must be non-negative: $\sqrt{g(x)} \implies g(x) \ge 0$.
- Strictly Positive Radicands in Denominators: $\frac{1}{\sqrt{g(x)}} \implies g(x) > 0$.
- Logarithms: Arguments must be strictly positive: $\ln(h(x)) \implies h(x) > 0$.
- Range ($\text{range } f$): The set of all resulting output values $y = f(x)$ generated as $x$ varies across the entire domain.
- For quadratics $f(x) = a(x - h)^2 + k$: if $a > 0$, range is $y \ge k$; if $a < 0$, range is $y \le k$.
- For square roots $f(x) = \sqrt{g(x)}$: principal root is non-negative, so range is $y \ge 0$.
Step 1: Establish the condition for the domain.
Because the square root is in the denominator, the expression inside the root must be strictly positive (cannot be negative, and cannot equal zero):
$$2x - 6 > 0 \implies 2x > 6 \implies x > 3$$Domain: $\{x \in \mathbb{R} \mid x > 3\}$ or $(3, \infty)$.
Step 2: Determine the range.
As $x \to 3^+$, $\sqrt{2x - 6} \to 0^+$, so $g(x) = \frac{4}{0^+} \to \infty$.
As $x \to \infty$, $\sqrt{2x - 6} \to \infty$, so $g(x) = \frac{4}{\infty} \to 0^+$.
Since the numerator 4 is strictly positive and the denominator is always positive and never zero, $g(x)$ can take any value strictly greater than 0, but never 0 or negative.
Range: $\{y \in \mathbb{R} \mid y > 0\}$ or $(0, \infty)$.
Perpendicular Bisectors & Point-Slope Formulation
(a) Find the coordinates of $M$, the midpoint of the line segment $[AB]$. [2 marks]
(b) Find the gradient of $[AB]$. [1 mark]
(c) Hence, find the equation of the perpendicular bisector of $[AB]$, writing your answer in the standard form $ax + by + d = 0$ where $a, b, d \in \mathbb{Z}$ and $a > 0$. [4 marks]
Part (a): Midpoint coordinates:
$$M = \left( \frac{-2 + 4}{2}, \frac{5 + 1}{2} \right) = \left( \frac{2}{2}, \frac{6}{2} \right) = (1, 3) \quad [M1][A1]$$Part (b): Gradient of $[AB]$:
$$m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 5}{4 - (-2)} = \frac{-4}{6} = -\frac{2}{3} \quad [A1]$$Part (c): Perpendicular bisector equation:
The perpendicular gradient satisfies $m_\perp \times m_{AB} = -1$ [M1]:
$$m_\perp = -\frac{1}{-2/3} = \frac{3}{2} \quad [A1]$$Apply point-slope formula through midpoint $M(1, 3)$:
$$y - 3 = \frac{3}{2}(x - 1) \quad [M1]$$Multiply both sides by $2$ to clear fractions:
$$2(y - 3) = 3(x - 1) \implies 2y - 6 = 3x - 3$$ $$3x - 2y + 3 = 0 \quad [A1]$$Answers: (a) $M(1, 3)$; (b) $m_{AB} = -\frac{2}{3}$; (c) $3x - 2y + 3 = 0$
10 Progressive Practice Exercises
Tier 1 tests core line algebra and function substitution; Tier 2 tackles perpendicular bisectors and domain constraints; Tier 3 challenges composite geometry and multi-step applications.
Tier 1 Core Fluency (Problems 1–3)
Find the equation of the line passing through $(-4, 1)$ with gradient $\frac{2}{3}$. Give your answer in the general form $ax + by + d = 0$, where $a, b, d \in \mathbb{Z}$ and $a > 0$.
View Full Worked Solution
Step 1: Use the point-gradient form $y - y_1 = m(x - x_1)$ with $(x_1, y_1) = (-4, 1)$ and $m = \frac{2}{3}$:
$$y - 1 = \frac{2}{3}(x - (-4)) \implies y - 1 = \frac{2}{3}(x + 4)$$Step 2: Clear the denominator by multiplying both sides by 3:
$$3(y - 1) = 2(x + 4) \implies 3y - 3 = 2x + 8$$Step 3: Rearrange into general form with $a > 0$:
$$2x - 3y + 8 + 3 = 0 \implies 2x - 3y + 11 = 0$$Answer: $2x - 3y + 11 = 0$.
Find the equation of the line parallel to $3x - 4y = 8$ that passes through the point $(2, -5)$. Give your answer in the form $y = mx + c$.
View Full Worked Solution
Step 1: Find the gradient of $3x - 4y = 8$ by rearranging to slope-intercept form:
$$4y = 3x - 8 \implies y = \frac{3}{4}x - 2$$The gradient is $m = \frac{3}{4}$.
Step 2: Since the line is parallel, its gradient is also $m = \frac{3}{4}$.
Step 3: Substitute into point-gradient form with point $(2, -5)$:
$$y - (-5) = \frac{3}{4}(x - 2) \implies y + 5 = \frac{3}{4}x - \frac{3}{2}$$ $$y = \frac{3}{4}x - \frac{3}{2} - 5 = \frac{3}{4}x - \frac{13}{2}$$Answer: $y = \frac{3}{4}x - \frac{13}{2}$ (or $y = 0.75x - 6.5$).
Let $f(x) = 3x^2 - 5x + 2$.
(a) Find $f(-1)$.
(b) Find the values of $x$ for which $f(x) = 0$.
View Full Worked Solution
Part (a): Substitute $x = -1$:
$$f(-1) = 3(-1)^2 - 5(-1) + 2 = 3(1) + 5 + 2 = 10$$Part (b): Solve $3x^2 - 5x + 2 = 0$ by factoring:
We need factors of $3 \times 2 = 6$ that sum to $-5$, which are $-2$ and $-3$:
$$(3x - 2)(x - 1) = 0$$ $$3x - 2 = 0 \implies x = \frac{2}{3} \quad \text{or} \quad x - 1 = 0 \implies x = 1$$Answer: (a) $f(-1) = 10$; (b) $x = \frac{2}{3}$ or $x = 1$.
Tier 2 Exam Modeling (Problems 4–7)
Line $L_1$ has equation $2x + 5y = 10$. Line $L_2$ is perpendicular to $L_1$ and passes through $(4, -1)$. Find the equation of line $L_2$ in the form $y = mx + c$.
View Full Worked Solution
Step 1: Find the gradient of line $L_1$:
$$5y = -2x + 10 \implies y = -\frac{2}{5}x + 2 \implies m_1 = -\frac{2}{5}$$Step 2: Find the perpendicular gradient $m_2$ using $m_1 m_2 = -1$:
$$m_2 = -\frac{1}{-2/5} = \frac{5}{2}$$Step 3: Write the equation of $L_2$ through $(4, -1)$:
$$y - (-1) = \frac{5}{2}(x - 4) \implies y + 1 = \frac{5}{2}x - 10 \implies y = \frac{5}{2}x - 11$$Answer: $y = \frac{5}{2}x - 11$.
The coordinates of two points are $P(1, 7)$ and $Q(5, -1)$. Find the equation of the perpendicular bisector of the line segment $[PQ]$ in the form $y = mx + c$.
View Full Worked Solution
Step 1: Find the midpoint of $[PQ]$.
$$M = \left(\frac{1 + 5}{2}, \frac{7 + (-1)}{2}\right) = \left(\frac{6}{2}, \frac{6}{2}\right) = (3, 3)$$Step 2: Find the gradient of $[PQ]$.
$$m_{PQ} = \frac{-1 - 7}{5 - 1} = \frac{-8}{4} = -2$$Step 3: Find the perpendicular gradient $m_\perp$.
$$m_\perp = -\frac{1}{-2} = \frac{1}{2}$$Step 4: Write the equation of the line through $(3, 3)$ with gradient $\frac{1}{2}$.
$$y - 3 = \frac{1}{2}(x - 3) \implies y = \frac{1}{2}x - \frac{3}{2} + 3 = \frac{1}{2}x + \frac{3}{2}$$Answer: $y = \frac{1}{2}x + \frac{3}{2}$ (or $y = 0.5x + 1.5$).
State the largest possible domain in $\mathbb{R}$ for the function $f(x) = \frac{\sqrt{3x + 12}}{x - 1}$.
View Full Worked Solution
Condition 1: Square root radicand must be non-negative.
$$3x + 12 \ge 0 \implies 3x \ge -12 \implies x \ge -4$$Condition 2: Denominator cannot be zero.
$$x - 1 \ne 0 \implies x \ne 1$$Combine both restrictions:
$$x \ge -4 \quad \text{and} \quad x \ne 1$$Answer: $\{x \in \mathbb{R} \mid x \ge -4, x \ne 1\}$ or $[-4, 1) \cup (1, \infty)$.
A function is defined by $f(x) = x^2 - 4x + 7$ for the restricted domain $1 \le x \le 5$. Find the exact range of $f$.
View Full Worked Solution
Step 1: Check for turning points (vertex) inside the domain.
Complete the square: $f(x) = (x - 2)^2 - 4 + 7 = (x - 2)^2 + 3$.
The vertex occurs at $x = 2$, which lies within the interval $[1, 5]$.
Since the parabola opens upwards ($a = 1 > 0$), the minimum value occurs at the vertex:
$$f_{\min} = f(2) = 3$$Step 2: Evaluate the function at the boundary endpoints.
$$f(1) = 1^2 - 4(1) + 7 = 4$$ $$f(5) = 5^2 - 4(5) + 7 = 25 - 20 + 7 = 12$$The maximum value occurs at $x = 5$, giving $f_{\max} = 12$.
Answer: The range is $[3, 12]$ (or $\{y \in \mathbb{R} \mid 3 \le y \le 12\}$).
Tier 3 Complex Applications & Synthesis (Problems 8–10)
Two perpendicular lines $L_1$ and $L_2$ intersect at point $P(2, 3)$. Line $L_1$ has an $x$-intercept at $(-4, 0)$.
(a) Find the equation of line $L_1$.
(b) Find the equation of line $L_2$.
(c) Find the coordinates of the $y$-intercept of line $L_2$.
View Full Worked Solution
Part (a): Equation of $L_1$
Line $L_1$ passes through $(-4, 0)$ and $(2, 3)$:
$$m_1 = \frac{3 - 0}{2 - (-4)} = \frac{3}{6} = \frac{1}{2}$$ $$y - 0 = \frac{1}{2}(x - (-4)) \implies y = \frac{1}{2}x + 2$$Part (b): Equation of $L_2$
Since $L_2 \perp L_1$, $m_2 = -\frac{1}{1/2} = -2$.
Line $L_2$ also passes through the intersection point $P(2, 3)$:
$$y - 3 = -2(x - 2) \implies y - 3 = -2x + 4 \implies y = -2x + 7$$Part (c): $y$-intercept of $L_2$
Set $x = 0$ in the equation of $L_2$: $y = -2(0) + 7 = 7$.
Answer: (a) $y = \frac{1}{2}x + 2$; (b) $y = -2x + 7$; (c) $(0, 7)$.
The points $A(-1, k)$, $B(2, 5)$, and $C(k+3, 11)$ lie on the same straight line (they are collinear). Find the possible values of the constant $k$.
View Full Worked Solution
Step 1: Set the gradients $m_{AB}$ and $m_{BC}$ equal.
$$m_{AB} = \frac{5 - k}{2 - (-1)} = \frac{5 - k}{3}$$ $$m_{BC} = \frac{11 - 5}{(k + 3) - 2} = \frac{6}{k + 1} \quad (k \ne -1)$$Step 2: Equate gradients and solve the resulting quadratic equation:
$$\frac{5 - k}{3} = \frac{6}{k + 1}$$ $$(5 - k)(k + 1) = 3 \times 6 = 18$$ $$5k + 5 - k^2 - k = 18 \implies -k^2 + 4k + 5 = 18$$ $$k^2 - 4k + 13 = 0$$Step 3: Check discriminant:
$$\Delta = (-4)^2 - 4(1)(13) = 16 - 52 = -36 < 0$$Because the discriminant is negative, there are no real solutions for $k$.
(If point $C$ is $(k+3, 10)$, then $(5-k)(k+1) = 15 \implies k^2 - 4k + 10 = 0$; if $(5-k)(k+1) = 8 \implies k^2 - 4k + 3 = 0 \implies k = 1$ or $k = 3$).
Answer: There are no real values of $k$ for which these three points are collinear (discriminant $\Delta = -36 < 0$).
A courier service calculates shipping costs $C(x)$ (in dollars) based on package distance $x$ (in km) using the rule:
$$C(x) = \begin{cases} 12 + 1.50x, & 0 \le x \le 20 \\ 22 + 1.00x, & x > 20 \end{cases}$$
(a) Show that $C(x)$ is continuous at $x = 20$.
(b) Find the range of $C(x)$ for deliveries up to $100\text{ km}$.
(c) A customer was billed $\$78.00$. Determine the distance of their delivery.
View Full Worked Solution
Part (a): Continuity at $x = 20$
For $0 \le x \le 20$: $C(20) = 12 + 1.50(20) = 12 + 30 = 42$.
As $x \to 20^+$: $\lim_{x \to 20^+} C(x) = 22 + 1.00(20) = 22 + 20 = 42$.
Since the left-hand value equals the right-hand limit ($42 = 42$), $C(x)$ is continuous at $x = 20$.
Part (b): Range for $0 \le x \le 100$
At $x = 0$: $C(0) = 12$.
At $x = 100$: $C(100) = 22 + 1.00(100) = 122$.
Since both pieces are strictly increasing linear functions, the range is $[12, 122]$.
Part (c): Distance for cost $\$78.00$
Since $\$78 > \$42$, the delivery distance falls in the second tier ($x > 20$):
$$22 + 1.00x = 78 \implies x = 78 - 22 = 56\text{ km}$$Answer: (a) Both pieces evaluate to $42$ at $x = 20$; (b) $[12, 122]$; (c) $56\text{ km}$.
Straight Lines & Coordinate Geometry
- Gradient Formula: $$m = \frac{y_2 - y_1}{x_2 - x_1}$$
- Forms of Linear Equations: $$y = mx + c$$ $$y - y_1 = m(x - x_1)$$ $$ax + by + d = 0$$
- Parallel & Perpendicular Slopes: $$\text{Parallel: } m_1 = m_2$$ $$\text{Perpendicular: } m_1 m_2 = -1 \iff m_2 = -\frac{1}{m_1}$$
Midpoint, Distance & Domain Boundaries
- Midpoint of Segment $[AB]$: $$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$
- Euclidean Distance: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
- Universal Domain Restrictions: $$\frac{1}{g(x)} \implies g(x) \ne 0$$ $$\sqrt{g(x)} \implies g(x) \ge 0$$ $$\frac{1}{\sqrt{g(x)}} \implies g(x) > 0$$
Critical IB Examination Pitfalls
- Direction Inconsistency in Gradient Calculation: Mixing subtraction order like $\frac{y_2 - y_1}{x_1 - x_2}$ inverts the sign of the gradient. Maintain identical point ordering in numerator and denominator.
- Fractional General Form: If an exam question asks for $ax + by + d = 0$ with $a, b, d \in \mathbb{Z}$, leaving fractions such as $\frac{2}{3}x - y + 4 = 0$ loses the final answer mark ($A0$). Clear denominators!
- Confusing Domain Boundary for Denominator Roots: For $f(x) = \frac{1}{\sqrt{x - 3}}$, the condition is $x - 3 > 0$ ($x > 3$), NOT $x \ge 3$. Including $x = 3$ causes division by zero.
- Evaluating Range Only at Domain Endpoints: If finding the range of a quadratic or non-linear function over an interval $[a, b]$, always check if the local minimum or maximum (vertex) lies between $a$ and $b$.
IB AA SL Topic Check: Straight Lines & Functions
Section A tests instant diagnostic recognition with automated feedback; Section B assesses multi-mark Paper 1 and Paper 2 style exam problems with full markscheme breakdown.
Section A: Diagnostic Concept Check
Score: 0 / 4Question 1: Line $L$ has equation $3x - 6y = 5$. What is the gradient of any line perpendicular to $L$?
Question 2: What are the coordinates of the midpoint of the line segment joining $A(3, -4)$ and $B(-7, 8)$?
Question 3: What is the largest possible real domain of $f(x) = \frac{1}{\sqrt{5 - x}}$?
Question 4: What is the range of the function $f(x) = (x - 3)^2 + 2$ for $x \in \mathbb{R}$?
Section B: Exam-Style Free Response Questions
Question 5: Perpendicular Bisector
The coordinates of two points are $A(1, 3)$ and $B(7, -1)$.
(a) Find the coordinates of the midpoint of $[AB]$. [2 marks]
(b) Find the equation of the perpendicular bisector of $[AB]$ in the form $ax + by + d = 0$, where $a, b, d \in \mathbb{Z}$ and $a > 0$. [4 marks]
View Official Markscheme & Model Solution
Part (a) [2 Marks]:
$$M = \left(\frac{1 + 7}{2}, \frac{3 + (-1)}{2}\right) \quad \text{\bf (M1)}$$ $$M = (4, 1) \quad \text{\bf (A1)}$$Part (b) [4 Marks]:
Gradient of $[AB]$:
$$m_{AB} = \frac{-1 - 3}{7 - 1} = \frac{-4}{6} = -\frac{2}{3} \quad \text{\bf (A1)}$$Perpendicular gradient:
$$m_\perp = -\frac{1}{-2/3} = \frac{3}{2} \quad \text{\bf (M1)}$$Equation of line passing through $(4, 1)$:
$$y - 1 = \frac{3}{2}(x - 4) \implies 2(y - 1) = 3(x - 4) \quad \text{\bf (M1)}$$ $$2y - 2 = 3x - 12 \implies 3x - 2y - 10 = 0 \quad \text{\bf (A1)}$$Answer: $3x - 2y - 10 = 0$.
Question 6: Rational Function Domain & Line Intersection
Consider the function $f(x) = \frac{6}{2x - 5}$.
(a) State the largest possible domain of $f$. [1 mark]
(b) Find the range of $f$ for the restricted domain $3 \le x \le 8$. [3 marks]
(c) A line $L$ with gradient $m = -3$ intersects the graph of $f(x)$ at the point where $x = 4$. Find the equation of line $L$ in the form $y = mx + c$. [3 marks]
View Official Markscheme & Model Solution
Part (a) [1 Mark]:
$$2x - 5 \ne 0 \implies x \ne 2.5 \quad (\text{or } x \ne 5/2) \quad \text{\bf (A1)}$$Part (b) [3 Marks]:
For $x \ge 3$, the denominator $2x - 5 > 0$ and is strictly increasing. Therefore, $f(x)$ is strictly decreasing on $[3, 8]$. (M1)
$$f(3) = \frac{6}{2(3) - 5} = \frac{6}{1} = 6 \quad \text{\bf (A1)}$$ $$f(8) = \frac{6}{2(8) - 5} = \frac{6}{11} \approx 0.545 \quad \text{\bf (A1)}$$Range is $\left[\frac{6}{11}, 6\right]$.
Part (c) [3 Marks]:
At $x = 4$:
$$y = f(4) = \frac{6}{2(4) - 5} = \frac{6}{3} = 2 \quad \text{\bf (A1)}$$Line $L$ has gradient $m = -3$ and passes through $(4, 2)$:
$$y - 2 = -3(x - 4) \quad \text{\bf (M1)}$$ $$y - 2 = -3x + 12 \implies y = -3x + 14 \quad \text{\bf (A1)}$$Answer: $y = -3x + 14$.
Topic 2 Curriculum
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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.