DP Maths AA SL - Geometry and Trigonometry - Coordinate Geometry

DP Maths AA SL - Geometry and Trigonometry - Coordinate Geometry - Full notes

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DP Maths AA SL | Coordinate Geometry

Coordinate Geometry

This topic contains 1 note:

2.1 Lines and Functions

DP MATHEMATICS AA · TOPIC 2: FUNCTIONS

2.1

Lines and Functions

Full notes for SL · Syllabus SL 2.1, 2.2, 2.3, 2.4

Contents

1	Gradient and equations of straight lines
2	Parallel and perpendicular lines
3	Distance, midpoint and intersections of lines
4	The concept of a function
5	Domain and range
6	Graphs of functions: drawing and sketching
7	Key features of graphs
8	Solving equations graphically
9	Forms of the equation of a line
10	Parallel and perpendicular lines in practice
11	Distances and midpoints in coordinate geometry
12	Perpendicular bisectors and circles
13	Linear models in context
14	Functions: notation, domain and range
15	Graphs and key features with technology
16	Solving equations graphically
17	Mixed review: worked problems
18	Additional exam-style practice
19	The angle a line makes with the axes
20	Exam-style questions with markscheme
21	Common misconceptions and links
22	Check your understanding
23	Summary and topic map
24	Answers to practice exercises

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and work through every worked example before attempting the Quick check.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
find the gradient of a line and interpret it as a rate of change
write the equation of a line in gradient-intercept, point-gradient and general forms
use gradients to identify parallel and perpendicular lines
find distances, midpoints and points of intersection
define a function, use function notation and apply the vertical line test
state the domain and range of a function using correct notation
draw and sketch graphs, identifying intercepts, turning points, asymptotes and symmetry
solve equations graphically, including with technology

This booklet covers SL 2.1, 2.2, 2.3, 2.4, which is common to SL and HL.

Guiding questions

How can relationships between quantities be described with functions? What features of a graph tell us most about a function?

How this topic is examined

Lines and basic function concepts are tested throughout both papers, often as the opening parts of longer questions. Domain and range notation must be precise, and sketches must show key features clearly.

When sketching, label intercepts, turning points and asymptotes, and show the correct shape and end behaviour. On Paper 2, give answers read from a GDC to 3 significant figures.

Command term	What the examiner wants
Write down	an answer with little or no working; it should be obtainable by inspection
Find / Calculate / Determine	obtain the answer showing relevant working
Solve	find the values of the unknowns, using algebra or technology as allowed
Show that	obtain the given result; every step must be shown and the result must not be assumed
Hence / Hence or otherwise	use the previous result (or another valid method)
Sketch	a diagram showing the general shape and key features, labelled
Prove	a formal, rigorous argument from accepted facts
Justify / Explain	give valid reasons or evidence

How to use these notes

Each section teaches one part of the syllabus. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or proof of a key result, so you understand where formulas come from
Key point	a result or idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
GDC tip	how to use your graphic display calculator efficiently on Paper 2
In the real world	where the mathematics is used outside the classroom
Worked example	a full solution with IB-style marks (M1 method, A1 answer, R1 reasoning, AG answer given); examples are graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet
Quick check	short questions to check the key ideas

1 Gradient and equations of straight lines

Gradient and a point fix a line.

Think about it

A taxi charges a fixed $4 plus $2.50 per km. What do the 4 and the 2.50 represent on a graph of cost against distance? What would the graph look like if the fixed charge were removed?

The gradient of the line through (x1, y1) and (x2, y2) is the change in y divided by the change in x:

A positive gradient slopes upward from left to right; negative slopes downward; zero is horizontal; a vertical line has undefined gradient. In context, the gradient is a rate of change: in C = 2.5d + 4, the cost rises by $2.50 per km.

There are three standard forms of the equation of a line:

Gradient-intercept form y = mx + c shows the gradient m and the y-intercept c. Point-gradient form is quickest when you know a point and the gradient. General form ax + by + d = 0 (usually with integer a, b, d) is often requested in exam questions. Vertical lines have equations x = k; horizontal lines y = k.

WORKED EXAMPLE 1 Equation through two points (basic)

Find the equation of the line through (2, −1) and (6, 7), in the form ax + by + d = 0.

Solution

m = (7 − (−1)) ÷ (6 − 2) = 2A1

y + 1 = 2(x − 2)M1

2x − y − 5 = 0A1

WORKED EXAMPLE 2 Interpreting a linear model (standard)

The temperature T (°C) of an oven t minutes after switching on is modelled by T = 18t + 20 for 0 ≤ t ≤ 10. Interpret 18 and 20, and find when T = 155.

Solution

18: the temperature rises 18 °C per minute; 20: initial (room) temperatureA1 A1

18t + 20 = 155, t = 7.5 minutesA1

WORKED EXAMPLE 3 Gradient from general form (standard)

Find the gradient and intercepts of 3x + 4y = 24.

Solution

4y = −3x + 24, so y = −0.75x + 6: m = −3/4A1

y-intercept (0, 6); x-intercept (8, 0)A1

Practice exercise 1

1. Find the gradient of the line through (−3, 5) and (1, −3).

2. Find the equation of the line with gradient 2/3 through (3, −1), in the form ax + by + d = 0.

3. Find the x- and y-intercepts of 5x − 2y = 20.

4. A plumber charges $45 call-out plus $30 per hour. Write a formula and find the cost of 3.5 hours.

5. Find k if the line through (1, k) and (4, 10) has gradient 3.

2 Parallel and perpendicular lines

Equal gradients; negative reciprocal gradients.

Parallel lines have equal gradients: m1 = m2. Perpendicular lines have gradients whose product is −1: m1m2 = −1, so m2 = −1/m1 (the negative reciprocal). A horizontal line is perpendicular to a vertical line.

Why it works

Rotating a line through 90° turns a 'run' of 1 and 'rise' of m into a 'run' of −m and 'rise' of 1. The new gradient is 1/(−m) = −1/m, so the product of the gradients is −1.

WORKED EXAMPLE 4 Perpendicular bisector (standard)

Find the equation of the perpendicular bisector of A(1, 3) and B(5, 1).

Solution

Midpoint (3, 2); gradient AB = −2/4 = −½M1

Perpendicular gradient 2A1

y − 2 = 2(x − 3): y = 2x − 4A1

WORKED EXAMPLE 5 Showing a right angle (exam level)

Show that the triangle with vertices P(0, 0), Q(4, 2) and R(1, 8) has a right angle.

Solution

mPQ = 1/2; mQR = 6/(−3) = −2; mPR = 8M1

mPQ × mQR = −1A1

So PQ ⊥ QR: the right angle is at QR1

Practice exercise 2

1. Find the gradient of a line perpendicular to y = −3x + 1.

2. Are 2x + 4y = 7 and y = −½x + 3 parallel?

3. Find the equation of the line through (2, 5) perpendicular to 3x − y = 4.

4. For what k are the lines y = kx + 1 and 4x − 2y = 3 perpendicular?

3 Distance, midpoint and intersections of lines

Coordinate geometry tools.

The distance formula is Pythagoras' theorem applied to the horizontal and vertical changes. The midpoint is the average of the coordinates. The intersection of two lines is found by solving their equations simultaneously.

WORKED EXAMPLE 6 Intersection (standard)

Find where y = 2x − 1 meets 3x + 2y = 12.

Solution

3x + 2(2x − 1) = 12, so 7x = 14M1

x = 2, y = 3: (2, 3)A1

Practice exercise 3

1. Find the distance between (−2, 3) and (4, −5).

2. Find the midpoint of (−1, 7) and (5, −3).

3. Find the intersection of x + y = 5 and 2x − y = 1.

4 The concept of a function

One input, exactly one output.

A function is a rule that assigns to each input exactly one output. We write f(x) = 2x + 3, or f: x ↦ 2x + 3. The value f(4) = 11 is the output when the input is 4; f(a + 1) = 2a + 5 means substituting a + 1 for x.

A mapping diagram shows inputs linked to outputs. A relation is a function if no input is linked to two outputs. On a graph, the vertical line test checks this: a graph represents a function if every vertical line meets it at most once. The circle x2 + y2 = 4 is not a function, because x = 0 gives y = ±2.

A function is one-to-one if each output comes from exactly one input (f(x) = 2x + 3), and many-to-one if some outputs come from several inputs (f(x) = x2, since f(2) = f(−2) = 4). This distinction matters for inverse functions (2.5).

WORKED EXAMPLE 7 Function notation (basic)

Given f(x) = 3x2 − 1, find (a) f(−2); (b) f(2a); (c) x such that f(x) = 26.

Solution

(a) 3(4) − 1 = 11A1

(b) 3(4a2) − 1 = 12a2 − 1A1

Practice exercise 4

1. If g(x) = 5 − 2x, find g(−3) and solve g(x) = −7.

2. If f(x) = x2 − 4x, find f(x + 1) and simplify.

3. Is y2 = x a function of x? Explain.

4. Is f(x) = x3 one-to-one or many-to-one?

5 Domain and range

Allowed inputs and possible outputs.

The domain is the set of input values for which the function is defined (or which are allowed in context). The range is the set of output values the function actually produces.

The natural domain is restricted by: division by zero (1/x needs x ≠ 0); square roots of negatives (√x needs x ≥ 0); logarithms of non-positive numbers (ln x needs x > 0); and context (time t ≥ 0, number of items must be a whole number).

To find the range, consider the smallest and largest outputs: use the vertex of a quadratic, the asymptote of a rational or exponential function, or a sketch.

Notation. Inequality: x ∈ ℝ, x ≥ 2. Interval: [2, ∞), where square brackets include the endpoint and round brackets exclude it. Set-builder: {x | x > 0}. Ranges are written in terms of the output: f(x) ≥ 3 or y ≥ 3.

Function	Domain	Range
f(x) = 2x + 1	x ∈ ℝ	f(x) ∈ ℝ
f(x) = x2	x ∈ ℝ	f(x) ≥ 0
f(x) = √(x − 3)	x ≥ 3	f(x) ≥ 0
f(x) = 1/x	x ≠ 0	f(x) ≠ 0
f(x) = ln x	x > 0	f(x) ∈ ℝ
f(x) = ex	x ∈ ℝ	f(x) > 0
f(x) = (x − 1)2 + 4	x ∈ ℝ	f(x) ≥ 4

WORKED EXAMPLE 8 Domain and range of a root function (standard)

Find the domain and range of f(x) = 3 + √(5 − x).

Solution

5 − x ≥ 0: domain x ≤ 5A1

√(5 − x) ≥ 0, so range f(x) ≥ 3A1

WORKED EXAMPLE 9 Restricted domain (exam level)

f(x) = x2 − 4x + 1 for 0 ≤ x ≤ 5. Find the range.

Solution

Vertex at x = 2: f(2) = −3 (minimum)M1

Endpoints: f(0) = 1, f(5) = 6A1

Range −3 ≤ f(x) ≤ 6A1

With a restricted domain, check the vertex and both endpoints.

Common mistake

Giving the range in terms of x. The range is a set of output values: write f(x) ≥ 3 or y ≥ 3, not x ≥ 3.

Practice exercise 5

1. State the domain of f(x) = 1/(x + 4).

2. State the domain and range of g(x) = √(2x − 6) + 1.

3. State the domain and range of h(x) = ln(x − 2).

4. Find the range of f(x) = 2x + 3 for −1 ≤ x < 4.

5. Find the range of f(x) = 5 − (x + 2)2.

6 Graphs of functions: drawing and sketching

Draw versus sketch.

To draw a graph means to plot it accurately, with labelled scales on both axes, either by plotting points or by copying a GDC display with a suitable window. To sketch means to show the general shape and the key features, labelled, without needing exact scale.

On Paper 2, use the GDC to graph functions and find features. On Paper 1 you must sketch using algebra: find intercepts by setting x = 0 and y = 0, find turning points by completing the square or calculus, and identify asymptotes from the equation.

Figure 1. A sketch shows intercepts and turning points.

GDC tip

Choose a window that shows all key features. If a graph looks like a straight line or disappears, zoom out; if turning points merge, zoom in. Use the 'zero', 'maximum', 'minimum' and 'intersect' functions rather than tracing by eye.

7 Key features of graphs

What to look for and label.

Intercepts: the y-intercept is f(0); the x-intercepts (zeros or roots) are the solutions of f(x) = 0.

Turning points: local maxima and minima, where the graph changes from increasing to decreasing or vice versa.

Asymptotes: lines that the graph approaches ever more closely. Vertical asymptotes occur where the function is undefined and the output grows without bound (x = 0 for 1/x). Horizontal asymptotes describe the value approached as x → ±∞ (y = 0 for e−x).

Symmetry: about the y-axis (even functions such as x2) or about the origin (odd functions such as x3). End behaviour: what happens as x → ±∞.

Figure 2. Vertical asymptote x = 2 and horizontal asymptote y = 1.

WORKED EXAMPLE 10 Identifying features (standard)

For f(x) = (x − 1)(x + 3), find the intercepts, the axis of symmetry and the vertex.

Solution

x-intercepts (1, 0) and (−3, 0); y-intercept (0, −3)A1

Axis x = (1 + (−3))/2 = −1A1

Vertex (−1, f(−1)) = (−1, −4)A1

Practice exercise 6

1. Write down the asymptotes of y = 3/(x + 1) − 2.

2. Find the intercepts of y = x2 − x − 6.

3. Use a GDC to find the coordinates of the turning points of y = x3 − 6x2 + 9x + 1.

8 Solving equations graphically

Where two graphs meet.

The solutions of f(x) = g(x) are the x-coordinates of the points of intersection of y = f(x) and y = g(x). Equivalently, they are the zeros of y = f(x) − g(x). On Paper 2, equations that cannot be solved algebraically, such as ex = x + 2, are solved with the GDC's intersect function.

WORKED EXAMPLE 11 Solving with technology (standard)

Solve ex = x + 2.

Solution

Graph y = ex and y = x + 2; find intersectionsM1

x = −1.84 or x = 1.15 (3 s.f.)A1 A1

Figure 3. Two points of intersection.

WORKED EXAMPLE 12 Number of solutions (exam level)

By sketching, determine how many solutions x3 − 3x = k has for k = 1 and for k = 3.

Solution

y = x3 − 3x has max (−1, 2) and min (1, −2)M1

k = 1 lies between −2 and 2: three solutionsA1

k = 3 > 2: one solutionA1

Practice exercise 7

1. Solve ln x = 3 − x.

2. Solve x2 = 2x for x < 0.

3. For what values of k does x3 − 3x = k have exactly two solutions?

9 Forms of the equation of a line

Choosing the most useful form.

The gradient-intercept form y = mx + c shows the gradient and y-intercept; the point-gradient form y − y1 = m(x − x1) is quickest from a point and a gradient; the general form ax + by + d = 0 (integers a, b, d) is often required in answers. Converting between them is routine algebra.

WORKED EXAMPLE 13 Converting (basic)

Write y − 3 = −(2/5)(x + 1) in the form ax + by + d = 0.

Solution

5y − 15 = −2x − 2M1

2x + 5y − 13 = 0A1

WORKED EXAMPLE 14 Intercepts (standard)

Find the intercepts and gradient of 3x − 4y + 12 = 0.

Solution

x-intercept (−4, 0); y-intercept (0, 3)A1

Gradient 3/4A1

Practice exercise 8

1. Find the line with gradient −3 through (2, −1) in general form.

2. Find the gradient and y-intercept of 2x + 7y = 14.

10 Parallel and perpendicular lines in practice

Gradients tell the story.

Parallel lines have equal gradients; perpendicular lines have gradients whose product is −1. A horizontal line (gradient 0) is perpendicular to a vertical line (undefined gradient).

WORKED EXAMPLE 15 Perpendicular through a point (standard)

Find the line through (4, 1) perpendicular to 2x + 3y = 6.

Solution

Gradient of given line −2/3; perpendicular 3/2A1

y − 1 = (3/2)(x − 4): 3x − 2y − 10 = 0M1 A1

WORKED EXAMPLE 16 Unknown coefficient (standard)

Find k if kx + 2y = 5 is perpendicular to y = 4x − 1.

Solution

Gradient −k/2 = −1/4M1

k = 1/2A1

Practice exercise 9

1. Find the line through (−2, 5) parallel to y = −3x + 1.

2. Show that A(1, 1), B(4, 5), C(8, 2) form a right angle at B.

11 Distances and midpoints in coordinate geometry

Pythagoras on a grid.

WORKED EXAMPLE 17 Isosceles triangle (standard)

Show that P(1, 2), Q(5, 4), R(3, 8) form an isosceles triangle.

Solution

PQ = √20, QR = √20, PR = √40A1 A1

PQ = QR, so isoscelesR1

Practice exercise 10

1. Find the perimeter of the triangle with vertices (0, 0), (6, 0), (6, 8).

2. Find the point dividing (2, 3) to (8, 9) in the ratio 1 : 2.

12 Perpendicular bisectors and circles

Equidistant points.

The perpendicular bisector of AB passes through the midpoint and is perpendicular to AB; every point on it is equidistant from A and B. The perpendicular bisectors of the sides of a triangle meet at the centre of the circle through the three vertices.

WORKED EXAMPLE 18 Circumcentre (exam level)

Find the centre of the circle through (0, 0), (8, 0) and (0, 6).

Solution

Bisector of the first side: x = 4; of the second: y = 3M1 A1

Centre (4, 3), radius 5A1

Practice exercise 11

1. Find the perpendicular bisector of (−1, 4) and (5, 2).

2. Find the centre of the circle through (1, 1), (5, 1), (1, 7).

13 Linear models in context

Gradient and intercept interpreted.

A linear model y = mx + c describes a quantity that changes at a constant rate m from an initial value c. Interpret both in context, with units, and state a sensible domain.

WORKED EXAMPLE 19 Phone plan (standard)

A plan costs $15 per month plus $0.08 per minute. Write the monthly cost and interpret the gradient.

Solution

C = 15 + 0.08mA1

Each extra minute costs 8 centsR1

WORKED EXAMPLE 20 Temperature conversion (standard)

Water freezes at 0 °C = 32 °F and boils at 100 °C = 212 °F. Find F in terms of C and convert 37 °C.

Solution

Gradient 180/100 = 1.8: F = 1.8C + 32M1 A1

37 °C = 98.6 °FA1

Practice exercise 12

1. A candle is 24 cm long and burns 1.5 cm per hour. Write its length after t hours and the time it burns out.

2. Find where the plans C = 15 + 0.08m and C = 25 + 0.04m cost the same.

14 Functions: notation, domain and range

Inputs and outputs.

A function maps each input in its domain to exactly one output. State domains by considering division by zero, square roots of negative numbers and logarithms of non-positive numbers. Find ranges from the graph, including turning points and asymptotes.

Function	Domain	Range
√(4 − x)	x ≤ 4	y ≥ 0
1/(x + 2)	x ≠ −2	y ≠ 0
x2 − 6x + 10	x ∈ ℝ	y ≥ 1
ln(x − 1)	x > 1	y ∈ ℝ

WORKED EXAMPLE 21 Range on a domain (standard)

Find the range of f(x) = 2x2 − 8x + 3 for 0 ≤ x ≤ 5.

Solution

Vertex (2, −5)A1

f(0) = 3, f(5) = 13A1

−5 ≤ f(x) ≤ 13A1

Practice exercise 13

1. State the domain and range of f(x) = 3 + √(x + 1).

2. Find the range of g(x) = 1/x for x ≥ 2.

15 Graphs and key features with technology

Intercepts, extrema and asymptotes.

Use the GDC to find zeros, maxima, minima and intersections, and to sketch graphs. When you transfer a graph to paper, label the axes and the key points with coordinates and draw asymptotes as dashed lines.

Figure 4. A cubic with a local maximum and minimum.

WORKED EXAMPLE 22 GDC features (standard)

Find the zeros of y = x3 − 3x + 1.

Solution

x = −1.88, 0.347, 1.53A2

Practice exercise 14

1. Find the intersections of y = x2 and y = 2x.

2. Find the maximum of y = xe−x.

16 Solving equations graphically

Where graphs meet.

The solutions of f(x) = g(x) are the x-coordinates of the intersections of y = f(x) and y = g(x). This is the main method for equations that cannot be solved algebraically, such as ex = 3 − x or ln x = x − 2.

WORKED EXAMPLE 23 Graphical solution (standard)

Solve ln x = x − 2.

Solution

Graph both sidesM1

x = 0.159 and x = 3.15A1 A1

Practice exercise 15

1. Solve ex = 3 − x.

2. Solve x3 = 2x + 1.

17 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 24 Mixed problem (exam level)

The line L passes through A(2, 5) and B(6, −3). (a) Find its equation. (b) Find the line through C(0, −2) perpendicular to L. (c) Find the intersection.

Solution

(a) gradient −2: y = −2x + 9A1

(b) y = x/2 − 2A1

Practice exercise 16

1. Find the shortest distance from C(0, −2) to L in the example.

2. Find k if (k, 3) lies on the perpendicular bisector of (1, 1) and (5, 5).

18 Additional exam-style practice

Questions in the style of Papers 1 and 2.

WORKED EXAMPLE 25 Paper 1 question (exam level)

f(x) = (2x − 1)/(x + 3). (a) State the domain. (b) Find the intercepts. (c) Find the asymptotes.

Solution

(a) x ≠ −3A1

(b) (0.5, 0) and (0, −1/3)A1 A1

Practice exercise 17

1. Find the area of the triangle formed by 2x + y = 8 and the axes.

2. Find the angle between y = 2x and the x-axis.

19 The angle a line makes with the axes

Gradient as tan θ.

The gradient m of a line equals tan θ, where θ is the angle the line makes with the positive x-axis. The acute angle between two lines with gradients m1 and m2 follows from the difference of these angles.

WORKED EXAMPLE 26 Angle between lines (exam level)

Find the acute angle between y = 2x + 1 and y = −x + 4.

Solution

Angles with the x-axis: 63.4° and 135°M1 A1

Difference 71.6°: acute angle 71.6°A1

Practice exercise 18

1. Find the gradient of a line at 30° to the x-axis.

2. Find the acute angle between y = 3x and y = x/3.

20 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 1 style) (about 15 minutes)

Points A(−2, 5) and B(4, −1). The function f(x) = 2 + √(x + 4).

(a) Find the equation of line AB. [2]

(b) Find the equation of the line perpendicular to AB through B. [2]

(d) Find the x-coordinate where y = f(x) meets line AB. [4]

Markscheme

Part	Answer	Marks
(a)	m = −1	A1
	y = −x + 3	A1
(b)	gradient 1	A1
	y = x − 5	A1
(c)	x ≥ −4	A1
	f(x) ≥ 2	A1
(d)	2 + √(x + 4) = 3 − x	M1
	x + 4 = (1 − x)2: x2 − 3x − 3 = 0	A1
	x = (3 ± √21)/2	A1
	x = (3 − √21)/2 only, since 1 − x ≥ 0	R1

Examiner's comments. In (d), squaring introduces a spurious root that must be rejected with a reason.

Question 2 (Paper 2 style) (about 10 minutes)

Consider f(x) = x3 − 4x2 + x + 6 and g(x) = 2x.

(a) Find the zeros of f. [2]

(b) Find the coordinates of the local maximum of f. [2]

Markscheme

Part	Answer	Marks
(a)	x = −1, 2, 3	A2
(b)	(0.131, 6.06)	A2
(c)	intersections by GDC	M1
	x = −0.956, 1.35, 4.69	A2

Examiner's comments. Answers from a GDC should be given to 3 s.f.; listing all intersections in (c) was essential.

21 Common misconceptions and links

Students often think…	Actually…
perpendicular gradients are reciprocals	negative reciprocals
the range is given in terms of x	the range describes outputs
a sketch must be to scale	a sketch shows shape and key features
every graph is a function	use the vertical line test

How 2.1 connects to the rest of the course

Topic	Connection
2.5 Composite and inverse functions	domain and range
2.6 Quadratics	key features
3.1 Geometry	distance and midpoint in 3D
5.1 Differentiation	tangents and normals

22 Check your understanding

Section A: multiple choice

1. The gradient of 3x + 2y = 6 is

A. −3/2 B. 3/2 C. 3 D. −2/3

2. Perpendicular to gradient 4:

A. 1/4 B. −4 C. −1/4 D. 4

3. The domain of ln(x − 1) is

A. x ≥ 1 B. x > 1 C. x > 0 D. all reals

4. The range of x2 + 3 is

A. f(x) ≥ 0 B. all reals C. f(x) > 3 D. f(x) ≥ 3

5. A vertical line meets a function's graph

A. exactly twice B. at most once C. never D. infinitely often

6. The y-intercept of y = 2x − 7 is

A. −7 B. 2 C. 7 D. 3.5

Section B: structured

7. Show that the lines 2x − 3y = 1 and 3x + 2y = 4 are perpendicular. [2]

8. Find the domain and range of f(x) = 1/(x + 3) − 2. [2]

9. Sketch y = x(x − 3)2, labelling intercepts. [3]

Answers

1. A 2. C 3. B 4. D 5. B 6. A

7. Gradients 2/3 and −3/2: product −1.

8. Domain x ≠ −3; range f(x) ≠ −2.

9. Intercepts (0, 0) crossing and (3, 0) touching; positive cubic shape.

23 Summary and topic map

Idea	Remember
Lines	y = mx + c; y − y1 = m(x − x1); ax + by + d = 0
Parallel / perpendicular	m1 = m2; m1m2 = −1
Distance, midpoint	Pythagoras; averages
Function	each input has exactly one output
Domain and range	inputs and outputs; restrictions
Key features	intercepts, turning points, asymptotes, symmetry
Intersections	solve f(x) = g(x)

Before the exam, make sure you…

•give ranges in terms of f(x)
•label key features on sketches
•use technology for intersections on Paper 2
•check for spurious roots

Lines and Functions on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 5. Topic map for Lines and Functions.

24 Answers to practice exercises

Check your working against these answers. Full methods are expected in the exam.

Practice exercise 1

1. −2

2. 2x − 3y − 9 = 0

3. (4, 0) and (0, −10)

4. C = 30h + 45; $150

5. k = 1

Practice exercise 2

1. 1/3

2. yes: both have gradient −½

3. y = −x/3 + 17/3

4. k = −½

Practice exercise 3

1. 10

2. (2, 2)

3. (2, 3)

Practice exercise 4

1. 11; x = 6

2. x2 − 2x − 3

3. no: x = 4 gives y = ±2

4. one-to-one

Practice exercise 5

1. x ≠ −4

2. x ≥ 3; g(x) ≥ 1

3. x > 2; h(x) ∈ ℝ

4. 1 ≤ f(x) < 11

5. f(x) ≤ 5

Practice exercise 6

1. x = −1, y = −2

2. (−2, 0), (3, 0), (0, −6)

3. max (1, 5), min (3, 1)

Practice exercise 7

1. x = 2.21

2. x = −0.767

3. k = ±2

Practice exercise 8

1. 3x + y − 5 = 0

2. −2/7; 2

Practice exercise 9

1. y = −3x − 1

2. gradients 4/3 and −3/4: product −1

Practice exercise 10

1. 24

2. (4, 5)

Practice exercise 11

1. y = 3x − 3

2. (3, 4)

Practice exercise 12

1. L = 24 − 1.5t; 16 hours

2. m = 250 minutes

Practice exercise 13

1. x ≥ −1; f(x) ≥ 3

2. 0 < g(x) ≤ ½

Practice exercise 14

1. x = −0.767, 2, 4

2. (1, 0.368)

Practice exercise 15

1. 0.792

2. x = −1, −0.618, 1.62

Practice exercise 16

1. distance from C to (4.4, 0.2) = √24.2 = 4.92

2. x + y = 6: k = 3

Practice exercise 17

1. 16

2. 63.4°

Practice exercise 18

1. √3/3

2. 53.1°

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