CSEC Mathematics cheat sheet, Cheat Sheet of Mathematics

Download CSEC Mathematics cheat sheet and more Cheat Sheet Mathematics in PDF only on Docsity! MATHEMATICS STUDY GUIDE Curriculum Planning and Development Division The booklet highlights some salient points for each topic in the CSEC Mathematics syllabus. At least one basic illustration/example accompanies each salient point. The booklet is meant to be used as a resource for “last minute” revision by students writing CSEC Mathematics. Produced by: Patrick Ramdath, Curriculum Officer- Mathematics Revised by: Nicole Harris-Knudsen, Acting Curriculum Coordinator- Mathematics Curriculum Planning and Development Division - 3 Number Theory Operations with Decimals Points to Remember Illustration/ Example Find the product of 3.77 x 2.8 =? 1. Line up the numbers on the right, 2. multiply each digit in the top number by each digit in the bottom number (like whole numbers), 3. add the products, 4. and mark off decimal places equal to the sum of the decimal places in the numbers being multiplied. Find the product of 3.77 × 2.8 3.77 (2 decimal places) 2.8 (1 decimal place) 754 3016 10.556 (3 decimal places) When dividing, if the divisor has a decimal in it, make it a whole number by moving the decimal point to the appropriate number of places to the right. If the decimal point is shifted to the right in the divisor, also do this for the dividend. Fractions can always be written as decimals. For example: 2 5 = 0.4 1 2 = 0.5 3 4 = 0.75 1 4 = 0.25 3 5 = 0.6 3 4 = 0.75 Curriculum Planning and Development Division - 4 Number Theory Significant figures Points to Remember Illustration/ Example The rules for significant figures: 1) ALL non-zero numbers (1,2,3,4,5,6,7,8,9) are ALWAYS significant. 2) ALL zeroes between non-zero numbers are ALWAYS significant. 3) ALL zeroes which are SIMULTANEOUSLY to the right of the decimal point AND at the end of the number are ALWAYS significant. 4) ALL zeroes which are to the left of a written decimal point and are in a number >= 10 are ALWAYS significant. A helpful way to check rules 3 and 4 is to write the number in scientific notation. If you can/must get rid of the zeroes, then they are NOT significant. Number Number of Significant Figures Rule(s) 48,923 5 1 3.967 4 1 900.06 5 1,2,4 0.0004 (= 4 E-4) 1 1,4 8.1000 5 1,3 501.040 6 1,2,3,4 3,000,000 (= 3 E+6) 1 1 10.0 (= 1.00 E+1) 3 1,3,4 Number Theory Binary Numbers Points to Remember Illustration/ Example Each digit "1" in a binary number represents a power of two, and each "0" represents zero. 0001 is 2 to the zero power, or 1 0010 is 2 to the 1st power, or 2 0100 is 2 to the 2nd power, or 4 1000 is 2 to the 3rd power, or 8 Binary numbers can be added 10001 +11101 101110 Binary numbers can be subtracted Curriculum Planning and Development Division - 5 Number Theory Computation – Fractions Points to Remember Illustration/ Example When the numerator stays the same, and the denominator increases, the value of the fraction decreases 3 1 , 4 1 , 5 1 4 3 , 5 3 , 6 3 7 3 When the denominator stays the same, and the numerator increases, the value of the fraction increases. 2 7 , 2 8 , 2 9 Equivalent fractions are fractions that may look different, but are equal to each other. 2 1 , 4 2 , 6 3 8 4 Equivalent fractions can be generated by multiplying or dividing both the numerator and denominator by the same number. 2 1 = 22 21 x x = 4 2 5 3 = 25 23 x x = 10 6 Fractions can be simplified when the numerator and denominator have a common factor in them 10 6 = 25 23 x x = 5 3 Fractions with different denominators, can be converted to a set of fractions that have the same denominator 4 3 , 3 2 is the same as 12 9 , 12 8 Addition and subtraction of fractions are similar to adding and subtracting whole numbers if the fractions being added or subtracted have the same denominator 12 9 – 12 8 = 12 1 When multiplying fractions, multiply the numerators together and then multiply the denominators together and simplify the results. 6 5 × 3 2 = 18 10 = 9 5 Number Theory Prime Numbers Points to Remember Illustration/ Example A prime number is a number that has only two factors: itself and 1e.g. 5 can only be divided evenly by 1 or 5, so it is a prime number. Numbers that are not prime numbers are referred to as composite numbers Curriculum Planning and Development Division - 8 Angles formed by a Transversal Crossing two Parallel Lines Vertical Angles are the angles opposite each other when two lines cross. Vertically opposite angles are equal a = d f = g b = c e = h The angles in matching corners are called Corresponding Angles. Corresponding Angles are equal a = e c = g b = f d = h The pairs of angles on opposite sides of the transversal but inside the two lines are called Alternate Angles. Alternate Angles are equal d = e c = f The pairs of angles on one side of the transversal but inside the two lines are called Consecutive Interior Angles. Consecutive Interior Angles are supplementary (add up to 180°) d + f c + e Illustration of all angles mentioned on a single diagram. The transversal crosses two Parallel Lines Curriculum Planning and Development Division - 9 Triangles Pythagoras' Theorem Points to Remember Illustration/ Example Pythagoras' Theorem states that the square of the hypotenuse is equal to the sum of the squares on the other two sides c 2 = a 2 + b 2 The Hypotenuse is c Find c c 2 = 5 2 + 12 2 = 25 + 144 = 169 c= √169 = 13 units Triangles Similar Triangles & Congruent Triangles Points to Remember Illustration/ Example Definition: Triangles are similar if they have the same shape, but can be different sizes. (They are still similar even if one is rotated, or one is a mirror image of the other). There are three accepted methods of proving that triangles are similar: If two angles of one triangle are equal to two angles of another triangle, the triangles are similar. If angle A = angle D and angle B = angle E Then ∆ABC is similar to ∆DEF Show that the two triangles given beside are similar and calculate the lengths of sides PQ and PR. Solution: ∠A = ∠P and ∠B = ∠Q, ∠C = ∠R(because ∠C = 180 - ∠A - ∠B and ∠R = 180 - ∠P - ∠Q) Therefore, the two triangles ΔABC and ΔPQR are similar. Curriculum Planning and Development Division - 10 Triangles Similar Triangles & Congruent Triangles Points to Remember Illustration/ Example 1) If the three sets of corresponding sides of two triangles are in proportion, the triangles are similar. If 𝐴𝐵 𝐷𝐸 = 𝐴𝐶 𝐷𝐹 = 𝐵𝐶 𝐸𝐹 Then ∆ABC is similar to ∆DEF 2) If an angle of one triangle is equal to the corresponding angle of another triangle and the lengths of the sides including these angles are in proportion, the triangles are similar. If angle A = angle D and 𝐴𝐵 𝐷𝐸 = 𝐴𝐶 𝐷𝐹 Then ∆ABC is similar to ∆DEF Consequently: 𝐴𝐵 𝑃𝑄 = 𝐵𝐶 𝑄𝑅 = 𝐴𝐶 𝑃𝑅 implies 𝐴𝐵 𝑃𝑄 = 𝐵𝐶 𝑄𝑅 Substituting known lengths give: 4 𝑃𝑄 = 6 12 or 6PQ = 4 × 12 Therefore PQ = 12 × 4 6 = 8 Also, 𝐵𝐶 𝑄𝑅 = 𝐴𝐶 𝑃𝑅 Substituting known lengths give: 6 12 = 7 𝑃𝑅 or 6PR = 12 × 7 Therefore PR = 12 × 7 6 = 14 Find the length AD (x) The two triangles ΔABC and ΔCDE appear to be similar since AB || DE and they have the same apex angle C. It appears that one triangle is a scaled version of the other. However, we need to prove this mathematically. AB || DE, CD || AC and BC || EC ∠BAC = ∠EDC and ∠ABC = ∠DEC Considering the above and the common angle C, we may conclude that the two triangles ΔABC and ΔCDE are similar. Curriculum Planning and Development Division - 13 Mensuration Areas & Perimeters Points to Remember Illustration/ Example Area of triangle, given two sides and the angle between them Either Area = ½ ab sin C or Area = ½ bc sin A or Area = ½ ac sin B Or in general, Area = ½ × side 1 × side 2 × sine of the included angle First of all we must decide what we know. We know angle C = 25º, and sides a = 7 and b = 10. Start with: Area= (½)ab sin C Put in the values we know: Area=½ × 7 × 10 × sin(25º) Do some calculator work: Area = 35 × 0.4226 =14.8 units 2 (1dp) Area of Parallelogram, given two sides and an angle The diagonal of a parallelogram divides the parallelogram into two congruent triangles. Consequently, the area of a parallelogram can be thought of as doubling the area of one of the triangles formed by a diagonal. This gives the trig area formula for a parallelogram: Either Area = ab sin C or Area = bc sin A or Area = ac sin B Find the area of the parallelogram: Area = ab sin C = (8)(6)sin 120 o = 41.569 =41.57 square units Curriculum Planning and Development Division - 14 Mensuration Areas & Perimeters Points to Remember Illustration/ Example Area of Trapezium = ½(a+b) × h = ½(sum of parallel sides) × h h = vertical height Perimeter = a + b + c + d Find the area of the trapezium A = ½ (a+b) × h = ½(10 + 8) × 4 =½ × (18) × 4 = 36 cm 2 Perimeter = a + b + c + d = 10 + 8 + 4.3 + 4.1 = 26.4 cm Area of Parallelogram = base × height b = base h= vertical height Find the area of a parallelogram with a base of 12 centimeters and a height of 5 centimeters. Area of parallelogram= b × h = 12cm × 5cm = 60 cm 2 Perimeter of parallelogram = a + b + a + b = 2 (a + b) = 12cm + 7cm + 12cm + 7cm = 38cm c d b a Curriculum Planning and Development Division - 15 Mensuration Surface Area and Volumes Points to Remember Illustration/ Example Volume of Cylinder = Area of Cross Section x Height = π r 2 h Surface Area of Cylinder = 2π r 2 + 2πrh = 2πr (r + h) Find the volume and total surface area of a cylinder with a base radius of 5 cm and a height of 7 cm. Volume = π r 2 h = 7 22 × 5 2 × 7 =22 × 25 cm 3 = 550cm 3 Conversion to Litres: 1000 cm 3 = 1L 550 cm 3 = 1000 550 L= 0.55 L Surface Area = 2π(5)(7) + 2π(5) 2 = 70π + 50π = 120π cm 2 ≈ 376.99 cm 2 * A prism is a three-dimensional shape which has the same shape and size of cross-section along the entire length i.e. a uniform cross-section Prism- Since a cylinder is closely related to a prism, the formulas for their surface areas are related Volume of Prism = area of cross section × length = A l Example: What is the volume of a prism whose ends have an area of 25 m 2 and which is 12 m long Answer: Volume = 25 m 2 × 12 m = 300 m 3 Volume of irregular prism = Ah Surface Area of irregular prism = 2A + (perimeter of base × h) Curriculum Planning and Development Division - 18 Mensuration Surface Area and Volumes Points to Remember Illustration/ Example = 3 1 × [8 × 6] × 5 = 80 cm 3 Curriculum Planning and Development Division - 19 Geometry Sum of all interior angles of a regular polygon Points to Remember Illustration/ Example The sum of interior angles of a polygon having n sides is (2n – 4) right angles = (2n-4) x 90 . Each interior angle of the polygon = (2n – 4)/n right angles. e.g. What is the sum of the interior angles of a triangle Find the sum of all interior angkes in i) Pentagon ii) Hexagon iii) Heptagon iv) Octagon Curriculum Planning and Development Division - 20 Geometry Sum of all interior angles of any polygon Illustration/ Example Name Figure No. of Sides Sum of interior angles (2n - 4) right angles Triangle 3 (2n - 4) right angles = (2 × 3 - 4) × 90° = (6 - 4) × 90° = 2 × 90° = 180° Quadrilateral 4 (2n - 4) right angles = (2 × 4 - 4) × 90° = (8 - 4) × 90° = 4 × 90° = 360° Pentagon 5 (2n - 4) right angles = (2 × 5 - 4) × 90° = (10 - 4) × 90° = 6 × 90° = 540° Name Figure No. of Sides Sum of interior angles (2n - 4) right angles Hexagon 6 (2n - 4) right angles = (2 × 6 - 4) × 90° = (12 - 4) × 90° = 8 × 90° = 720° Heptagon 7 (2n - 4) right angles = (2 × 7 - 4) × 90° = (14 - 4) × 90° = 10 × 90° = 900° Octagon 8 (2n - 4) right angles = (2 × 8 - 4) × 90° = (16 - 4) × 90° = 12 × 90° = 1080° Curriculum Planning and Development Division - 23 Geometry Circle Geometry Points to Remember Illustration/ Example Area = π × r 2 Circumference = 2 × π × r = D × π Area of Circle= = π × r 2 = 7 22 × 14m × 14m = 7 22 × 196 m 2 = 616 m 2 Perimeter (Circumference) of circle = 2 × π × r = 2 × 7 22 × 14 = 87.976 cm = 87.98 cm to (2 dp) Area of Sector AOB = π × r 2 × 360  Length of Arc AB = 2π r × 360  Perimeter = BO + OA + arc AB Area of Sector = π × r 2 × 360  = 7 22 × 12 × 12 × 360 45 = 56.55 units 2 Arc length AB = 2πr × 360  = 2 × 7 22 × 12 × 360 45 = 9.428 = 9.43 units (2dp) Perimeter of sector ABC = BC + CA + Arc length AB = 12 + 12 + 9.43 = 33.43 units Curriculum Planning and Development Division - 24 Algebra Simplifying algebraic expressions Points to Remember Illustration/ Example Algebraic expressions are the phrases used in algebra to combine one or more variables, constants and the operational (+ - x / ) symbols. Algebraic expressions don't have an equals = sign. Letters are used to represent the variables or the constants 1) Nine increased by a number x 9 + x 2) Fourteen decreased by a number p 14 – p 3) Seven less than a number t t – 7 4) The product of nine and a number, decreased by six 9m – 6 5) Three times a number, increased by seventeen 3a + 17 6) Thirty-two divided by a number y 32 ÷ y 7) Five more than twice a number 2n + 5 8) Thirty divided by seven times a number 30 ÷ 7n Algebra Substitution Points to Remember Illustration/ Example In Algebra "Substitution" means putting numbers where the letters are 1) If x = 5, then what is 10 x + 4 10 5 + 4 = 2 + 4 = 6 2) If x = 3 and y = 4, then what is x 2 + xy 3 2 + 3×4 = 9 + 12 = 21 3) If x = −2, then what is 1 − x + x 2 1 − (−2) + (−2) 2 = 1 + 2 + 4 = 7 Curriculum Planning and Development Division - 25 Algebra Binary Operations Points to Remember Illustration/ Example A binary operation is an operation that applies to two numbers, quantities or expressions e.g. a*b = 3a + 2b Commutative Law Let * be a binary operation. * is said to be commutative if , a * b = b * a Associative Law Let * be a binary operation. * is said to be an associative if , a * (b * c) = (a * b) * c An operation * is defined by a * b = 3a + b. Determine: i) 2*4 ii) 4*2 iii) (2*4)*1 iv) 2* (4*1) v) Is * associative? vi) Is * communicative? i) 2*4 = 3(2) + 4 = 10 ii) 4*2 = 3(4) + 2 = 14 iii) (2*4)*1= 10*1= 3(10) + 1 = 31 iv) 2* (4*1)= 2 * [3(4)+1]= 2*13= 3(2) +13=19 v) Since (2*4)*1 ≠ 2* (4*1), * is not associative. That is (a*b)*c, ≠ a* (b*c) vi) Since 2*4 *≠ 4*2, * is not commutative. That is a*b ≠ b*a Algebra Solving Linear Equations Points to Remember Illustration/ Example An equation shows the link between two expressions 1)Solve 2x + 6 = 10 2x + 6 = 10 2x = 10-6 2x = 4 x = 4 2 x = 2 2) Solve 5x – 6 = 3x – 8 5x – 6 = 3x – 8 5x – 3x = -8 + 6 2x = -2 x = −2 2 x = -1 Curriculum Planning and Development Division - 28 Algebra Solving Simultaneous Equations (both Linear) Points to Remember Illustration/ Example *Simultaneous means “at the same time” *There are two methods used for solving systems of equations: the elimination method and substitution method * The elimination method is most useful when one variable from both equations has the same coefficient in both equations, or the coefficients are multiples of one another. * To use the substitution method, two conditions must be met: there must be the same number of equations as variables; and one of the equations must be easily solved for one variable Solve simultaneously using the elimination method : 3x – y = 1 ….eq. (1) 2x + 3y = 8 .…eq. (2) Eq.(1) x 3 … 9x – 3y = 3 Eq. (2) x 1 … 2x + 3y = 8 Add both equations 11x = 11 x = 11 11 x = 1 Substitute x=1 into Eq. (1) 3(1) – y = 1 3 – 1 = y 2 = y Solution (1, 2) Using the substitution method: Make y the subject of the formula in Eq. (1) 3x – y = 1 3x – 1 = y Substitute y= 3x – 1 into Eq. (2) 2x + 3(3x – 1) = 8 2x + 9x – 3 = 8 11x – 3 = 8 11x = 8 + 3 11x = 11 x = 11 11 x = 1 Substitute x=1 into Eq. (1) 3(1) – y = 1 3 – 1 = y 2 = y Solution (1, 2) Curriculum Planning and Development Division - 29 Algebra Solving Simultaneous Equations (Linear and Quadratic) Points to Remember Illustration/ Example Solve simultaneously: 2x + y = 7 Eq. (1) x 2 – xy = 6 Eq. (2) From Eq.(1), 2x + y = 7 y = 7 – 2x Substituting this value of y into Eq. (2) x 2 – x (7 – 2x) – 6 = 0 x 2 – 7x + 2x 2 – 6 = 0 3x 2 – 7x - 6 = 0 (3x + 2) (x – 3) = 0 x = 3 2 or x = 3 Using Eq. 1, when x = 3 2 y = 7 – 2 ( 3 2 ) = 7 + ( 3 4 ) = 3 25 When x = 3 y = 7 – 2(3) = 7 – 6 = 1 Solutions: ( 3 2 , 3 25 ) or (3, 1) Solving a pair of equations in two variables (linear and quadratic) Use graphs to find solutions to simultaneous equations y = x 2 – 5x + 7 … eq. (1) y = 2x + 1 … eq. (2) Set them equal to each other x 2 – 5x + 7 = 2x + 1 x 2 – 5x – 2x + 7 – 1 = 0 x 2 – 7x + 6 = 0 (x – 1) (x – 6) = 0 x = 1 and x = 6 Substitute into eq. (2) When x = 1 ; y = 2(1) + 1 = 3 x = 6 ; y = 2(6) + 1 = 13 Solutions (1, 3) and (6, 13) Curriculum Planning and Development Division - 30 Algebra Indices Points to Remember Illustration/ Example The laws of indices are used to simplify expressions and numbers with indices e.g. 2 5 = 2 x 2 x 2 x 2 x 2 Laws of Indices a 0 = 1 a m x a n = a m+n a m ÷ a n = a m-n (a b ) c = =a bc = a cb 𝑎−𝑛 = 1 𝑎𝑛 Use of a 0 = 1 2 0 = 1 Use of a m x a n = a m+n Use of a m ÷ a n = a m-n Use of (a b ) c = =a bc = a cb Use of 𝒂−𝒏 = 𝟏 𝒂𝒏 Use of Curriculum Planning and Development Division - 33 Relations, Functions and Graphs Relations and Functions Points to Remember Illustration/ Example * A relation is a set of ordered pairs in which the first set of elements is called the domain and the second set of elements the range or co-domain * Relations can be expressed in three ways: as expressions; as maps or diagrams; or as graphs *A function is a mathematical operation that assigns to each input number or element, exactly one output number or value * Maximum and minimum points on a graph are found where the slope of the curve is zero Given the graph of a relation, if you can draw a vertical line that crosses the graph in more than one place, then the relation is not a function. Here are a couple examples: This graph shows a function, because there is no vertical line that will cross this graph twice. This graph does not show a function, because any number of vertical lines will intersect this oval twice. For instance, the y- axis intersects (crosses) the line twice. This is a function. There is only one y for each x; there is only one arrow coming from each x. This is a function! There is only one arrow coming from each x; there is only one y for each x This one is not a function: there are two arrows coming from the number 1; the number 1 is associated with two different range elements. So this is a relation, but it is not a function. Each element of the domain has a pair in the range. However, what about that 16? It is in the domain, but it has no range element that corresponds to it! This is neither a function nor a relation Curriculum Planning and Development Division - 34 Relations, Functions and Graphs Composite Functions & Inverses Points to Remember Illustration/ Example "Function Composition" is applying one function to the results of another: The result of f( ) is sent through g( ) It is written: (g o f)(x) Which means: g(f(x)) Given f(x) = 2x + 3 and g(x) = –x 2 + 5, find ( f o g)(x). ( f o g)(x) = f (g(x)) = f (–x 2 + 5) = 2( ) + 3 … setting up to insert the input formula = 2(-x 2 + 5) + 3 = -2x 2 + 10 + 3 = -2x 2 + 13 Given f(x) = 2x + 3 and g(x) = -x 2 + 5, find (g o f )(x). (g o f )(x) = g( f(x)) = g(2x + 3) = - ( ) 2 + 5 … setting up to insert the input = - (2x + 3) 2 + 5 = - (4x 2 + 12x + 9) + 5 = - 4x 2 – 12x – 9 + 5 = - 4x 2 – 12x – 4 Find ( f o g)(2) using ( f o g)(x) = –2x 2 + 13 ( f o g)(2) = -2 (2) 2 + 13 = -8 +13 = 5 OR Find g(2) = -2 2 + 5 = -4 + 5 = 1 Then f[g(2)] = 2 (1) +3 = 5 The inverse of a function has all the same points as the original function, except that the x's and y's have been reversed. For instance, supposing your function is made up of these points: { (1, 0), (-3, 5), (0, 4) }. Then the inverse is given by this set of points: { (0, 1), (5, -3), (4, 0) } Given f (x) = 2x – 1 and g(x) = ( 𝟏 𝟐 )x + 4, find i) f –1 (x), ii) g –1 (x), iii) ( f o g) –1 (x), and iv) (g –1 o f –1 )(x). First, find f –1 (x), g –1 (x), and ( f o g) –1 (x): Inverting f (x): f (x) = 2x – 1 Let y = 2x – 1 Interchange x = 2y – 1 Make y the subject x + 1 = 2y x+1 2 = y Hence, f –1 (x) = x+1 2 Curriculum Planning and Development Division - 35 Relations, Functions and Graphs Composite Functions & Inverses Points to Remember Illustration/ Example Inverting g(x): g(x) = 𝟏 𝟐 x + 4 Let y = 1 2 x + 4 Interchange x = 1 2 y + 4 Make y the subject x – 4 = 1 2 y 2(x – 4) = y 2x – 8 = y Hence g –1 (x) = 2x – 8 Finding the composite function: ( f o g)(x) = f [g(x)] = f [ 1 2 x + 4] = 2[ 1 2 x + 4] – 1 = x + 8 – 1 = x + 7 Inverting the composite function: ( f o g)(x) = x + 7 Let y = x + 7 Interchange x = y + 7 Make y the subject x – 7 = y ( f o g) –1 (x) = x – 7 Now compose the inverses of f(x) and g(x) to find the formula for (g –1 o f –1 )(x): (g –1 o f –1 )(x) = g –1 [ f –1 (x)] = g –1( 𝑥+1 2 ) = 2( 𝑥+1 2 ) – 8 = (x + 1) – 8 Hence, (g –1 o f –1 )(x) = x – 7 The inverse of the composition ( f o g) –1 (x) gives the same result as does the composition of the inverses (g –1 o f –1 )(x). We therefore conclude that ( f o g) –1 (x) = (g –1 o f –1 )(x) Curriculum Planning and Development Division - 38 Relations, Functions and Graphs Introduction to Graphs Points to Remember Illustration/ Example Standard form A quadratic function is written as y = ax 2 + bx + c Roots Can be found by factorization. It can also be found using the quadratic formula which gives the location on the x-axis of the two roots and will only work if a is non-zero. 𝑥 = −𝑏 ± √𝑏2 − 4𝑎𝑐 2𝑎 Axis of symmetry x = a b 2  Completing the Square When f(x) is written in the form y= a(x - h) 2 + k (h, -k) is the maximum or minimum point The y-intercept is found by asking the question: When x = 0, what is y? By Calculation: 1) The values of x for which f(x) = 0 a = 1, b = -1, c = -12 x = −𝑏 ± √𝑏2 − 4𝑎𝑐 2𝑎 x = −1 ± √(−1)2 − 4(1)(−12) 2(1) x = −1 ± √1 + 48 2 = −1 ± √49 2 = −1 ± 7 2 x = −1 + 7 2 = 6 2 = 3 or x = −1−7 2 = −8 2 = −4 2) The value of x for which f(x) is minimum x = a b 2  = )1(2 )1( = 2 1 3) The minimum value of f(x) - complete the square or i.e. write f(x) in the form y = a(x – h) 2 + k y = x 2 – x – 12 y = x 2 – x + ( 1 2 ) 2 – 12 – ( 1 2 ) 2 y = (x − 1 2 ) (x − 1 2 ) – 12 – 4 1 y = (x − 1 2 ) 2 – 4 49 a(x – h) 2 + k = (x − 1 2 ) 2 – 4 49 This implies that h = 2 1 and k = 4 49 So the minimum value of f(x) is 4 49 or -12.25 The minimum point is ( 2 1 , 4 49 ) 4) The y-intercept : When x = 0 y = 0 2 – 0 – 12 = -12 Curriculum Planning and Development Division - 39 Relations, Functions and Graphs Non-Linear Relations Points to Remember Illustration/ Example Exponential functions involve exponents, where the variable is now the power. We encounter non-linear relations in the growth of population with time and the growth of invested money at compounded interest rates Draw the graph of y= 2 x Curriculum Planning and Development Division - 40 Relations, Functions and Graphs Direct & Inverse Variation Points to Remember Illustration/ Example Direct Variation The statement " y varies directly as x ," means that when x increases, y increases by the same factor. y α x Introducing the constant of proportionality, k y = kx . Other examples of direct variation: The circumference of a circle is directly proportional to its radius. Inverse Variation Two quantities are inversely proportional if an increase in one quantity leads to a reduction in the other. Example of Direct Variation: If y varies directly as x, and y = 15 when x = 10, then what is y when x = 6? Find the constant of proportionality: y α x y = kx use (10, 15) 15 = k (10) 15 10 = k 3 2 = k Therefore the equation becomes y = 3 2 x Substitute x = 6 y = 3 2 (6) y = 9 Solution ( 6, 9) If y varies inversely as x, and y = 10 when x = 6 , then what is y when x = 15 ? y α 1 𝑥 y = 𝑘 𝑥 10 = 𝑘 6 k = 60 Therefore, the equation becomes y = 60 𝑥 When x = 15 y = 60 15 = 4 Solution (6, 4) Curriculum Planning and Development Division - 43 Relations, Functions and Graphs Linear Programming Points to Remember Illustration/ Example Linear programming is the process of taking various linear inequalities relating to some situation, and finding the "best" value obtainable under those conditions. A typical example would be taking the limitations of materials and labor, and then determining the "best" production levels for maximal profits under those conditions. Suppose that the three (3) inequalities are related to some situation. These inequalities can be represented on a graph: To draw the line y= - 𝟏 𝟐 x + 7: When x = 0 , y = 7 and when y=0, x = 14 Therefore, coordinates on the line are (0, 7) and (14,0) To draw the line y = 3x When x = 0, y = 0 and say when x = 2, y =6 Coordinates on the line are (0,0) and (2,6) To draw the line y= x - 2 When x = 0, y = -2 and when y = 0, x = 2 Coordinates on the line are (0, -2) and (2,0) Suppose the profit is given by the equation "P = 3x + 4y" To find maximum profit: The corner points are (2, 6), (6, 4), and (–1, –3). For linear systems like this, the maximum and minimum values of the equation will always be on the corners of the shaded region. So, to find the solution simply plug these three points into "P = 3x + 4y". (2, 6): P = 3(2) + 4(6) = 6 + 24 = 30 (6, 4): P = 3(6) + 4(4) = 18 + 16 = 34 (–1, –3): P = 3(–1) + 4(–3) = –3 – 12 = –15 Then the maximum of P = 34 occurs at (6, 4), and the minimum of P = –15 occurs at (–1, –3). Curriculum Planning and Development Division - 44 Relations Functions and Graphs Distance - Time Graphs Points to Remember Illustration/ Example Speed, Distance and Time The following is a basic but important formula which applies when speed is constant (in other words the speed doesn't change) Speed = 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑇𝑖𝑚𝑒 If speed does change, the average (mean) speed can be calculated: Average speed = Total Distance Total Time Taken Distance - Time Graphs These have the distance from a certain point on the vertical axis and the time on the horizontal axis. The velocity can be calculated by finding the gradient of the graph. If the graph is curved, this can be done by drawing a chord and finding its gradient (this will give average velocity) or by finding the gradient of a tangent to the graph (this will give the velocity at the instant where the tangent is drawn). Example a) Jane runs at an average speed of 12.5 m/s in a race journey of 500 metres. How long does she take to complete the race? To find a time, we need to divide distance by speed. 500 metres ÷ 12.5 m/s = 40 secs b) Chris cycles at an average speed of 8 km/h. If he cycles for 6½ hours, how far does he travel? To find a distance, we need to multiply speed by time. 8 km/h × 6.5 hours = 52 km a) Change 15km/h into m/s. 15 km/h = 15 𝑘𝑚 1 ℎ𝑜𝑢𝑟 = 15 𝑘𝑚 60 𝑚𝑖𝑛 = 15000 𝑚 3600 𝑠𝑒𝑐𝑠 = 4 1 6 m/s Curriculum Planning and Development Division - 45 Units When using any formula, the units must all be consistent. For example speed could be measured in m/s, in terms of distance in metres and time in seconds or in km/h in terms of distance in kilometres and time in hours. In calculations, units must be consistent, so if the units in the question are not all the same e.g. m/s, and km/h, then you must first convert all to the same unit at the start of solving the problem. b) Example If a car travels at a speed of 10m/s for 3 minutes, how far will it travel? i. Firstly, change the 3 minutes into 180 seconds, so that the units are consistent. ii. Now rearrange the first equation to get distance = speed × time. iii. Therefore distance travelled = 10m × 180 = 1800m = 1.8 km c) A car starts from rest and within 10 seconds is travelling at 10m/s. What is its acceleration? Acceleration = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑡𝑖𝑚𝑒 = 10 10 = 1 m/s2 d) What is the speed represented by the steeper line? Speed = 10−0 2− 0 = 10 2 = 5ms-1 Relations Functions and Graphs Velocity - Time Graphs Points to Remember Illustration/ Example Velocity and Acceleration Velocity is the speed of a particle and its direction of motion (therefore velocity is a vector quantity, whereas speed is a scalar quantity). When the velocity (speed) of a moving object is increasing we say that the object is accelerating. If the velocity decreases it is said to be decelerating. Acceleration is therefore the rate of change of velocity (change in velocity / time) and is measured in m/s². Example Consider the motion of the object whose velocity-time graph is given in the diagram. a) What is the acceleration of the object between times and ? b) What is the acceleration of the object between times and ? c) What is the net displacement of the object between times and ? Curriculum Planning and Development Division - 48 Statistics Displaying data: Pie Chart, Bar Graph, Histogram and Line Graph Points to Remember Illustration/ Example * There are three basic types of Bar Charts: simple charts, proportionate charts and chronological charts Example :Bar Graph .* Histograms are a great way to show results of continuous data, such as:  weight  height  how much time  etc. The horizontal axis is continuous like a number line: The histogram below shows the heights (in cm) distribution of 30 people. a) How many people have heights between 159.5 and 169.5 cm? b) How many people have heights less than 159.5 cm? c) How many people have heights more than 169.5cm? d) What percentage of people have heights between 149.5 and 179.5 cm? a) 7 people b) 9 + 6 = 15 people c) 5 + 2 + 1 = 8 people d) (9 + 7 + 5)/ 30 x 100 = 70% Curriculum Planning and Development Division - 49 Statistics Displaying data: Pie Chart, Bar Graph, Histogram and Line Graph Points to Remember Illustration/ Example Line Graph - A graph that shows information that is connected in some way (such as change over time) You are learning facts about dogs, and each day you do a short test to see how good you are. Draw a line graph to represent the information: Facts I got Correct Day 1 Day 2 Day 3 Day 4 3 4 12 15 Example: Line Graph Statistics Frequency Distribution Displaying data on the Bar Graph Measure of Central Tendency – Mean, Median and Mode Points to Remember Illustration/ Example The frequency of a particular data value is the number of times the data value occurs A frequency table is constructed by arranging collected data values in ascending order of magnitude with their corresponding frequencies The Mean is the average of the numbers. Add up all the numbers, then divide by how many numbers there are mean = Ʃ 𝑥𝑓 Ʃ𝑓 Rick did a survey of how many games each of 20 friends owned, and got this: 9, 15, 11, 12, 3, 5, 10, 20, 14, 6, 8, 8, 12, 12, 18, 15, 6, 9, 18, 11 a) Find the Mode b) Find the Median c) Show this data in a frequency table d) Calculate the mean e) Draw a histogram to represent the data Curriculum Planning and Development Division - 50 Statistics Frequency Distribution Displaying data on the Bar Graph Measure of Central Tendency – Mean, Median and Mode Points to Remember Illustration/ Example To find the Median, place the numbers in value order and find the middle number (or the mean of the middle two numbers). To find the Mode, or modal value, place the numbers in value order then count how many of each number. The Mode is the number which appears most often (there can be more than one mode): a) Mode is 12 (occurs most often) b) To find the median, first order the data then find the mean of the 10 th and 11 th values: 3, 5, 6, 6, 8, 8, 9, 9, 10, 11, 11, 12, 12, 12, 14, 15, 15, 18, 18, 20 Median = 11+11 2 = 22 2 = 11 c) Frequency table for the number of games owned. Number of games (x) Tally Frequency (f) xf 3 | 1 3 5 | 1 5 6 || 2 12 8 || 2 16 9 || 2 18 10 | 1 10 11 || 2 22 12 ||| 3 36 14 | 1 14 15 || 2 30 18 || 2 36 20 | 1 20 Ʃf = 20 Ʃxf = 222 d) mean = Ʃ 𝑥𝑓 Ʃ𝑓 = 222 20 = 11.1 e) 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 1 2 3 4 5 6 7 8 9 1011121314151617181920 F re q u e n c y ( f) Number of Games Frequency (f) Curriculum Planning and Development Division - 53 Statistics Cumulative Frequency Curve (Ogive) Interquartile Range and Semi-Interquartile Range Points to Remember Illustration/ Example A Cumulative Frequency Graph is a graph plotted from a cumulative frequency table. A cumulative frequency graph is also called an ogive or cumulative frequency curve The total of the frequencies up to a particular value is called the cumulative frequency The lower quartile or first quartile (Q1) is the value found at a quarter of the way through a set of data The median or second quartile (Q2) is the value found at half of the way through a set of data The upper quartile (Q3) is the value found at three quarters of the way through a set of data The Interquartile range is the difference between the upper and lower quartile: Q3 – Q1 Semi-interquartile range = ½ (Q3 – Q1) We need to add a class with 0 frequency before the first class and then find the upper boundary for each class interval Length (cm ) Frequency Upper Class Boundary Length (x cm) Cumulative Frequency 6 – 10 0 10.5 x ≤ 10.5 0 11 – 15 2 15.5 x ≤ 15.5 2 16 – 20 4 20.5 x ≤ 20.5 6 21 – 25 8 25.5 x ≤ 25.5 14 26 – 30 14 30.5 x ≤ 30.5 28 31 – 35 6 35.5 x ≤ 35.5 34 36 – 40 4 40.5 x ≤ 40.5 38 41 – 45 2 45.5 x ≤ 45.5 40 Ʃ f = 40 Q1 = 23.5 Q2 = 27.5 Q3 = 31.5 Interquartile Range = Q3 – Q1 = 31.5- 23.5= 8 Semi-interquartile Range = ½ (Q3 – Q1) = ½ (8) = 4 Curriculum Planning and Development Division - 54 Consumer Arithmetic Ready Reckoner Points to Remember Illustration/ Example A ready reckoner is a table of numbers used to facilitate simple calculations, especially one for applying rates of discount, interest, charging, etc., to different sums The table shows an extract from a ready reckoner giving the price of N articles at 27 cents each. N N N N 21 5.67 63 17.01 105 28.35 500 135.00 22 5.94 64 17.28 106 28.62 525 141.75 23 6.21 65 17.55 107 28.89 550 148.50 24 6.48 66 17.82 108 29.16 600 162.00 25 6.75 67 18.09 109 29.43 625 168.75 26 7.02 68 18.36 110 29.70 650 175.50 27 7.29 69 18.63 111 29.97 700 189.00 28 7.56 70 18.90 112 30.24 750 202.50 29 7.83 71 19.17 113 30.51 800 216.00 30 8.10 72 19.44 114 30.78 900 243.00 Use the table to find the cost of: 1) 23 articles at 27 cents each 2) 571 articles at 27 cents each 3) 61/4 m of material at 27 cents each 4) 72.9 kg of foodstuff at 27 cents per kilogram 1) Directly from the table, the cost is $6.21 2) From the table: Cost of 500 articles = $135.00 Cost of 71 articles = $ 19.17 Cost of 571 articles = $154.17 3) 6 1 4 = 6.25. To use the tables we find the cost of 625 m to be: $162.00 + $6.75 = $168.75. Hence the cost of 6.25 m is $ 168.75 100 = $ 1.69 4)The cost of 729 kg at 27 cents each is: $189.00 + $7.83 = $196.83 Hence the cost of 72.9kg is $ 196.83 10 = $ 19.68 Curriculum Planning and Development Division - 55 Consumer Arithmetic Foreign Exchange Rates Points to Remember Illustration/ Example Foreign exchange, is the conversion of one country's currency into that of another Amelia is going on a holiday to Italy, so she will have to purchase some euros (€). How many euros will she get for £375 if the exchange rate is £1 = €1.2769? Give your answer to the nearest euro. £1 = €1.2769 £375 = 1.2769 1 x 375 = $ 478.8375 Change US$80 to TT$, given that TT$1.00 = US$6.35 US $6.35 = TT$ 1.00 US $1.00 = TT$ 1.00 6.35 US $ 6.35 = TT$ 1.00 6.35 x 80 = TT$ 12.59 Consumer Arithmetic Hire Purchase Points to Remember Illustration/ Example Goods purchased on hire purchase are paid for at regular intervals over a specified period of time. Sometimes the purchaser may pay a deposit, then the remainder (cash price- deposit + interest) is repaid at a number of regular intervals. A bicycle can be bought for $160.00 cash or it can be bought on hire purchase by depositing 25% of the cash price, then paying the balance + 10% interest per annum (p.a.) on the balance in 12 monthly instalments. If the bicycle was sold on hire purchase determine the monthly repayments. Cash Price = $160.00 Deposit = 25 100 x 160 = $40.00 Balance = $160.00 - $40.00 = $120.00 Interest on Balance = 10 100 x 120 = $12.00 Total amount still to be paid = $120.00 + $12.00 = $132.00 Monthly repayment = 132 12 = $ 11.00 Curriculum Planning and Development Division - 58 Consumer Arithmetic Mortgages Points to Remember Illustration/ Example A mortgage is a loan to finance the purchase of real estate, usually with specified payment periods and interest rates. The borrower (mortgagor) gives the lender (mortgagee) a right of ownership on the property as collateral for the loan. 1) Tim bought a house for 250,000. He makes a down payment of 15% of the purchase price and takes a 30- year mortgage for the balance. a) What is Tim’s down payment? b) What is Tim’s mortgage? Downpayment = Percent Down × Purchase Price = 15 100 x $250,000 = $37500 Amount of Mortgage = Purchase Price − Down Payment = 250,000 − 37500 = 212500 2) If your monthly payment is 1200 dollars, what is the total interest charged over the life of the loan? Total Monthly Payment = Monthly payment × 12 × Number of years = $1200 × 12 × 30 = $432000 Total Interest Paid = Total Monthly Payment − Amount of Mortgage = $432000 − $212500 = $219500 Consumer Arithmetic Rates and Taxes Points to Remember Illustration/ Example Taxes are ‘calculated’ sums of money paid to a government by to meet national expenditures  e.g. schools, hospitals, salaries, road networks Gross Salary is the figure before making other deductions.  Tax-free allowance - Working people do not pay tax on all their income. Part of their earnings is not taxed. A tax-free allowance is made for each dependent. Examples of dependents are : a wife, a young child, old father.  Taxable income is obtained after the tax-free allowance is subtracted from the gross salary  Net salary is the take home salary of the employee after paying taxes Mr. Salandy’s salary is $22 000 per year. He has a personal allowance of $2000, a marriage allowance of $1000, a child allowance of $800, national insurance of $400 and an insurance allowance of $300. A flat rate of 18% is paid on income tax. Determine his net salary. Personal allowance = 2000 Marriage allowance = 1000 Child allowance = 800 National insurance = 400 Insurance allowance = 300 Total Allowance = 4500 Taxable income = $22000 - $4500 = $17 500 Income Tax = 18% of $17500 = 18 100 x 17,500 = $3150. Net Salary = $ 22 000 - $3150 = $18 850 Curriculum Planning and Development Division - 59 Consumer Arithmetic Wages Points to Remember Illustration/ Example *Basic Week- Number of hours worked per week *Basic Rate- Amount of money paid per hour *Workers are paid wages and salaries. Wages can be paid fortnightly, weekly or daily. *Overtime- The money earned for extra hours beyond the basic week 1) A man works a basic week of 38 hours and his basic rate is $13.75 per hour. Calculate his total wage for the week Total wage for week= Basic Rate x Time = 13.75 x 38 = $522.50 2)John Williams works a 42 hour week for which he is paid a basic wage of $113.40. He works 6 hours overtime at time and a half and 4 hours at double time. Calculate his gross wage for the week. Basic hourly rate = $113.40 42 = $2.70 Overtime rate at time and a half = 1 ½ x $2.70 = 4.05 For 6 hours at time and a half, Mr. William will earn $4.05 x 6 = $24.30 Overtime rate at double time = 2 x $2.70 = $5.40 For 4 hours at double time, Mr. William will earn $5.40 x 4 = $21.60 Gross Wage = $113.40 + 24.30 + 21.60 = $159.30 Curriculum Planning and Development Division - 60 Trigonometry Cosine Rule Points to Remember Illustration/ Example When a triangle does not have a right angle, we can find the missing sides or angles using either the sine rule or the cosine rule It is used when two sides and an angle between them are given or all three sides are given The Cosine Rule is very useful for solving triangles: a, b and c are sides C is the angle opposite side c This following examples will cover how to:  Use the Cosine Rule to find unknown sides and angles  Use the Sine Rule to find unknown sides and angles  Combine trigonometry skills to solve problems a 2 = b 2 + c 2 − 2bc cosA a 2 = 5 2 + 7 2 − 2 × 5 × 7 × cos(49°) a 2 = 25 + 49 − 70 × cos(49°) a 2 = 74 − 70 × 0.6560... a 2 = 74 − 45.924... = 28.075 a = √28.075... a = 5.298... a = 5.30 to 2 decimal places Trigonometry Sine Rule Points to Remember Illustration/ Example The Sine Rule is also very useful for solving triangles: B b sin = A a sin When two angles and any side are given or when two sides and an angle not between them are given B b sin = A a sin Bsin 5 = 49sin 298.5 sin B = (sin(49°) × 5) / 5.298... sin B = 0.7122... B = sin −1 (0.7122...) B = 45.4° to one decimal place C = 180° − 49° − 45.4° C = 85.6° to one decimal place Curriculum Planning and Development Division - 63 Sets Definitions and Notation Points to Remember Illustration/ Example *A set is a collection of identifiable elements or members that are connected in some way * There are two types of sets: finite and infinite *The symbol ∈ is used to show that an item is an element or member of a set * A subset is represented by the symbol: ⊂ and is used to present part of a set separately. * A universal set is made up of all elements from which all subsets will be pulled and is represented by the symbol ɛ or U * Venn Diagrams are used to represent sets and the relationship between sets (Describe each region). * Complements of a set B are represented by Bʹ and show members of a set that are NOT part of B. * The intersection of two or more sets consists of those elements that are common to those sets *The union of two or more sets consists of those elements that make up those sets. Set Notation Description Meaning A U B “A union B” everything that is in either of the sets “A intersect B” only the things that are in both of the sets A ' "A complement" or "not A" everything in the universal set outside of A B' “B complement" everything in A except for anything in its overlap with B (A ∪ B)' "not (A union B)" everything outside A and B ( )' "not (A intersect B)" everything outside of the overlap of A and B Examples of sets are: a collection of coins; a pack of cards; all vowels in the English alphabet etc. Examples of finite sets: A= {All odd numbers between 1 and 10} = {3, 5, 7, 9} V= {the vowels in the alphabet} = {a, e, i, o, u} Examples of infinite sets: A= {All natural numbers} = {1, 2, 3 …} B= {All whole numbers} = {0, 1, 2, 3, 4 …} Use of symbol ∈ : {2} ∈ {1, 2, 3, 4} Use of symbol ⊂ : {2, 3} ∈ {1, 2, 3, 4} Set Notation: Set notation Venn diagram Set A U B {1, 2, 3} {2} A ' {3, 4} B' {1} (A U B)' {4} ( )' Curriculum Planning and Development Division - 64 Circle Geometry Circle Theorems Points to Remember Illustration/ Example The angle which an arc of a circle subtends at the centre of a circle is twice the angle it subtends at any point on the remaining part of the circumference. What is the size of Angle POQ? (O is circle's center) Angle POQ = 2 × Angle PRQ = 2 × 62° = 124° The angle in a semicircle is a right angle. What is the size of Angle BAC? The Angle in the Semicircle Theorem tells us that Angle ACB = 90° Now use the sum of angles of a triangle equals 180° to find Angle BAC: Angle BAC + 55° + 90° = 180° Angle BAC = 35° A tangent of a circle is perpendicular to the radius of that circle at the point of contact. Curriculum Planning and Development Division - 65 Circle Geometry Circle Theorems Points to Remember Illustration/ Example Angles in the same segment of a circle and subtended by the same arc are equal. What is the size of Angle CBX? Angle ADB = 32° = Angle ACB. Angle ACB = Angle XCB. So in triangle BXC we know Angle BXC = 85°, and Angle XCB = 32° Now use sum of angles of a triangle equals 180° : Angle CBX + Angle BXC + Angle XCB = 180° Angle CBX + 85° + 32° = 180° Angle CBX = 63° The line joining the centre of a circle to the midpoint of a chord is perpendicular to the chord. Given that OQ is perpendicular to PR and units, determine the value of PQ = QR= 4 (perpendicular from centre bisects chord) In : Curriculum Planning and Development Division - 68 Symmetry Lines of Symmetry Points to Remember Illustration/ Example A rhombus has 2 lines of symmetry Non-example The scalene triangle does not have any lines of symmetry. A rectangle has 2 lines of symmetry An equilateral triangle has 3 lines of symmetry Curriculum Planning and Development Division - 69 Transformations Translation Points to Remember Illustration/ Example * In a translation, all points in a line or object are changed in the same direction so there is no change in shape The points A(2,4), B(4,4), C(5,2), D(2,1) were translated under ( 7 −3 ). Find the image A', B', C', D' A' (-5, 1), B' (-3, 1), (-2, -1), (-5, -2) Curriculum Planning and Development Division - 70 Transformations Reflection Points to Remember Illustration/ Example A transformation is a change made to a figure * The main types of transformations are: reflections, translations, rotations, and enlargements * A reflection, or flip, is a transformation that creates a symmetry on the coordinate plane * Matrices can be used to conduct transformations Reflect the points (-1,7), (6,5), (-2,2) on the x-axis Reflect the points (2,3), (10,0), (3,-2) on the y-axis Reflect the points A(1,2), B(1,5) and C(3,2) on the line x=-1 Curriculum Planning and Development Division - 73 Transformations Enlargement Points to Remember Illustration/ Example * An enlargement is a transformation that changes the size of a figure What is a scale factor? Enlarging a shape by a positive scale factor means changing the size of a shape by a scale factor from a particular point, which is called the centre of enlargement. Fractional scale factors If we 'enlarge' a shape by a scale factor that is between -1 and 1, the image will be smaller than the object) If you enlarge it by a positive number greater than 1, the shape will get bigger. Enlargement with scale factor greater than 1 Enlarge the object ABC by scale factor 2 Enlargement with scale factor between -1 and +1 Fractional Scale Factors Enlarge triangle ABC with a scale factor 1 /2, centred about the origin The scale factor is ½, so: OA' = ½OA, OB' = ½OB, OC' = ½OC Since the centre is the origin, we can in this case multiply each coordinate by ½to get the answers. A = (2, 2), so A' will be (1, 1). B = (2, 6), so B' will be (1, 3). C = (4, 2), so C' will be (2, 1). Curriculum Planning and Development Division - 74 Transformations Enlargement Points to Remember Illustration/ Example Negative Scale Factors An enlargement using a negative scale factor is similar to an enlargement using a positive scale factor, but this time the image is on the other side of the centre of enlargement, and it is upside down to create you enlarged shape. Negative Scale Factors Enlarge the rectangle WXYZ using a scale factor of - 2, centred about the origin. The scale factor is -2, so multiply all the coordinates by -2. So OW' is 2OW. This time we extend the line WO beyond O, before plotting W'. In a similar way, we extend XO, YO and ZO and plot X',Y' and Z'. Can you see that the image has been turned upside down? Curriculum Planning and Development Division - 75 Transformations Glide-Reflection Points to Remember Illustration/ Example When a translation (a slide or glide) and a reflection are performed one after the other, a transformation called a glide reflection is produced. In a glide reflection, the line of reflection is parallel to the direction of the translation. It does not matter whether you glide first and then reflect, or reflect first and then glide. This transformation is commutative. ∆𝐴′𝐵′𝐶′ is the image of ∆𝐴𝐵𝐶 under a glide reflection that is a composition of a reflection over the line l and a translation through the vector v. Examine the graph below. Is triangle A"B"C" a glide reflection of triangle ABC? Answer: Yes, Triangle ABC is reflected on the x-axis to A'B'C' and then translated through 5 places to the left or ( −5 0 ) Vectors Scalar Quantities Points to Remember Illustration/ Example *Scalars are quantities that only have a magnitude, meaning they can be expressed with just a number. There are absolutely no directional components in a scalar quantity - only the magnitude of the medium e.g. Time - the measurement of years, months, weeks, days, hours, minutes, seconds, and even milliseconds; Volume - tons to ounces to grams, milliliters and micrograms Speed and - speed in miles or kilometers-per-hour, temperature Give some real life examples of scalar quantities: Answer: Height of a building, time taken for a trip, temperature outside, an avocado on the scale reading 87.9 grams, Curriculum Planning and Development Division - 78 Vectors Addition and Subtraction of Vectors Points to Remember Illustration/ Example * Vector addition is simply the sum of the two vector's components. We can add two vectors by simply joining them head-to-tail: Add ( 8 13 )+ ( 26 7 ) and illustrate this on a diagram ( 8 13 )+ ( 26 7 )= ( 34 20 ) We can also subtract one vector from another:  first we reverse the direction of the vector we want to subtract,  then add them as usual: If u = ( 2 1 ) and v = ( 3 −2 ) i) Illustrate u – v on a graph ii) Calculate the value of u – v i) ii) u – v = ( 2 1 ) - ( 3 −2 ) = ( 2 1 ) + ( −3 2 ) = ( −1 3 ) Vectors Curriculum Planning and Development Division - 79 Magnitude of a vector Points to Remember Illustration/ Example *The magnitude of a vector is shown by two vertical bars on either side of the vector: |a| We use Pythagoras’ Theorem to calculate it: |a| = √𝑥2 + 𝑦2 1) What is the magnitude of the vector b = ( 𝟔 𝟖 ) |b| = √( 6 2 + 8 2 ) = √( 36+64 ) = √100 = 10 Vectors Parallel Vectors Points to Remember Illustration/ Example One can use vectors to solve problems in Geometry e.g. to prove that two vectors are parallel. Two vectors are parallel if they have the same direction To prove that two vectors are parallel: If two vectors ?⃗? and 𝑣 are parallel, then one is a simple ratio of the other , or one is a multiple of the other 𝑣 = k?⃗? In the triangle ABC the points X and Y are the mid- points of AB and AC. Show that XY and BC are parallel. 𝑋𝑌⃗⃗⃗⃗ ⃗ = 𝑋𝐴⃗⃗⃗⃗ ⃗ + 𝐴𝑌⃗⃗⃗⃗ ⃗ = -a + b = b – a 𝐵𝐶⃗⃗⃗⃗ ⃗ = 𝐵𝐴⃗⃗⃗⃗ ⃗ + 𝐴𝐶⃗⃗⃗⃗ ⃗ = - 2a + 2b = 2b – 2a = 2 (b – a) This implies that one vector is a simple ratio of the other: 𝑋𝑌⃗⃗⃗⃗ ⃗ 𝐵𝐶⃗⃗ ⃗⃗ ⃗ = 𝑏−𝑎 2 (𝑏−𝑎) = 1 2 i.e. 𝑋𝑌⃗⃗⃗⃗ ⃗: 𝐵𝐶⃗⃗⃗⃗ ⃗ = 1:2 OR one is a scalar multiple of the other (cross multiply) 𝐵𝐶⃗⃗⃗⃗ ⃗ = 2 𝑋𝑌⃗⃗⃗⃗ ⃗ or 𝑋𝑌⃗⃗⃗⃗ ⃗ = 1 2 𝐵𝐶⃗⃗⃗⃗ ⃗ Hence, XY is parallel to BC and half its length. Curriculum Planning and Development Division - 80 Vectors Collinear Vectors Points to Remember Illustration/ Example Points that lie on the same line are called collinear points. To prove that two vectors are collinear: If two vectors are collinear, then one is a simple ratio of the other, or one is a multiple of the other 𝑣 = k?⃗? and they have a common point. The position vectors of points P, Q and R are vectors a + b, 4a - b and 10a – 5b respectively. Prove that P, Q and R are collinear. 𝑃𝑄⃗⃗⃗⃗ ⃗ = 𝑃𝑂⃗⃗⃗⃗ ⃗ + 𝑂𝑄⃗⃗⃗⃗⃗⃗ = (-a –b) + (4a – b) = 3a – 2b 𝑄𝑅⃗⃗⃗⃗ ⃗= 𝑄𝑂⃗⃗⃗⃗⃗⃗ + 𝑂𝑅⃗⃗⃗⃗ ⃗ = (-4a + b) + (10a – 5b) = 6a – 4b = 2 ( 3a – 2b) This implies that one vector is a simple ratio of the other and they have a common point Q 𝑃𝑄⃗⃗ ⃗⃗ ⃗ 𝑄𝑅⃗⃗⃗⃗⃗⃗ = 3𝑎−2𝑏 2 (3𝑎−2𝑏) = 1 2 i.e. 𝑃𝑄⃗⃗⃗⃗ ⃗: 𝑄𝑅⃗⃗⃗⃗ ⃗ = 1:2 OR one is a scalar multiple of the other (cross multiply) 𝑄𝑅⃗⃗⃗⃗ ⃗ = 2 𝑃𝑄⃗⃗⃗⃗ ⃗ or 𝑃𝑄⃗⃗⃗⃗ ⃗ = 1 2 𝑄𝑅⃗⃗⃗⃗ ⃗ Since 𝑄𝑅⃗⃗⃗⃗ ⃗ = 2 𝑃𝑄⃗⃗⃗⃗ ⃗ and they have a common point Q, then P, Q and R are collinear. Curriculum Planning and Development Division - 83 Matrices Singular Matrix Points to Remember Illustration/ Example A singular matrix is a square matrix that has no inverse A matrix is singular if and only if its determinant is zero i.e. ad – bc = 0 If the determinant of a matrix is 0, the matrix has no inverse Determine if the matrix A = ( 2 6 1 3 ) is singular Det A = ad- bc = (2)(3) – (6)(1) = 6 – 6 = 0 Matrices Simultaneous Equations Points to Remember Illustration/ Example One of the most important applications of matrices is to the solution of linear simultaneous equations Solve the simultaneous equation using a matrix method x + 2y = 4 3x – 5y = 1 This can be written in matrix form AX = B: ( 1 2 3 −5 ) ( 𝑥 𝑦 ) = ( 4 1 ) ( 𝑥 𝑦 ) = ( 1 2 3 −5 ) −1 ( 4 1 ) ( 𝑥 𝑦 ) = )32()51( 1 xx  ( −5 −2 −3 1 ) ( 4 1 ) = )32()51( 1 xx  ( −5 −2 −3 1 ) ( 4 1 ) = )6()5( 1  ( −5 −2 −3 1 ) ( 4 1 ) = 11 1 ( −22 −11 ) = ( 2 1 ) Curriculum Planning and Development Division - 84 Matrices Transformational matrices Points to Remember Illustration/ Example R= 90 o rotation about the origin, given the matrix. This transformation matrix rotates the point matrix 90 degrees anti-clockwise. When multiplying by this matrix, the point matrix is rotated 90 degrees anti- clockwise around (0,0). ( 0 −1 1 0 ) S= 180 o anticlockwise rotation about the origin, given the matrix. This transformation matrix creates a rotation of 180 degrees. When multiplying by this matrix, the point matrix is rotated 180 degrees around (0,0). This changes the sign of both the x and y co- ordinates. ( −1 0 0 −1 ) T= 270 o rotation about the origin, given the matrix ( 0 1 −1 0 ) I= 360 o rotation about the origin, given the matrix. This transformation matrix is the identity matrix. When multiplying by this matrix, the point matrix is unaffected and the new matrix is exactly the same as the point matrix ( 1 0 0 1 ) X= reflection on x axis, given the matrix. This transformation matrix creates a reflection in the x-axis. When multiplying by this matrix, the x co-ordinate remains unchanged, but the y co-ordinate changes sign ( 1 0 0 −1 ) Y= reflection on y axis, given the matrix This transformation matrix creates a reflection in the y-axis. When multiplying by this matrix, the y co-ordinate remains unchanged, but the x co-ordinate changes sign ( −1 0 0 1 ) W= reflection on y= x, given the matrix. This transformation matrix creates a reflection in the line R = 90 o anti-clockwise rotation about the origin ( 0 −1 1 0 ) ( 4 3 ) = ( (0 𝑥 4) + (−1 𝑥 3) (1 𝑥 4) + (0 𝑥 3) ) = ( −3 4 ) S = 180 o anti-clockwise rotation about the origin ( −1 0 0 −1 )( 4 3 ) = ( (−4 × 1) + (3 × 0) (4 × 0) + (3 × −1) ) = ( −4 −3 ) T = 270 o anti-clockwise rotation about the origin ( 0 1 −1 0 ) ( 4 3 ) = ( (4 × 0) + (3 × 1) (4 × −1) + (3 × 0) ) = ( 3 −4 ) I = Identity Matrix ( 1 0 0 1 ) ( 4 3 ) = ( (4 × 1) + (3 × 0) (4 × 0) + (3 × 1) ) = ( 4 3 ) X = Reflection on x axis ( 1 0 0 −1 ) ( 4 3 ) = ( (4 × 1) + (3 × 0) (4 × 0) + (3 × −1) ) = ( 4 −3 ) Y = Reflection on y axis ( −1 0 0 1 ) ( 4 3 ) = ( (4 × −1) + (3 × 0) (4 × 0) + (3 × 1) ) = ( −4 3 ) W= reflection on y = x Curriculum Planning and Development Division - 85 Matrices Transformational matrices Points to Remember Illustration/ Example y=x. When multiplying by this matrix, the x co- ordinate becomes the y co-ordinate and the y-ordinate becomes the x co-ordinate. ( 0 1 1 0 ) Reflection on y = - x, given the matrix. This transformation matrix creates a reflection in the line y=-x. When multiplying by this matrix, the point matrix is reflected in the line y=-x changing the signs of both co-ordinates and swapping their values. ( 0 −1 −1 0 ) ( 0 1 1 0 ) ( 4 3 ) = ( (4 × 0) + (1 × 3) (1 × 4) + (0 × 3) ) = ( 3 4 ) Reflection on y = - x, ( 0 −1 −1 0 )( 4 3 ) = ( (4 × 0) + (−1 × 3) (−1 × 4) + (0 × 3) ) = ( −3 −4 ) Matrices Combining Transformations Points to Remember Illustration/ Example We see that sometimes 2 transformations are equivalent to a single transformation. 1) The diagram shows how the original shape A is first reflected to B, and B is then reflected to C. . Solution: A rotation of 180° about the origin would take A straight to C. Curriculum Planning and Development Division - 88 Indices ..................................................................................................................................................... 30 Product of two brackets .......................................................................................................................... 31 Factorization of Simple expressions ....................................................................................................... 31 Solving quadratic inequalities ................................................................................................................. 32 Relations, Functions and Graphs ................................................................................................................ 33 Relations and Functions .......................................................................................................................... 33 Relations, Functions and Graphs ............................................................................................................ 34 Composite Functions & Inverses ............................................................................................................ 34 Introduction to Graphs ............................................................................................................................ 36 Non-Linear Relations .............................................................................................................................. 39 Direct & Inverse Variation ...................................................................................................................... 40 Coordinate Geometry .............................................................................................................................. 41 Linear Programming ............................................................................................................................... 43 Distance - Time Graphs .......................................................................................................................... 44 Velocity - Time Graphs .......................................................................................................................... 45 Statistics ...................................................................................................................................................... 47 Displaying data: Pie Chart, Bar Graph, Histogram and Line Graph ....................................................... 47 Frequency Distribution ........................................................................................................................... 49 Displaying data on the Bar Graph ........................................................................................................... 49 Measure of Central Tendency – Mean, Median and Mode ..................................................................... 49 Cumulative Frequency Curve (Ogive) .................................................................................................... 53 Interquartile Range and Semi-Interquartile Range ................................................................................. 53 Consumer Arithmetic .................................................................................................................................. 54 Ready Reckoner ...................................................................................................................................... 54 Foreign Exchange Rates ......................................................................................................................... 55 Hire Purchase .......................................................................................................................................... 55 Profit, Loss, Discount ............................................................................................................................. 56 Simple Interest ........................................................................................................................................ 57 Compound Interest .................................................................................................................................. 57 Mortgages ............................................................................................................................................... 58 Rates and Taxes ...................................................................................................................................... 58 Wages ...................................................................................................................................................... 59 Curriculum Planning and Development Division - 89 Trigonometry .............................................................................................................................................. 60 Cosine Rule ............................................................................................................................................. 60 Sine Rule ................................................................................................................................................. 60 Bearings .................................................................................................................................................. 61 Sets .............................................................................................................................................................. 63 Definitions and Notation ......................................................................................................................... 63 Circle Geometry .......................................................................................................................................... 64 Circle Theorems ...................................................................................................................................... 64 Symmetry .................................................................................................................................................... 67 Lines of Symmetry .................................................................................................................................. 67 Transformations .......................................................................................................................................... 69 Translation .............................................................................................................................................. 69 Reflection ................................................................................................................................................ 70 Rotation ................................................................................................................................................... 72 Enlargement ............................................................................................................................................ 73 Glide-Reflection ...................................................................................................................................... 75 Vectors ........................................................................................................................................................ 75 Scalar Quantities ..................................................................................................................................... 75 Vector Quantities .................................................................................................................................... 76 Vector Representation ............................................................................................................................. 76 Product of a Vector and a Scalar ............................................................................................................. 77 Position Vector........................................................................................................................................ 77 Addition and Subtraction of Vectors ...................................................................................................... 78 Magnitude of a vector ............................................................................................................................. 79 Parallel Vectors ....................................................................................................................................... 79 Collinear Vectors .................................................................................................................................... 80 Matrices ...................................................................................................................................................... 81 Introduction to Matrices .......................................................................................................................... 81 Addition and Subtraction of Matrices ..................................................................................................... 82 Multiplication of Matrices ...................................................................................................................... 82 Inverse of a Matrix .................................................................................................................................. 82 Singular Matrix ....................................................................................................................................... 83 Curriculum Planning and Development Division - 90 Simultaneous Equations .......................................................................................................................... 83 Transformational matrices ...................................................................................................................... 84 Combining Transformations ................................................................................................................... 85