National 5 Learning Checklist - Relationships

Size: px

Start display at page:

Download "National 5 Learning Checklist - Relationships"

Kimberly Warren
6 years ago
Views:

1 National 5 Learning Checklist - Relationships Topic Skills Extra Stud / Notes Straight Line Gradient Represented b m Measure of steepness of slope Positive gradient the line is increasing Negative gradient the line is decreasing Y-intercept Represented b c Shows where the line cuts the -axis Find b making x = 0 Find the gradient of a Know that gradient is represented b the letter m line joining two points Step : Select two coordinates Step : Label them (x, ) (x, ) Step 3: Substitute them into gradient formula Find equation of a line (from gradient and - intercept) Find equation of a line (from two points) Rearrange equation to find gradient and - intercept Sketch lines from their equations Solving Equations / Inequations Solving Equations e.g. (-, ), (, -8) ( 8) 3 m x x ( ) 6 x x Step : Find gradient m Step : Find -intercept c Step 3: Substitute into = mx + c (see above for definitions) Use this when there are onl two points (i.e. no -intercept) Step : Find gradient Step : Substitue into b = m(x a) where (a, b) are taken from either one of the points e.g x = 3 = -6x + = -x + m = -, c = Step : Rearrange equation to the form = mx + c (see note above) Step : Draw a table of points Step 3: Plot points on coordinate axes Use suitable method: e.g. 5(x + ) = (x 5) 5x + 0 = x 0 5x = x 30 3x = 30 x = 0 Solving inequations Solve the same wa as equations. NB: When dividing b a negative change the sign: e.g. -3x 5 x -5 Simultaneous Equations Solve b sketching lines Step : Rearrange lines to form = mx + c Step : Sketch lines using table of points (as above) Step 3: Find coordinate of point of intersection Solve b substitution This works when one or both equations are of the form = ax + b e.g. Solve 3x + = 7 = x + Sub equation into : 3x + (x + ) = 7 5x + = 7 x = 3 so = 3 + =

2 Simultaneous Equations Contd. Solve b Elimination Step : Scale equations to make one unknown equal with opposite sign. Step : Add Equations to eliminate equal term and solve. Step 3: Substitute number to find second term. Form Equations Change the Subject Linear Equations e.g. a + 3b = 7 a b = - x 8a + 6b = 3 x 3 6a 6b = a = -8 a = - substitute a = - into (-) + 3b = 7 3b = 5 b = 5 Ans. a = -, b = 5 Form equations from a variet of contexts to solve for unknowns Rearrange equations change the subject: Equations with powers or roots Quadratic Functions Quadratics and their equations e.g. e.g. D = C 3 [C] D + 3 = C D 3 C D 3 C V r h [r] V r h r V h 5 = x = -x = 5(z + 6) z z z 6 5 [z] = x = x + 5 = (x 3) = (x + ) - 3

3 Equations of quadratics = kx Sketching Quadratics = k(x + a) + b Sketching Quadratics (Harder) = (x + a)(x b) Solving Quadratics (finding roots) Algebraicall Solving Quadratics (finding roots) Graphicall Step : Identif coordinate from graph Step : Substitute into = kx Step 3: Solve to find k e.g. Coordinate: (, ) Substitution: = k() = k k = 0.5 Quadratic: = 0.5x Step : Identif shape, if k = then graph is +ve or if k = - then the graph is -ve Step : Identif turning point (-a, b) Step 3: Sketch axis of smmetr x = -a Step 5: Find -intercept (make x = 0) Step : Sketch information Step : Identif shape (+ve or -ve) Step : Identif roots (x-intercepts) x = -a, x = b Step 3: Find -intercept (make x = 0) Step : Identif turning point e.g. = (x + )(x ) +ve graph Minimum turning point Roots: x =, x = - -intercept: = (0 + )(0 ) = -8 Turning Point (-, -9) (see below) NB: Turning point is halfwa between roots. x-coord = ( + (-)) = - -coord = (- + )(- ) = -9 Step : Factorise quadratic Step : Set each factor equal to zero Step 3: Solve each factor to find roots e.g. = x + x = x 5x 6 x(x + ) = 0 (x 6)(x + ) = 0 x = 0 or x + = 0 x 6 = 0 or x + = 0 x = 0 or x = - x = 6, x = - Read roots from graph x =, x = - Solving Quadratics Quadratic Formula When asked to solve a quadratic to a number of decimal places use the formula: b b ac x a where = ax + bx + c

Discriminant Using the Discriminant e.g. Solve = x 6x + to d.p. a = b = -6 c = ( 6) x 6 x 8 ( 6) 6 8 x x x = 5.6 x = 0.

the quadratic = x + 5x + Solution: a =, b = 5, c = b ac = 5 = 5 6 = 9 Since b ac > 0 the quadratic has two real roots.

4 Discriminant Using the Discriminant e.g. Solve = x 6x + to d.p. a = b = -6 c = ( 6) x 6 x 8 ( 6) 6 8 x x x = 5.6 x = b ac where = ax + bx + c The discriminant describes the nature of the roots b ac > 0 two real roots b ac = 0 equal roots (tangent to axis) b ac < 0 Example : Determine the nature of the roots of the quadratic = x + 5x + Solution: a =, b = 5, c = b ac = 5 = 5 6 = 9 Since b ac > 0 the quadratic has two real roots. Properties of Shapes Circles Example : Determine p, where x + 8x + p has equal roots Solution: b ac = 0 8 p = 0 6 p = 0 6 = p P = 6 Pthagoras Use Pthagoras Theorem to solve problems involving circles and 3D shapes. e.g. Find the depth of water in a pipe of radius 0cm. r is the radius 0cm x 8cm x = 0-9 x = x =.cm Depth = 0. = 5.6cm Similar Shapes Linear Scale Factor Linear.Scale.Factor New.Length Original.Length

Length Trigonometr Trig Graphs Sine = asin bx + c Curve a = maxima and minima of graph b = no.

5 Area Scale Factor New.Length Area.Scale.Factor Original.Length Volume Scale Factor 3 New.Length Volume.Scale.Factor Original.Length Trigonometr Trig Graphs Sine = asin bx + c Curve a = maxima and minima of graph b = no. of waves between 0 and 360⁰ c = movement of graph verticall = sin x maxima and minima and -, period = 360⁰ - = sin x 80⁰ - = sin 3x 80⁰ - = sin x + 60⁰ 0⁰ x - = -sin x 80⁰ - = sin (x - 30⁰) 80⁰ - 30⁰ 0⁰

6 Trig Graphs Cosine Curve = acos bx + c a = maxima and minima of graph b = no. of waves between 0 and 360⁰ c = movement of graph verticall = cos x maxima and minima and -, period = 360⁰ - 80⁰ Trig Graphs Tan Curve The same transformations appl for Cosine as Sine (above) = tan x no maxima or minima, period = 80⁰ -90⁰ 90⁰ x Solving Trig Equations Know the CAST diagram Memor Aid: All Students Take Care Use the diagram above to solve trig equations: Example : Solve sinx = 0 sinx = sin x = ½ x = sin - (½) x = 30 ⁰, 80 ⁰ - 30 ⁰ x = 30 ⁰, 50 ⁰ Example : Solve tan x + 5 = 0 tan x = -5 tan x = -5/ NB: tan x is negative so there will be solutions in the second and fourth quadrant x = = tan - (5/) x acute = 5.3 ⁰ Trig Identities Know: sin x + cos x = sin x = - cos x and cos x = - sin x and tanx sinx cos x x = 80 ⁰ ⁰, ⁰ x = 8.7 ⁰, ⁰ Use the above facts to show one trig function can be another. Start with the left hand side of the identit and work through until it is equal to the right hand side.

National 5 Learning Checklist Expressions & Formulae

National 5 Learning Checklist Expressions & Formulae Topic Skills Extra Stud / Notes Rounding Round to decimal places 5.4 5. to d.p. 4.676 4.68 to d.p. Round to Significant Figures 76 00 to sig. figs.