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Objectives: Be able to….. • Distinguish between rational and irrational numbers. • Understand what is meant by a surd. • Simplify expressions involving surds. • Perform arithmetic with surds. Objectives: Be able to….. Use language of coordinate geometry. Find the distance between two given points. Find the mid points of a line segment joining two given points. Find, use and interpret the gradient of a line segment. Know the relationship between the gradients for parallel and perpendicular lines. Rearrange equation to show line in for Ax + By + C = 0 and vice versa. Find the equations of straight lines given – (a) the gradient and a point – (b) two points Verify, given their coordinates, that points lie on a line. Find the coordinates of a point of intersection of two lines. Find the fourth vertex of a parallelogram given the other three. Chapter 4: Quadratic functions and there graphs. Learning objectives: To be able to… Solve quadratic equations. Complete the square of a quadratic expression. Understand the effect of a translation on a quadratic graph. Find the equation of a parabola when translated by a given vector. Understand the term discriminant. What do these have in common? They all have similar curved flight through the air. The curve is called a parabola and it is generated by a quadratic function. A general function for this type of curve is given by:𝑓 𝑥 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 Quadratic expressions A quadratic expression is an expression in which the highest power of the variable is 2. For example: t2 x2 – 2 w2 + 3w + 1 4 – 5g2 2 The general form of a quadratic expression in x is: ax2 + bx + c (where a ≠ 0) x is a variable. a is the coefficient of x2. b is the coefficient of x. c is a constant term. Factorising quadratic expressions Factorising an expression is the inverse of expanding it. Expanding or multiplying out x2 + 3x + 2 (x + 1)(x + 2) factorising When we expand an expression we multiply out the brackets. When we factorise an expression we write it with brackets. Factorising Quadratics – The Process Step 1: Find factor pairs of 3x2 that sum -7 3x2 - 7x + 2 3x2 -1x -6x + 2 Step 2: Rewrite expression with factor pairs Step 3: Factorise the two halves x(3x -1) -2(3x-1) (x-2) (3x-1) Step 4: Put Into brackets 6 -1&-6 Factorising Quadratics – The Process Step 1: Find factor pairs of 4x-3 that sum +1 4x2 + x - 3 4x2 -3x +4x -3 Step 2: Rewrite expression with factor pairs Step 3: Factorise the two halves x(4x -3) +1(4x-3) (x+1) (4x-3) Step 4: Put Into brackets -12 -3&4 Factorising Quadratics – The Process -8x2 -14 x +15 -8x2 +6x -20x+15 Step 1: Find factor pairs of-8x15 that sum -14 Step 2: Rewrite expression with factor pairs Step 3: Factorise the two halves -2x(4x -3) -5(4x-3) (-2x-5) (4x-3) Step 4: Put Into brackets -120 -6&20 Significant points on a Quadratic Curve that crosses the x axis. Sketch the graph The ROOTS = the x intercepts. (Where Y = 0) Find from table and or graph. The Y INTERCEPT = C /constant. (Where X = 0) Find from equation. The VERTEX = the maximum or minimum point on the graph. Find by substituting the X value, that is the way between the roots, to find Y. -2 -1 0 1 2 3 4 5 x2 4 1 0 1 4 9 16 25 x2 -3x 10 4 0 -2 -2 0 4 10 x2 -3x – 4 6 0 -4 -6 -6 -4 0 6 y 6 0 -4 -6 -6 -4 0 6 4 5 7 6 5 4 3 2 1 0 -3 -2 -1 -1 0 y values y = x -3x – 4 2 x 1 2 -2 -3 -4 -5 -6 -7 x values 3 6 y = -x2 + 1 2 1 0 -4 -3 -2 -1 y values The quadratic curve -1 0 -2 -3 -4 -5 -6 -7 -8 -9 x values 1 2 3 4 Perfect squares Some quadratic expressions can be written as perfect squares. For example: x2 + 2x + 1 = (x + 1)2 x2 – 2x + 1 = (x – 1)2 x2 + 4x + 4 = (x + 2)2 x2 – 4x + 4 = (x – 2)2 x2 + 6x + 9 = (x + 3)2 x2 – 6x + 9 = (x – 3)2 Completing the square Adding 16 to the expression x2 + 8x to make it into a perfect square is called completing the square. We can write x2 + 8x = x2 + 8x + 16 – 16 If we add 16 we then have to subtract 16 so that both sides are still equal. By writing x2 + 8x + 16 we have completed the square and so we can write this as x2 + 8x = (x + 4)2 – 16 In general: 2 b b x bx x 2 2 2 2 Completing the square Complete the square for x2 – 10x. Compare this expression to (x – 5)2 = x2 – 10x + 25 x2 – 10x = x2 – 10x + 25 – 25 = (x – 5)2 – 25 Complete the square for x2 + 3x. Compare this expression to ( x + 32 )2 = x 2 + 3 x + 94 x 2 + 3 x = x 2 + 3 x + 94 94 = ( x + 32 )2 94 Completing the square How can we complete the square for x2 – 8x + 7? Look at the coefficient of x. This is –8 so compare the expression to (x – 4)2 = x2 – 8x + 16. x2 – 8x + 7 = x2 – 8x + 16 – 16 + 7 = (x – 4)2 – 9 In general: 2 2 b b x bx c x c 2 2 2 Completing the square When the coefficient of x2 is not 1, quadratic equations in the form ax2 + bx + c can be rewritten in the form a(x + p)2 + q by completing the square. Complete the square for 2x2 + 8x + 3. Take out the coefficient of x2 as a factor from the terms in x: 2x2 + 8x + 3 = 2(x2 + 4x) + 3 By completing the square, x2 + 4x = (x + 2)2 – 4 so 2x2 + 8x + 3 = 2((x + 2)2 – 4) + 3 = 2(x + 2)2 – 8 + 3 = 2(x + 2)2 – 5 Solving quadratics by completing the square Quadratic equations that cannot be solved by factorisation can be solved by completing the square. For example, the quadratic equation x2 – 4x – 3 = 0 can be solved by completing the square as follows: (x – 2)2 – 7 = 0 (x – 2)2 = 7 x–2= x=2+ x = 4.65 7 or 7 x=2– 7 x = –0.646 (to 3 s.f.) Sketching graphs by completing the square In general, when the quadratic function y = ax2 + bx + c is written in completed square form as a(x + p)2 + q 𝑏 𝑏 where: p= 2 𝑎𝑛𝑑 𝑞 = −(2)2 + c The coordinates of the vertex will be (–p, q). The axis of symmetry will have the equation x = –p. Also: If a > 0 (–p, q) will be the minimum point. If a < 0 (–p, q) will be the maximum point. Plotting the y-intercept, (0, c) will allow the curve to be sketched using symmetry. Sketch the graph of y = x2 + 4x – 1 by writing it in completed square form. x2 + 4x – 1 = (x + 2)2 – 5 The least value that (x + 2)2 can have is 0 because the square of a number cannot be negative. (x + 2)2 ≥ 0 Therefore (x + 2)2 – 5 ≥ – 5 The minimum value of the function y = x2 + 4x – 1 is therefore y = –5. When y = –5, we have, (x + 2)2 – 5 = –5 (x + 2)2 = 0 x = –2 The coordinates of the vertex are therefore (–2, –5). The equation of the axis of symmetry is x = –2. x = –2 y Also, when x = 0 we have y = x2 + 4x – 1 y = –1 0 (0, -1) So the curve cuts the y-axis at the point (0, -1). Using symmetry we can now sketch the graph. Solving (x + 2)2 – 5 = 0 gives the roots as x = ±√5 – 2. x (–2, –5) Translations of Parabola (Movement of Curve) Completed square = (x ± 𝑎) ± 𝑏 Movement of graph Vector +↔ − 𝑎 ±𝑏 +↔ − ± Translations of Parabola (Movement of Curve) Completed square = x±𝑎 2±𝑏 Movement of graph Vector +↔ − 𝑎 ±𝑏 +↔ − ± Quadratic Graphs • y = x2 x x2 y -2 4 4 -1 1 1 0 0 0 1 1 1 2 4 4 • y = x2+1 y y = x2 +1 5 4 x -2 -1 0 1 2 3 x2 4 1 0 1 4 2 x2+1 5 2 1 2 5 1 y 5 2 1 2 5 • y=x2-1 x -2 -1 0 1 2 x2 4 1 0 1 4 2-1 3the curves? 0 -1 oticexabout y 3 0 -1 0 3 0 3 -2 -1 0 -1 y = x2 y = x2 - 1 1 2 x What do you notice about where the lines cross the y-axis? Quadratic Graphs • y = x2 x x2 y -2 4 4 -1 1 1 0 0 0 1 1 1 2 4 4 • y = x2 8 7y 6 5 4 x -2 -1 0 1 2 3 x2 4 1 0 1 4 2 x2 2 0 2 1 y 2 0 2 • y=2x2 y = 2x2 -2 -1 0 -1 y = x2 y = x2 1 2 x x -2 -1 0 1 2 x2 4 1 0 1 4 2x2 8 2 0 2 8 What do you notice curves? y 8 2 0 2 8 What do you notice about where the lines cross the y-axis? Sketching a curve - process • Intercept • Roots (if they exist) • as 𝒙 → ∞, what happens to 𝒚 Objective • Practise and consolidate graph sketching. • Be able to translate cubic graphs. • Sketch curves of functions showing points of intersection. Key points in sketching quadratic graphs ax + bx + c • Shape a Positive Pothole / Minus Mountain • Y Intercept = c • X Intercept (Roots) from Discriminant / Solving – b2 – 4ac > 0 two roots (cuts x axis) factorise/formula/square – b2 – 4ac = 0 one repeated root (vertex touches x axis) – b2 – 4ac < 0 no real roots (vertex above/below x axis) • Vertex (turning point → positive/min or negative/max) – Symmetry / half way between roots for x then substitute for y. – Completing the square The quadratic formula • ax2 + bx + c • • • • • • • 𝑏 𝑐 + x+ 𝑎 𝑎 𝑏 2 x + x 𝑎 𝑏 2 𝑏 2 (x + ) – ( ) 2𝑎 2𝑎 𝑏 2 𝑏2 (x + ) - 2 2𝑎 4𝑎 𝑏 2 (x + ) 2𝑎 𝑏 2 (x + ) 2𝑎 𝑏 x + 2𝑎 x2 • x =0 =0 𝑐 =𝑎 𝑐 = 𝑎 𝑐 =𝑎 2 𝑏 𝑐 = 2 4𝑎 𝑎 2 −4𝑎𝑐 𝑏 = 4𝑎2 2 −4𝑎𝑐) ± (𝑏 = 4𝑎2 2 −4𝑎𝑐) −𝑏 ± (𝑏 = 2𝑎 Learning Objectives: Be able to… • Understand the vocabulary of polynomials • Add subtract and multiply polynomials. • Find, compare and equate coefficients of polynomials. Vocabulary, definitions and conventions. An expression comprising constants, variables and powers / indices. The powers can only be positive integers : x2 − 4x + 7 is a polynomial, but x2 − 4/x + 7x is not. The degree or order of the polynomial is given by the highest power of the variable, so; CONSTANT (Polynomial of degree 0) LINEAR (Polynomial in m of order 1) QUADRATIC (Polynomial in x of order 2) CUBIC (Polynomial in g of order 3) QUARTIC (Polynomial in f of order 4) =5 = 2m + 3 = 3x2 + 4x – 4 = 5g3 – 8g – 3 = f4 + 3f + 7 The normal convention is that the polynomial is written in descending powers of the variable. Polynomials can be added or subtracted by collecting like terms. They can be multiplied using a variety of methods. Coefficient means the constant preceding a variable. Adding and subtracting polynomials. Simply follow normal rules of column addition / subtraction and simplify by collecting like terms. Make sure that polynomials are lined up in correct order in columns. Be careful when handling negative numbers. + X4 4 + 3x + 7 2 3x + 4x – 4 2 X +3x + 7x + 3 - X4 4 + 3x + 7 2 3x + 4x – 4 2 X -3x - x + 11 Multiplying polynomials – using a grid. Multiply x² + 3x – 2 by 2x² – x + 4 x² 2x² -x 4 3x -2 Multiplying polynomials – using a grid. Multiply x² by 2x² x² 2x² -x 4 3x -2 Multiplying polynomials – using a grid. Multiply x² by 2x² x² 2x² -x 4 2x4 3x -2 Multiplying polynomials – using a grid. Multiply 3x by 2x² x² 2x² -x 4 2x4 3x -2 Multiplying polynomials – using a grid. Multiply 3x by 2x² 2x² -x 4 x² 3x 2x4 6x³ -2 Multiplying polynomials – using a grid. Fill in the rest of the table in the same way 2x² -x 4 x² 3x 2x4 6x³ -2 Multiplying polynomials – using a grid. Fill in the rest of the table in the same way 2x² -x 4 x² 3x -2 2x4 6x³ -4x² Multiplying polynomials – using a grid. Fill in the rest of the table in the same way x² 3x -2 2x² 2x4 6x³ -4x² -x -x³ 4 Multiplying polynomials – using a grid. Fill in the rest of the table in the same way x² 3x -2 2x² 2x4 6x³ -4x² -x -x³ -3x² 4 Multiplying polynomials – using a grid. Fill in the rest of the table in the same way x² 3x -2 2x² 2x4 6x³ -4x² -x -x³ -3x² 2x 4 Multiplying polynomials – using a grid. Fill in the rest of the table in the same way x² 3x -2 2x² 2x4 6x³ -4x² -x -x³ -3x² 2x 4 4x² Multiplying polynomials – using a grid. Fill in the rest of the table in the same way x² 3x -2 2x² 2x4 6x³ -4x² -x -x³ -3x² 2x 4 4x² 12x Multiplying polynomials – using a grid. Fill in the rest of the table in the same way x² 3x -2 2x² 2x4 6x³ -4x² -x -x³ -3x² 2x 4 4x² 12x -8 Multiplying polynomials – using a grid. First the term in x4 Now add up all the terms in the table x² 3x -2 2x² 2x4 6x³ -4x² -x -x³ -3x² 2x 4 4x² 12x -8 (x² + 3x – 2)(2x² - x + 4) = 2x4 Multiplying polynomials – using a grid. Now add up all the terms in the table x² 3x -2 2x² 2x4 6x³ -4x² -x -x³ -3x² 2x 4 4x² 12x -8 (x² + 3x – 2)(2x² - x + 4) = 2x4 + 5x³ then the terms in x³ Multiplying polynomials – using a grid. Now add up all the terms in the table x² 3x -2 2x² 2x4 6x³ -4x² -x -x³ -3x² 2x 4 4x² 12x -8 (x² + 3x – 2)(2x² - x + 4) = 2x4 + 5x³ then the terms in x² - 3x² Multiplying polynomials – using a grid. Now add up all the terms in the table x² 3x -2 2x² 2x4 6x³ -4x² -x -x³ -3x² 2x 4 4x² 12x -8 (x² + 3x – 2)(2x² - x + 4) = 2x4 + 5x³ then the terms in x - 3x² + 14x Multiplying polynomials – using a grid. Now add up all the terms in the table x² 3x -2 2x² 2x4 6x³ -4x² -x -x³ -3x² 2x 4 4x² 12x -8 (x² + 3x – 2)(2x² - x + 4) = 2x4 + 5x³ and finally the constant term - 3x² + 14x -8 Multiplying polynomials – using a grid. x² 3x -2 2x² 2x4 6x³ -4x² -x -x³ -3x² 2x 4 4x² 12x -8 (x² + 3x – 2)(2x² - x + 4) = 2x4 + 5x³ - 3x² + 14x - 8 Using a grid to find a specific coefficient in a term. Eg: Find the coefficient of x2 in (x² + 2x – 2)(6x² -3x + 4) x² 2x 6x² -12x² -3x 4 -2 -6x² 4x² - 14x² Equating coefficients: used to find either a specific coefficient in a term or a unknown polynomial. (x + 6)P(x) = x3 + 12x2 + 34x – 12 (x + 6)(ax2 + bx + c) = x3 + 12x2 + 34x – 12 ax3 + (6a +b)x2 + (6b + c)x – 6c = x3 + 12x2 + 34x – 12 → (equate coefficients of x3) a=1 → (equate coefficients of x2) 6a + b = 12 because a = 1 → b = 6 → (equate coefficients of x) 6b + c = 34 because b = 6 → c = -2 → (equate to constant) 6c = -12 → c = -2 The Factor Theorem This is the Factor Theorem: If (x – a) is a factor of a polynomial f(x) then f(a) = 0. The converse is also true: If f(a) = 0 then (x – a) is a factor of a polynomial f(x). The Factor Theorem Use the factor theorem to show that (x + 2) is a factor of f(x) = 3x2 + 5x – 2. Hence or otherwise, factorize f(x). (x + 2) is a factor of 3x2 + 5x – 2 if f(–2) = 0 f(–2) = 3(–2)2 + 5(–2) – 2 = 12 – 10 – 2 =0 We can write By inspection And so 3x2 + 5x – 2 = (x + 2)(ax + b) a = 3 and b = –1 3x2 + 5x – 2 = (x + 2)(3x – 1) as required. Factorizing polynomials The factor theorem can be used to factorize polynomials by systematically looking for values of x that will make the polynomial equal to 0. For example: Factorize the cubic polynomial x3 – 3x2 – 6x + 8. Let f(x) = x3 – 3x2 – 6x + 8. f(x) has a constant term of 8. So possible factors of f(x) are: (x ± 1), (x ± 2), (x ± 4) or (x ± 8) f(1) = 1 – 3 – 6 + 8 = 0 (x – 1) is a factor of f(x). f(–1) = –1 – 3 + 6 + 8 ≠ 0 (x + 1) is not a factor of f(x). f(2) = 8 – 12 – 12 + 8 ≠ 0 (x – 2) is not a factor of f(x). f(–2) = – 8 – 12 + 12 + 8 = 0 (x + 2) is a factor of f(x). f(4) = 64 – 48 – 24 + 8 = 0 (x – 4) is a factor of f(x). We have found three factors and so we can stop. The given polynomial can therefore be fully factorized as: x3 – 3x2 – 6x + 8 = (x – 1)(x + 2)(x – 4) The Factor Theorem Factorize f(x) = x3 + 1 f(x) has a constant term of 1 so the only possible factors of f(x) are (x – 1) or (x + 1). (x – 1) is not a factor of f(x). (x + 1) is a factor of f(x). f(1) = 1 + 1 ≠ 0 f(–1) = (–1)3 + 1 = 0 We don’t know any other factors but we do know that the expression x + 1 must be multiplied by a quadratic expression to give x3 + 1. We can therefore write x3 + 1 = (x + 1)(ax2 + bx + c) We can see immediately that a = 1 and c = 1 so x3 + 1 = (x + 1)(x2 + bx + 1) = x3 + bx2 + x + x2 + bx + 1 = x3 + (b + 1)x2 + (b + 1)x + 1 Equating coefficients of x2 gives b+1=0 b = –1 So x3 + 1 can be fully factorized as x3 + 1 = (x + 1)(x2 – x + 1) Dividing polynomials Suppose we want to divide one polynomial f(x) by another polynomial of lower order g(x). There are two possibilities. Either: g(x) will leave a remainder when divided into f(x). g(x) will divide exactly into f(x). In this case, g(x) is a factor of f(x) and the remainder is 0. We can use either of two methods to divide one polynomial by another. These are by: using long division, or writing an identity and equating coefficients. Quotient Divisor Dividend + Remainder ) Dividend ÷ Divisor = Quotient + Remainder Dividend = Divisor x Quotient + Remainder For polynomials Cubic = Linear x Quadratic + Remainder Remainder = 0 then linear is factor of cubic. (factor theorem) Remainder ≠0 then linear is not factor and has remainder. (remainder theorem) Dividing polynomials by long division Using long division The method of long division used for numbers can be applied to the division of polynomial functions. Let’s start by looking at the method for numbers. For example, we can divide 5482 by 15 as follows: 3 6 5 15 5 4 8 2 – 4 5 9 8 – 9 0 8 2 – 7 5 7 This tells us that 15 divides into 5482 365 times, leaving a remainder of 7. We can write 5482 ÷ 15 = 365 remainder 7 or 5482 = 15 × 365 + 7 The The The The = + × dividend divisor quotient remainder Dividing polynomials by long division We can use the same method to divide polynomials. For example: What is f(x) = x3 – x2 – 7x + 3 divided by x – 3? x2 x–3 + 2x This tells us that x3 – x2 – 7x + 3 divided by x – 3 is x2 + 2x – 1. –1 x3 – x2 – 7x + 3 x3 – The remainder is 0 and so x – 3 is a factor of f(x). 3x2 2x2 – 7x 2x2 – We can write 6x x3 x 2 7 x + 3 = x2 + 2x 1 x 3 –x +3 –x+3 or 0 x3 – x2 – 7x + 3 =(x – 3)(x2 + 2x – 1) Dividing polynomials by long division Here is another example: What is f(x) = 2x3 – 3x2 + 1 divided by x – 2? 2x2 x–2 + x +2 This tells us that 2x3 – 3x2 + 1 divided by x – 2 is 2x2 + x + 2 remainder 5. 2x3 – 3x2 + 0x + 1 There is a remainder and so x – 2 is not a factor of f(x). 2x3 – 4x2 x2 + 0x We can write x2 – 2x 2x + 1 2x – 4 5 2 x3 3 x 2 1 5 2 2x x 2 x2 x2 or 2x3 – 3x2 + 1 =(x – 2)(2x2 + x + 2) + 5 Objective: Sketching a cubic curve - steps • Slope: • Intercept: • Roots: +ve -ve where does it cross y axis (x = 0) where it crosses x axis (y = 0) • Factor Theorem • Does quadratic factorise • Quadratic Formula: Discriminant / Repeat roots • Quadrant: Sub for x close to y check slope. y = (2x + 3)(x + 4)(5 – 2x) • Slope: • Intercept: -ve y = 60 3 , 2 5 2 • Roots: x=- • Quadrant: sub x = 1, 75 > 60 -4 or y = (x + 1)(x2 –x + 1) • • • • Slope: Intercept: Roots: Quadrant: +ve y=1 x = -1 na y = (2x + 3)(x - 1)2 • Slope: • Intercept: +ve y=3 3 , 2 • Roots: x=- • Quadrant: sub x = -1, 4 > 3 1 repeated root