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# Core 1 revision notes a

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1. 1. www.mathsbox.org.ukCORE 1 Summary Notes1 Linear Graphs and Equationsy = mx + cgradient =increase in yincrease in xy interceptGradient Facts§ Lines that have the same gradient are PARALLEL§ If 2 lines are PERPENDICULAR then m1 ´ m2 = – 1 or m2 =1m1e.g. 2y = 4x – 8y = 2x – 4 gradient = 2gradient of perpendicular line = -½Finding the equation of a straight linee.g. Find the equation of the line which passes through (2,3) and (4,8)GRADIENT =y1 – y2x1 – x2GRADIENT =3 – 8=– 5=52 – 4 – 2 2Method 1y – y1 = m(x – x1)y – 3 =5(x – 2)2y =5x – 222y = 5x – 4Using the point (2,3)Method 2y = mx + c3 =52´ 2 + cc = – 2y =5x – 222y = 5x – 4Using the point (2,3)Finding the Mid-Point( x1 y1) and (x2 y2) the midpoint isæèçx1 + x22y1 + y2 ö2 ø÷Given the pointsFinding the point of IntersectionTreat the equations of the graphs as simultaneous equations and solveFind the point of intersection of the graphs y = 2x – 7 and 5x + 3y = 451
2. 2. Substituting y = 2x – 7 gives 5x + 3(2x –7) = 455x + 6x – 21 = 4511x = 66x = 6 y = 2 x 6 – 7y = 5Point of intersection = (6, 5)2 Surds§ A root such as √5 that cannot be written exactly as a fraction is IRRATIONAL§ An expression that involves irrational roots is in SURD FORM e.g. 3√5§ ab = a ´ b§ab=ab75 – 12= 5 ´ 5 ´ 3 – 2 ´ 2 ´ 3= 5 3 – 2 3= 3 3e.g§ RATIONALISING THE DENOMINATOR3 + √2  and  3  √2 is called a pair of CONJUGATESThe product of any pair of conjugates is always a rational numbere.g.  (3 + √2)(3  √2) = 9  3√2+3√22= 721 – 5Rationalise the denominator of21 – 5´1 + 51 + 5=2 + 2 51 – 5=2 + 2 5– 4=– 1 – 523. Quadratic Graphs and EquationsSolution of quadratic equations§ Factorisationx =x2– 3x – 4 = 0(x + 1)(x – 4) = 0– 1 or x = 42www.mathsbox.org.uk
3. 3. § Completing the squarex2– 4x – 3 = 0(x – 2)2– (2)2– 3 = 0(x – 2)2– 7 = 0(x – 2)2= 7x – 2 = ± 7x = 2 + 7 or x = 2 – 7§ Using the formula to solve ax2+ bx + c = 0x =– b ± b2– 4ac2aE.g Solve x2- 4x - 3 = 0x =– ( – 4) ± ( – 4)2– 4 ´ 1 ´ ( – 3)2 ´ 1=4 ± 282= 2 ± 7§ The graph of y = ax2+ bx + c crosses the y axis at y = cIt crosses or touches the x-axis if the equation has real solutionsThe DISCRIMINANT of ax2+ bx + c = 0 is the expression b2– 4acIf b2– 4ac >0 there are 2 real distinct rootsIf b2– 4ac = 0 there is one repeated rootIf b2– 4ac < 0 there are no real rootsGraphs of Quadratic Functions§ The graph of any quadratic expression in x is called a PARABOLA§ The graph of y – q = k(x - p)2is a TRANSLATION of the graph y = kx2In VECTOR notation this translation can be described as úûùêëé pqThe equation can also be written as y = k(x – p)2+ qThe VERTEX of the graph is (p,q)The LINE OF SYMMETRY is x = p32x2+ 4x + 5 = 2(x + 1)2+ 3Vertex (-1,3)Line of symmetry x = -1yx1 2 3 4 5– 1– 2– 3– 4– 512345678910– 1– 2– 3– 4Translation of y=2x2úûùêëé-13www.mathsbox.org.uk
4. 4. 4 Simultaneous Equations§ Simultaneous equations can be solved by substitution to eliminate one of thevariablesSolve the simultanoeus equations y = 2x – 7 and x2+ xy + 2 = 0y = 7 + 2xso x2+ x(7 + 2x) + 2 = 03x2+ 7x + 2 = 0(3x + 1)(x + 2) = 0x = –13y = 613or x = – 2 y = 3§ A pair of simultaneous equations can be represented as graphs and thesolutions interpreted as points of intersection.If they lead to a quadratic equation then the DISCRIMINANT tells you thegeometrical relationship between the graphs of the functionsb2– 4ac < 0 no points of intersectionb2– 4ac = 0 1 point of intersectionb2– 4ac > 0 2 points of intersection5 InequalitiesLinear Inequality§ Can be solved like a linear equation exceptMultiplying or dividing by a negative value reverses the directionof the inequality signe.g Solve – 3x + 10 £ 4x– 3x + 10 £ 4– 3x £ – 6³ 2Quadratic Inequality§ Can be solved by either a graphical or algebraic approach.e.g. solve the inequality x2+ 4x – 5 <0Algebraic x2+ 4x – 5 < 0 factorising gives (x + 5)(x –1) < 0Using a sign diagramx + 5 - - - 0 + + + + + + +x – 1 - - - - - - - 0 + + +(x + 5)(x – 1) + + + 0 - - - 0 + + +The product is negative for – 5 < x < 14yx1 2– 1– 2– 3– 4– 5– 6– 72– 2– 4– 6– 8– 10GraphicalThe curve lies below thex–axis for –5 < x < 1www.mathsbox.org.uk
5. 5. 6 PolynomialsTranslation of graphsTo find the equation of a curve after a translation of ú replace x with (x-p) andûùêëé pqreplace y with (y - p)e.g The graph of y = x3is translated by úûùêëé 3-1The equation for the new graph isy=(x - 3)3-1Polynomial FunctionsA polynomial is an expression which can be writtenin the form a + bx + cx2+ dx3+ ex4+ fx5(a, b, c..are constants)yx1 2– 13 4 5– 1– 2– 3– 4– 512345– 2– 3– 4– 5· Polynomials can be divided to give a QUOTIENT and REMAINDERx2-3x + 7x + 2 x3- x2+ x + 15x3+2x2-3x2+ x Qutoient-3x2- 6x7x + 157x + 141 Remainder· REMAINDER THEOREMWhen P(x) is divided by (x - a) the remainder is P(a)· FACTOR THEOREMIf P(a) = 0 then (x – a) is a factor of P(x)e.g. The polynomial f(x) = hx3-10x2+ kx + 26 has a factor of (x - 2)When the polynomial is divided by (x+1) the remainder is 15.Find the values of h and k.Using the factor theorem f(2) = 08h -40 + 2k + 26 = 08h +2k = 14Using the remainder theorem f(-1) = 15-h -10 –k + 26 = 14h + k = 2Solving simultaneously k = 2 –h 8h + 2(2 –h) = 146h + 4 = 145www.mathsbox.org.uk