Like this document? Why not share!

- Core 1 revision booklet edexcel by claire meadows-smith 10822 views
- Core 2 revision notes by claire meadows-smith 640 views
- 3.6 -curve sketching-c by Ron_Eick 232 views
- C2 june 2012 by anicholls1234 2911 views
- Grace Kennedy Group by Grace Kennedy 119 views
- Business Development Grace Kennedy ... by Lakesia Wright 1414 views

No Downloads

Total Views

1,042

On Slideshare

0

From Embeds

0

Number of Embeds

0

Shares

0

Downloads

31

Comments

0

Likes

2

No embeds

No notes for slide

- 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. 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. § 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 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. 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
- 6. · A circle with centre (a,b) and radius r has the equation (x - a)2+( y - b)2=r2e.g. A circle has equation x2+ y2+ 2x – 6y= 0Find the radius of the circle and the coordinates of its centre.x2+ 2x + y2– 6y = 0(x + 1)2– 1 + (y – 3)2– 9 = 0(x + 1)2+ (y – 3)2= 10Centre (1, 3) radius = √10· A line from the centre of a circle to where a tangent touches the circle isperpendicular to the tangent. A perpendicular to a tangent is called aNORMAL.e.g. C(-2,1) is the centre of a circle and S(-4,5) is a point on thecircumference. Find the equations of the normal and the tangent tothe circle at S.C(-2,1)S (-4,5)Gradient of SC is1 – 5– 2( – 4)=– 42= – 2Equation of SC y = – 2x + 7Gradient of the tangent = –1– 2=12Equation of y =1x + 72· Solving simultaneously the equations of a line and a circle results in aquadratic equation.b2- 4ac > 0 the line intersects the circleb2- 4ac = 0 the line is a tangent to the circleb2- 4ac < 0 the line fails to meet the circle8 Rates of Change· The gradient of a curve is defined as the gradient of the tangentGradient is denoteddxdyif y is given as a function of xGradient is denoted by f ’(x) if the function is given as f(x)· The process of findingdydxor f’(x) is known as DIFFERENTIATING· Derivativesf(x) = xnf (x) = nxn – 1f(x) = a f (x) = 06www.mathsbox.org.ukEquation of a Circle· A circle with centre (0,0) and radius r has the equation x2+y2=r2
- 7. dxy = x3+ 4x2– 3x + 6dy= 3x2+ 8x – 39 Using Differentiation· If the value ofdydxis positive at x = a, then y is increasing at x = a· If the value ofdydxis negative at x = a, then y is decreasing at x = a· Points wheredydx= 0 are called stationary pointsMinimum and Maximum Points (Stationary Points)Stationary points can be investigatedyx1 2 3 4 5– 1– 2– 3– 4– 512345– 1– 2– 3– 4– 5+ ve- veGRADIENTGRADIENT0- ve+ veLocal MinimumLocal Maximum0· by calculating the gradients close to the point (see above)· by differentiating again to findd2or f “(x)2ydxod2> 0 then the point is a local minimum2ydxod22ydx< 0 then the point is a local maximumOptimisation ProblemsOptimisation means getting the best result. It might mean maximising (e.g. profit) orminimising (e.g. costs)10 Integration§ Integration is the reverse of differentiation7www.mathsbox.org.uk
- 8. ò xndx =xn + 1n + 1+ cConstant of integratione.g. Given thatf (x) = 8x3– 6x and that f(2) = 9 find f(x)f(x) =ò 8x3– 6x dx=8x–6x+ c4422= 2x4– 3x2+ cTo find c use f(2) = 9So f(x) = 2x4– 3x2–1132 – 12 + c = 9c = – 1111 Area Under a Graph§ The are under the graph of y = f(x) between x= a and x = b is found byevaluating the definite integralòbf(x) dxae.g. Calculate the area under the graph of y = 4x – x3between thelines x = 0 and x = 2 yx1 2 3 4 5– 112345– 1ò024x – x3dx == 2x2–x44= (8 – 4) – (0 – 0)= 4§ An area BELOW the x–axis has a NEGATIVE VALUE8www.mathsbox.org.uk

Be the first to comment