y intercept Gradient Facts Lines that have the same gradient are PARALLEL

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1 CORE Summar Notes Linear Graphs and Equations = m + c gradient = increase in increase in intercept Gradient Facts Lines that have the same gradient are PARALLEL If lines are PERPENDICULAR then m m = or m = - e.g. = 8 = gradient = gradient of perpendicular line = -½ m Finding the equation of a straight line e.g. Find the equation of the line which passes through (,) and (,8) GRADIENT = GRADIENT = 8 = = Method = m( ) Using the point (,) Method = m + c Using the point (,) = ( ) = = = + c Finding the Mid-Point c = = = Given the points ( ) and ( ) the midpoint is æ è ç + ' + ö ø Finding the point of Intersection Treat the equations of the graphs as simultaneous equations and solve Find the point of intersection of the graphs = 7 and + =

2 Substituting = 7 gives + ( 7) = + 6 = = 66 = 6 = 6 7 = Point of intersection = (6, ) Surds A root such as that cannot be written eactl as a fraction is IRRATIONAL An epression that involves irrational roots is in SURD FORM e.g. ab = a b a b = a b e.g 7 = = = RATIONALISING THE DENOMINATOR + and is called a pair of CONJUGATES The product of an pair of conjugates is alwas a rational number e.g. ( + )( ) = 9 + = 7 Rationalise the denominator of + + = + = + =. Quadratic Graphs and Equations Solution of quadratic equations Factorisation = 0 ( + )( ) = 0 = or =

3 Completing the square = 0 ( ) () = 0 ( ) 7 = 0 ( ) = 7 = ± 7 = + 7 or = 7 Using the formula to solve a + b + c = 0 = E.g Solve - - = 0 b ± b ac a = ( ) ± ( ) ( ) = ± 8 = ± 7 The graph of = a + b + c crosses the ais at = c It crosses or touches the -ais if the equation has real solutions The DISCRIMINANT of a + b + c = 0 is the epression b ac If b ac >0 there are real distinct roots If b ac = 0 there is one repeated root If b ac < 0 there are no real roots Graphs of Quadratic Functions The graph of an quadratic epression in is called a PARABOLA The graph of q = k( - p) is a TRANSLATION of the graph = k In VECTOR notation this translation can be described as The equation can also be written as = k( p) + q The VERTEX of the graph is (p,q) The LINE OF SYMMETRY is = p é pù ê ú ëq û + + = ( + ) + Verte (-,) Line of smmetr = Translation of = é-ù ê ú ë û

4 Simultaneous Equations Simultaneous equations can be solved b substitution to eliminate one of the variables Solve the simultanoeus equations = 7 and + + = 0 = 7 + so + (7 + ) + = = 0 ( + )( + ) = 0 = = 6 or = = A pair of simultaneous equations can be represented as graphs and the solutions interpreted as points of intersection. If the lead to a quadratic equation then the DISCRIMINANT tells ou the geometrical relationship between the graphs of the functions b ac < 0 no points of intersection b ac = 0 point of intersection b ac > 0 points of intersection Inequalities Linear Inequalit Can be solved like a linear equation ecept Multipling or dividing b a negative value reverses the direction of the inequalit sign e.g Solve ³ Quadratic Inequalit Can be solved b either a graphical or algebraic approach. e.g. solve the inequalit + <0 Algebraic + < 0 factorising gives ( + )( ) < 0 Using a sign diagram ( + )( ) The product is negative for < < Graphical 7 6 The curve lies below the ais for < < 6 8 0

5 6 Polnomials Translation of graphs To find the equation of a curve after a translation of replace with ( - p) e.g The graph of = is translated b é ù ê ú ë- û é pù ê ëqû ú replace with (-p) and The equation for the new graph is =( - ) - Polnomial Functions A polnomial is an epression which can be written in the form a + b + c + + e + f (a, b, c..are constants) Polnomials can be divided to give a QUOTIENT and REMAINDER Qutoient Remainder REMAINDER THEOREM When P() is divided b ( - a) the remainder is P(a) FACTOR THEOREM If P(a) = 0 then ( a) is a factor of P() e.g. The polnomial f() = h -0 + k + 6 has a factor of ( - ) When the polnomial is divided b (+) the remainder is. Find the values of h and k. Using the factor theorem f() = 0 8h -0 + k + 6 = 0 8h +k = Using the remainder theorem f(-) = -h -0 k + 6 = h + k = Solving simultaneousl k = h 8h + ( h) = 6h + =

6 Equation of a Circle A circle with centre (0,0) and radius r has the equation + =r A circle with centre (a,b) and radius r has the equation ( - a) +( - b) =r e.g. A circle has equation + + 6= 0 Find the radius of the circle and the coordinates of its centre = 0 ( + ) + ( ) 9 = 0 ( + ) + ( ) = 0 Centre (, ) radius = 0 A line from the centre of a circle to where a tangent touches the circle is perpendicular to the tangent. A perpendicular to a tangent is called a NORMAL. e.g. C(-,) is the centre of a circle and S(-,) is a point on the circumference. Find the equations of the normal and the tangent to the circle at S. Gradient of SC is ( ) = = S (-,) Equation of SC = + 7 Gradient of the tangent = = C(-,) Equation of = + 7 Solving simultaneousl the equations of a line and a circle results in a quadratic equation. b - ac > 0 the line intersects the circle b - ac = 0 the line is a tangent to the circle b - ac < 0 the line fails to meet the circle 8 Rates of Change The gradient of a curve is defined as the gradient of the tangent Gradient is denoted d if is given as a function of Gradient is denoted b f () if the function is given as f() The process of finding d Derivatives f() = n f '() = n n f() = a f '() = 0 or f () is known as DIFFERENTIATING 6

7 = d = Using Differentiation If the value of d is positive at = a, then is increasing at = a If the value of d is negative at = a, then is decreasing at = a Points where d = 0 are called stationar points Minimum and Maimum Points (Stationar Points) Local Maimum 0 + ve - ve GRADIENT Local Minimum - ve + ve 0 GRADIENT Stationar points can be investigated b calculating the gradients close to the point (see above) b differentiating again to find d o d o d or f () > 0 then the point is a local minimum < 0 then the point is a local maimum Optimisation Problems Optimisation means getting the best result. It might mean maimising (e.g. profit) or minimising (e.g. costs) 0 Integration Integration is the reverse of differentiation 7

8 ò n = n + n + + c Constant of integration e.g. Given that f '() = 8 6 and that f() = 9 find f() f() = ò 8 6 = c = + c To find c use f() = 9 + c = 9 c = So f() = Area Under a Graph The are under the graph of = f() between = a and = b is found b evaluating the definite integral ò b f() a e.g. Calculate the area under the graph of = between the lines = 0 and = ò = 0 = = (8 ) (0 0) = An area BELOW the ais has a NEGATIVE VALUE 8