Form 3 Mathematics – FUNCTION msomimaktaba, November 13, 2018August 17, 2024 FUNCTIONA function is a set of ordered pairs which relates two sets such that to each element of one set there is only one element of the second setExampleRepresent the following set of ordered pairs in a pictorial diagram. (1,4), (2,3), (-1,2), (-2,-3).SolutionA function whose graph is such that any line drawn parallel to the x – axis at any point cuts it at only one point is one-to-one function.Examples.1. Write the expression of a function ‘ double plus one’ Solutionf:x 2x +1b) 2.Given the function G(x) = 4x-1. Find the value of G(-2). SolutionG(x) = 4x -1G(-2) = 4(-2) -1G(-2) = -8 -1G (-2) = -9Exercise 2.1 1. Write each of the following function in the form f: x f(x) use any functions symbol to represent the functions(a)Divide by 5 and add 2. (b)Subtract 7 and square (b)Cube and then double Solution(a) F:x f(x)F:x + 2(b) F: x (x-7)2(c) F:x x3+2x2. Find the value of the function for each given value of x.(a) f(x)= 2x+3; when (i) x=1 (ii) x= -2 (iii) x =a Solution (a) (i) when x=1f(x)= 2x+3 f(1)= 2(1)+3f(1)= 2+3∴ f(1)= 5.(ii)when x= -2.f(x) = 2(-2) +3f(-2) = -4+3∴f(-2) = -1(iii) when x=a f(x) = 2(a) +3f(a) =2a+3∴f(a) = 2a+3(b) C(x) = x3; when (i) x=1 (ii) x = -1 (iii)x= 0 (iv)x=bSolution(i) C(x) =x3C (1) =13C (1) =1C (1) =1(ii)C(x) =x3C (-1) = (-1)3C (-1) = -1(iii)C(x) =x3C (0) =03C (0) =0(iv)C(x) =x3C(b) = b3(c) K(x) = 3-x; when (i) x= -1 (ii)x=7Solution (i) K(x) = 3-xK (-1) = 3- (-1)K (-1) = 4 (ii) k(x)=3 – x k(7) = 3-7 k(7) = -4 DOMAIN AND RANGE OF FUNCTIONSExample 1.1. Find the domain and range of f(x) =2x+1let f(x) = y y=2x+1 Solution Domain = {x: xR}Range – make x the subject y=2x+1 y-1=2x (y-1)/2 =x ∴x=(y-1)/2 Range={y: yR}b Example 2 If y =4x+7 and its domain ={x : -10 < x< 10} find the range.Solution Domain = {x: -10 < x< 10}Range ={y : -33 < y < 47}y = 4x + 7 Table of values c Example 3. Y= √x and domain is -5 < x < 5, Find its range.Solution Domain = {x: -5 < x < 5}Range =yTable of value of x = √(y) (y)2= (√x)2y2= xx= y2√5= √y2y = √5Range = {y: o < y < √5}d Example 4. Given F(x) = find the domain and range.Solution Let f(x)=y Domain of y= For real values of y: 1-x2 > 0-x2 > 0-1-x2 > -1x2 < 1√x2< √1X <√1X <1∴Domain {x : X <1 }let F(x) = yy=To get the range, make x the subjecty2 = ()2y2= 1-x2X2=1-y2Therefore = X=For real value of x: 1- y2 > 0 y2 < 1 y < √1 y < 1 ...Range ={y : y <1} LINEARFUNCTIONS Is the function with form f(x) = mx + c.Where:f(x)= y m and c are real numbersm is called gradient[ slope].c is called y – intercept. Example1. Find the linear function f(x) given the slope of -2 and f( -1)=3 SolutionGiven: m(slope)=-2 x=-1 f(-1)=3 from; f(x) =mx +cf(-1) =( -2x-1)+c3 =2+c3-2 =cc=1∴f(x) = -2x+1Example 2. Find the linear function f(x) when m=3 and it passes through the points (2, 1)Solutionf(x) =mx +cf(2) = 3(2)+ c1= 6+c1-6 =c-5 = cc= -5f(x) = 3x + -5∴ f(x) =3x-5 Example 3 Find the linear function f (x) which passes through the points (-1, 1) and (0, 2 )Solution Slope = =1m = 1f(x) = mx +cf(-1) = 1x(-1) +c1 =-1+c c=2f(x) = 1x +2∴f(x) = x +2.4.Draw the graph of h(x) = 3x-4Table of values of functionX-101h(x)-7-4-1 Exercise In problems 1 to 3 find the equation of a linear function f(x) which satisfies the given properties. In each case, m dissolves the gradient.1). m =-3, f(1) = 3 2). m=2, f(0) =5 3). f(1) =2, f(-1) =34. Givenm= -4 , f(3) = -4 Find f(x) In the problem 5 to 9 draw the graphs of each of the given functions without using the table of values 5) f(x)= +6) f(x) =4 Solution 1. f(x) = mx + cf(x) = -3(1)+cf(x) =-3 +c3= -3 +c3+3 =c6=cC=6f(x) = -3x +62. f(x) =mx +cf(0) = (2 x 0)+c5= 0+cc=5f(x) =2x+53) 3.f(1) =2, f(-1) =3 Alternativelym = f(-1) = m(-1) + cM= 3= -m + c…………………..(i)f(1)=2 f(1) = m(1) + cf(x)= mx +c 2= m + c(ii)f(1)= –x 1+c Solve (i) and (ii) Simultaneouslyf(1) = –+c -m + c = 32=+c + m + c = 22+= c C=5/2C=2 put c in (i)f(x) =+ 2 3 = -m + 5/2m = 5/2 – 3 m= -1/2 ...f(x) =+ 24. f(x)= mx + cf(x) = -4(3)+c-4 =-12+c-4+12= c8=cC=8f(x) = -4x+85. f(x)= +y- intercept, x=0f(0)= 2/5[0] +1/5f(0)= 0+1/5y=1/5{0,0.2}x- intercept, y=00=2/5[x] +1/5x intercept = – ½ ( -1/2, 0)y intercept = 1/5 ( 0,0.2) 6. f(x) =4From f(x) = yy = 4 QUADRATIC FUNCTIONSA quadratic function is any function of the form;f(x)=ax2+bx+cWhere a ≠ 0a ,b and c are real numbers.When a = 1, b=0, and c= 0X-3-2-10123 f(x)9410149 The shape of the graph of f(x)=ax2+bx+c is a parabola•The line that divides the curve into two equal parts is called a line of symmetry [axis of symmetry]•Point (0,0) in f(x)=x2 called the turning point (vertex).If “a” is positive the turning point is called minimum point (least value).If “a” is negative the turning point is called maximum point.PROPERTIES OF QUADRATIC FUNCTIONSThink of f(x)= ax2+bx+cy=a(x2+bx/a)+cy= a( x2+bx/a+b2/4a2)+ c-b2/4a2=a(x + ) 2 +x> 0 thena(x+)2 > 0y = This is when x = –The turning point of the quadratic function is )Example Find the minimum or maximum point and line of symmetry f(x) = x2 – 2x-3. Draw the graph of f(x)Solution:Turning point = )= ( – )Maximum point = (1, -4)Line of symmetry is x=1x-5-2-101234 f(x) 1550-3-4-305 Exercise 1. Draw the graph of the function y=x2-6x+5 find the least value of this function and the corresponding value of xSolutionx-3-2-1012345y32211250-3-4-30 Least value.y= – 4 where x=32. Draw the graph of the function y =x2-4x+2 find the maximum function and the corresponding value of x use the curve to solve the following equationsa) X2-4x-2 = 0b) X2-4x-2 = 3b)Solution Table of values of y= x2-4x+2x-3-2-10123456y231472-12-12714 Find the maximum value of the function.;Maximum value=Y= ( ) = – 2 Find the maximum value of the function;Maximum value= )Maximum value= ) Maximum value[2, -2]The maximum value is = (-2,-2)a) x2-4x-2 = 0 add 4 both sidesx2-4x + 2 = 4 but x2-4x+2 = y ... y = 4 Draw a line y = 4 to the graph above. The solution from the graph isx1 = -1/2 , x2 = 9/2 (x1,x2) = (-1/2,9/2 )(b) x2-4x – 2 = 3add 4 both sides x2-4x – 2 +4 = 3+ 4 x2-4x + 2 = 7but x2-4x+2 = y ... y = 7 Draw a line y = 7 to the graph above. The solution from the graph isx1 = -1 , x2 = 5 (x1,x2) = (-1,5 ) 3. In the problem 3 to 5 write the function in the form f(x)= a( x + b)2 +c where a, b, c are constantsf(x) =5-x-9x2Solutionf(x) = -9x2 – x + 5= -9( x2 – ) + 5= -9( x2 – + ) + 5 + = -9(x – )2 + 4. In the following functions find:a) The maximum valueb) The axis of the symmetryf(x)= x2– 8x+18Solution) = )( 4, 2)Maximum value =2 where the axis of symmetry x=4.5. f(x) = 2x2+3x+1SolutionMaximum value ))The turning point of the graph is )The minimum value of the graph is y = – axis of symmetry = POLYNOMIAL FUNCTIONS The polynomial functions are the functions of the form, ” P(x) =an xn + a n-1 xn-1 + an-2 xn-2 + . . . + a1 x1+a0 x0 “. Where n is non negative integer and an, an-1, an-2 . . . a0 are real numbers. The degree of a polynomial function is the highest power of that polynomial function.Example a) f(x) =5x4 -7x3 +8x2– 2x+3 is a degree of 4b) H(x)=6x-8x2+9x9-6 is a degree of 9c) G(x) =16x-7 is a degree of 1d) M(x) =6 degree is 0 =6x0GRAPHS OF POLYNOMIAL FUNCTIONS EXAMPLEDraw the graph of f(x) = x3-2x2-5x+6SolutionTable of valuesx-3-2-101234 F[x]-240860-4018 STEP FUNCTIONS EXAMPLE1. If f is a function such that;Draw the graph and find its domain and range. Domain = {x: xR, except -3< x < -2}Range = {1,4, -2} 2. The function is defined bya) Sketch the graph of f(x) use the graph to determine the range and the domain. Find the value of f(-6) , f(0).State if it is a one to one functionDomain= {x : XR}Range{ y:y >1}f(-6) = -2f(0)= 2It is not a one to one function.EXERCISE1. Draw the graph of the following defined as indicated Solution: Solution 2. Given that f(x) = a) On the same set of axes sketch the graphs of f(x )and the inverse of f(x),From your graphs in [a] above determine;· (a)The domain and range of f(x) (b)The domain and range if the inverse of f(x) (c)Find f(- 5)and f(5) (d)Is f (x)a one to one? (e)Is the inverse of f(x) a function? (a) Domain of f(x ) = { x: x }(b) •Range of f(x) = { y : y } •Domain of f-1(x ) = { x : x }(c) Range of f-1(x ) = { y: x }(d) f(-5) = 1 and f(5) = 6(e)Yes the inverse of f(x) is a function and f(x) is one-to-one function ABSOLUTE VALUE FUNCTIONS The absolute value function is defined by f(x) Example1. Draw the graph of f(x) = | x |SolutionTable of values f(x) = | x |X-3-2-10123f(x)3210123 THE INVERSE OF A FUNCTIONGiven a functon y= f(x), the inverse of f(x) is denoted as f-1(x). The inverse of a function can be obtained by interchanging y with x (interchanging variables) and then make y the subject of the formula.Example1. Find the inverse of f(x) = 2x+3SolutionY= 2x+3X=2y+32y=x-3Y =∴ f-1(x EXPONENTIAL FUNCTIONSAn exponential function is the function of the form f(x) = nx where n is called base and x is called exponent. Example 1. Draw the graph of f(x)=2xSolution :Table of values X-3-2-10123 f(x)1/81/4½1248 2. Draw the graph of f (x)=2-xSolution:Table of values X-3-2-10123 f(X)84211/2¼1/8 Properties of exponential functionWhen x increases without bound, the function values increase without boundWhen x decreases, the function values decreases toward zeroThe graph of any exponential function passes through the point (0,1).The domain of the exponential function consists of all real numbers whereas the range consist of all positive values. LOGARITHMIC FUNCTIONLogarithmic function is any function of the form f(x)= read as function of logarithm x under base a or f(x) is the logarithm x base aExample Draw the graph of f(x)=SolutionTable of valuesx1/8¼1/21248f(x)-3-2-10123 THE INVERSE OF EXPONENTIAL AND LOGARITHMIC FUNCTIONThe inverse of the exponential function is the relation of logarithmic in the line y=xEXAMPLE 1. Draw the graph of the inverse of f(x) =2x and f(x)= under the same graph.Solution(i) y= 2xApply log on both sides log x=log 2y log x =y log 2 y = log (x- 2) f-1(x) = log (x- 2) (ii) f(x)= y= x= y= 2x Table of values X1/81/41/21248 f(x)-3-2-10123 2. Draw the graph of the function f(x)=3xTable of valuesx-3-2-10123f(X)1/271/91/3139273. Draw the graph of the function f(X)=8xSolutionTable of values X-3-2-10123 f(X)1/5121/641/81864512 Exercise1. Find the graph of y =2x and given that ¾ =2-0.42 draw the graph of f(x)= (3/4)xTable of values if f(x) =(3/4)xX-3-2-101232x1/81/4½1248-0.42x1.260.840.420-0.42-0.841.262-0.42x64/2716/94/313/49/1627/64 Copy and complete the following table and hence draw the graph of f(x)=(1/4)xX-3-2-101232x-6-4-20246-2x6420-2-4-62-2x6416401/41/161/64[1/4]x6416401/42/4¾ OPERATION OF POLYNOMIAL FUNCTIONADDITIONAL AND SUBTRACTIONThe sum of two polynomials is found by adding the coefficients of terms of the same degree or like terms, and subtraction can be found by subtracting the coefficient of like terms.EXAMPLEIf p(x) = 4x2– 3x+7, Q(x) =3x+2 and r(x)= 5x3-7x2+9SolutionSum =p(x)+q(x)+r(x)(4x2-3x+7)+(3x+2)+(5x3-7x2+9) 4x2-3x+7+3x+2+5x3-7+94x2+5x2+7x3+3x+7+2+95x3-3x2+18Alternatively the sum can be obtained by arranging them vertically4x2-3x+70x2+3x+25x3-7x2+0+95x3-3x2+0+18P(x)+q(x)+r(x)=5x3-3x2+18Subtract -5x2+9+5 from 3x2+7x-2Solution(3x2+7x-2)-(-5x2+9x+5)3x2+7x-2+5x2-9x-53x2+5x2+7x-9x-2-58x2-2x-7 answerMULTIPLICATION The polynomial R(x) and S(x) can be multiplied by forming all the product of terms from R(x) and terms from S(x) and then summing all the products by collecting the like terms. The product can be denoted by RS(X) example If p(x)=2x2-x+3 and q(x)=3x3-x find the product p(x)q(x)SolutionP(x) q(x)= (2x2-x+3) (3x3-x)6x5-2x3-3x4+x2+9x3-3x6x5-3x4+7x3+x2-3xDIVISION The method used in dividing one polynomial by another polynomial of equal or lower degree is the same to the one used for the long division of numberExample: 1. Given p[x] =x3-3x2+4x+2 and q[x]=x-1 find Solutionx3-x20-2x2+4x-2x2+2x0+2x+22x-20+4Therefore = x2 – 2x + 2 + This meansX3– 3x2+4x+2 is dividend x-1 is divisorX2-2x+2 is the quotient and 4 is the remainderThenX3-3x2+4x+2 = (x2-2x+2) (x-1) + 4Divided quotient x divisor + remainder 2. Divide p(x) = x3-8 by q(x) = x-2Solution:-x3 -2x22x2-8-2x2-4x4x-84x-8 – – In the example [2] above there is a remainder i.e. the remainder is zero so dividing one function by another function is one of the way of finding the factors of the polynomial thus if we divide p(x) by q(x) is one of the factors of p(x) other factors can be obtained by fractionazing the quotient q(x).3. Given p(x) = x3-7x+6 and q(x) = x+3 determine whether or not d(x) is one of the factors of p(x) and hence find the factors if p(x)SolutionX3+3x2-3x2-7x-3x2-9x2x+6-2x+6– – Since the remainder is zero d(x) =x+3 is one of the factors of p(x)To factorizeX2-3x+2(x2-x) – (2x+2)x(x-1)-2(x-1)(x-2) (x-1)Therefore other factors are (x-2) and (x-1) THE REMAINDER THEOREM The remainder theorem is the method of finding the remainder without using long divisionExample 1. If p(x) = (x-2) and q(x)+8. Dividend divisor quotient remainder, Then by taking x-2=0 we findx-2= 0 x=2subtracting x=2 in p(x)p(2) =2-2 q(x) +rp(2)= 0 x q(x)+rp(2)= 0+rp(2) =rSo the remainder r is the value of the polynomial p(x) when x=22. Give p(x) = x3-3x2+6x+5 is divided by d(x) = x-2 find the remainder using the remainder theoremSolutionLet d(x)=0x-2=0X-2+2 =0+2X=2Subtracting x=2 in p(x)P(2)= 23-3[2]2+6[2]+5P(2)=8-12+12+5P(2)=8+5P(2)=13The remainder is 13 3. The remainder theorem states that if the polynomial p[x] is divided by [x-a] then the remainder ‘r’ is given by p[a]P(x) =(x-a) q(x)+r henceP(a) =(a-a) q(x)+rP(a)=o(q(x))+rP(a)=0+rP(a)=r More Examples1. By using the remainder theorem, Find the remainder when p[x] =4x2-6x+5 is divided by d[x] =2x-1Solutiond (x)=02x-1=02x-1+1=0+1X = SubstitutingX= in p(x)P() =4 x -6()+5P()=1 – 6 x +5P()=1 -3+5P()= 3The remainder is 32. P(x)= 3x2-5x+5 is divided by d(x) =x+4SolutionLet d(x)=0X+4=0X+4=0-4X=-4Subtracting x=-4 in p(x)P(-4)=3(-4)2-=5(-4)+5P(-4)=48+20+5P(-4)=68+5P(-4)= 73The remainder is 73 3. P(x)= x3+2x2-x+4 is divided by d(x) = x+3Solution Let d(x)=0X+3=0-3X=-3Substituting x=-3 in p(x)P(-3)= -33+2(-3)2-(-3)+4P(-3)=-27+18+3+4P(-3)=-27+21+4P(-3)=-27+25P(-3)=-2The remainder is -24. Find the value of ‘a’ if x3-3x2+ax+5 has the remainder of 17 when divided by x-3SolutionBy using remainder theoremLet x-3 =0x-3=0+3x=3Substituting x=3 we havex3-3x2+ax+5=17(3)3-3(3)2+a(3)+5=1727-27+3a+5=173a+5=17-53a=12a=45. If ax2+3x-5 has a remainder -3 when divided by x-2. Find the value of a.SolutionBy using the remainder theoremLet x-2=0x-2=0+2x=2subtracting x=2 we havea(2)2+3(2)-5 =-34a+6-5= -34a+1= -34a= -4a = -1Exercise1. Divide p(x) by d(x) in the followingP(x)=2x2+3x+7 d(x)=x2+4Solution Use remainder theorem to find the remainder when;1. P(x)=x3-2x2+5x-4 is divide by d(x)=x-2SolutionLet d[x]=0x-2 + 2=0+2x=2substituting x=2 in p(x)p(2)=23-2[2]2+5[2]-4p(2)=8-8+10-4p(2)=10-4p(2)=10-4p(2)=10-4Therefore the remainder is 62. p(x)= 2x4+ x3+x- is divided by d(x)=x+2solutionlet d(x)=0x+2 – 2 = 0-2x = -2Substituting x = – 2 in p(x)p(-2)=2(-2)4+(-2)3+(-2)( –)p(-2)= 2(16)+(-8)-2( –)p(-2)=32-8-2 ( –)p(-2)=24-2( –)p(-2)=21Therefore the remainder is 21 ALL NOTES FOR ALL SUBJECTS QUICK LINKS:AGRICULTURE O LEVEL PURE MATHEMATICS A LEVELBAM NOTES A LEVELBASIC MATH O LEVELBIOLOGY O/A LEVELBOOK KEEPING O LEVELCHEMISTRY O/A LEVELCIVICS O LEVELCOMPUTER(ICT) O/A LEVELECONOMICS A LEVELENGLISH O/A LEVELCOMMERCE O/A LEVELACCOUNTING A LEVELGENERAL STUDIES NOTESGEOGRAPGY O/A LEVELHISTORY O/A LEVELKISWAHILI O/A LEVELPHYSICS O/A LEVELMOCK EXAMINATION PAPERSNECTA PAST PAPERS Basic Mathematics Study Notes Form 3 Basic Mathematics Study Notes FORM 3MATHEMATICSPost navigationPrevious postNext postRelated Posts Basic Mathematics Study Notes MATHEMATICS FORM 1 – APPROXIMATIONS November 11, 2018August 17, 2024APPROXIMATIONS Is the process of rounding off a number in the given places in rounding makes numbers easier to deal with but at the same time reduces their accuracy. The methods used for approximation or rounding off numbers are decimal places and significant figures. ROUNDING OFF PROCEDURES i) If the… Read More Basic Mathematics Study Notes Form 2 Mathematics – CONGRUENCE OF SIMPLE POLYGON November 13, 2018August 17, 2024 CONGRUENCE OF SIMPLE POLYGON The triangles above are drawn such that CB= ZY AC=XZ B=YX Corresponding sides in the triangles are those sides which are opposite to the equal angles i.e. If the corresponding sides are equal i.e. In general, polygons are congruent if corresponding sides and corresponding… Read More Basic Mathematics Study Notes Form 2 Mathematics – TRIGONOMETRY November 13, 2018August 17, 2024TRIGONOMETRY Trigonometric ratio Introduction: TRI – is the Greek word which means three. -Trigonometry is the branch of mathematics which deals with measurement. -A Trigonometric ratio consists of three parts that is – Hypotenuse, Adjacent and opposite. Consider the diagram below which is the right angled triangle For example… Read More Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment *Name * Email * Website Save my name, email, and website in this browser for the next time I comment. Δ
Basic Mathematics Study Notes MATHEMATICS FORM 1 – APPROXIMATIONS November 11, 2018August 17, 2024APPROXIMATIONS Is the process of rounding off a number in the given places in rounding makes numbers easier to deal with but at the same time reduces their accuracy. The methods used for approximation or rounding off numbers are decimal places and significant figures. ROUNDING OFF PROCEDURES i) If the… Read More
Basic Mathematics Study Notes Form 2 Mathematics – CONGRUENCE OF SIMPLE POLYGON November 13, 2018August 17, 2024 CONGRUENCE OF SIMPLE POLYGON The triangles above are drawn such that CB= ZY AC=XZ B=YX Corresponding sides in the triangles are those sides which are opposite to the equal angles i.e. If the corresponding sides are equal i.e. In general, polygons are congruent if corresponding sides and corresponding… Read More
Basic Mathematics Study Notes Form 2 Mathematics – TRIGONOMETRY November 13, 2018August 17, 2024TRIGONOMETRY Trigonometric ratio Introduction: TRI – is the Greek word which means three. -Trigonometry is the branch of mathematics which deals with measurement. -A Trigonometric ratio consists of three parts that is – Hypotenuse, Adjacent and opposite. Consider the diagram below which is the right angled triangle For example… Read More