Functions and Their Graphs. Jackie Nicholas Janet Hunter Jacqui Hargreaves

Transcription

1 Mathematics Learning Centre Functions and Their Graphs Jackie Nicholas Janet Hunter Jacqui Hargreaves c 997 Universit of Sdne

2 Contents Functions. What is a function? Definition of a function..... The Vertical Line Test Domain of a function Range of a function Specifing or restricting the domain of a function The absolute value function Eercises... 8 More about functions. Modifing functions b shifting..... Vertical shift Horizontal shift.... Modifing functions b stretching Modifing functions b reflections Reflection in the -ais Reflection in the -ais Other effects Combining effects Graphing b addition of ordinates Using graphs to solve equations Eercises Even and odd functions Increasing and decreasing functions Eercises... 3 Piecewise functions and solving inequalities 7 3. Piecewise functions Restricting the domain Eercises Inequalities Eercises... 35

3 Mathematics Learning Centre, Universit of Sdne 3 Polnomials 36. Graphs of polnomials and their zeros Behaviour of polnomials when is large Polnomial equations and their roots Zeros of the quadratic polnomial Zeros of cubic polnomials Polnomials of higher degree Eercises.... Factorising polnomials..... Dividing polnomials The Remainder Theorem The Factor Theorem Eercises Solutions to eercises 5

4 Mathematics Learning Centre, Universit of Sdne Functions In this Chapter we will cover various aspects of functions. We will look at the definition of a function, the domain and range of a function, what we mean b specifing the domain of a function and absolute value function.. What is a function?.. Definition of a function A function f from a set of elements X to a set of elements Y is a rule that assigns to each element in X eactl one element in Y. One wa to demonstrate the meaning of this definition is b using arrow diagrams. X f Y 5 X g Y f : X Y is a function. Ever element in X has associated with it eactl one element of Y. g : X Y is not a function. The element in set X is assigned two elements, 5and6inset Y. A function can also be described as a set of ordered pairs (, ) such that for an -value in the set, there is onl one -value. This means that there cannot be an repeated -values with different -values. The eamples above can be described b the following sets of ordered pairs. F = {(,5),(3,3),(,3),(,)} is a function. G={(,5),(,),(,3),(3,3),(,6)} is not a function. The definition we have given is a general one. While in the eamples we have used numbers as elements of X and Y, there is no reason wh this must be so. However, in these notes we will onl consider functions where X and Y are subsets of the real numbers. In this setting, we often describe a function using the rule, = f(), and create a graph of that function b plotting the ordered pairs (, f()) on the Cartesian Plane. This graphical representation allows us to use a test to decide whether or not we have the graph of a function: The Vertical Line Test.

5 Mathematics Learning Centre, Universit of Sdne.. The Vertical Line Test The Vertical Line Test states that if it is not possible to draw a vertical line through a graph so that it cuts the graph in more than one point, then the graph is a function. This is the graph of a function. All possible vertical lines will cut this graph onl once. This is not the graph of a function. The vertical line we have drawn cuts the graph twice...3 Domain of a function Forafunction f : X Y the domain of f is the set X. This also corresponds to the set of -values when we describe a function as a set of ordered pairs (, ). If onl the rule = f() isgiven, then the domain is taken to be the set of all real for which the function is defined. For eample, = has domain; all real. This is sometimes referred to as the natural domain of the function... Range of a function Forafunction f : X Y the range of f is the set of -values such that = f() for some in X. This corresponds to the set of -values when we describe a function as a set of ordered pairs (, ). The function = has range; all real. Eample a. State the domain and range of = +. b. Sketch, showing significant features, the graph of = +.

6 Mathematics Learning Centre, Universit of Sdne 3 Solution a. The domain of = +is all real. We know that square root functions are onl defined for positive numbers so we require that +, ie. We also know that the square root functions are alwas positive so the range of = +is all real. b. 3 3 The graph of = +. Eample a. State the equation of the parabola sketched below, which has verte (3, 3) b. Find the domain and range of this function. Solution a. The equation of the parabola is = 6 3. b. The domain of this parabola is all real. The range is all real 3. Eample Sketch + =6and eplain wh it is not the graph of a function. Solution + =6isnot a function as it fails the vertical line test. For eample, when = = ±.

7 Mathematics Learning Centre, Universit of Sdne The graph of + = 6. Eample Sketch the graph of f() =3 and find a. the domain and range b. f(q) c. f( ) d. f(+h) f() h, h. Solution 3 The graph of f() =3. a. The domain is all real. The range is all real where.5. b. f(q) =3q q

8 Mathematics Learning Centre, Universit of Sdne 5 c. f( )=3( ) ( ) =3 d. f( + h) f() h = (3( + h) ( + h) ) (3() () ) h = 6+3h (h +h +) h = h h h = h Eample Sketch the graph of the function f() =( ) +and show that f(p) =f( p). Illustrate this result on our graph b choosing one value of p. Solution 6 The graph of f() =( ) +. f( p) = (( p) ) + = ( p) + = (p ) + = f(p)

9 Mathematics Learning Centre, Universit of Sdne 6 6 The sketch illustrates the relationship f(p) =f( p) for p =. p = ( ) = 3, and f( ) = f(3). If p = then. Specifing or restricting the domain of a function We sometimes give the rule = f() along with the domain of definition. This domain ma not necessaril be the natural domain. For eample, if we have the function = for then the domain is given as. The natural domain has been restricted to the subinterval. Consequentl, the range of this function is all real where. illustrate this b sketching the graph. We can best The graph of = for.

10 Mathematics Learning Centre, Universit of Sdne 7.3 The absolute value function Before we define the absolute value function we will review the definition of the absolute value of a number. The Absolute value of a number is written and is defined as = if or = if <. That is, =since is positive, but =since isnegative. We can also think of geometricall as the distance of from on the number line. = = More generall, a can be thought of as the distance of from a on the numberline. a = a a Note that a = a. The absolute value function is written as =. We define this function as + if = if < From this definition we can graph the function b taking each part separatel. The graph of = is given below. = < = The graph of =.

11 Mathematics Learning Centre, Universit of Sdne 8 Eample Sketch the graph of =. Solution For = we have +( ) when or = ( ) when < or < That is, for = + for < Hence we can draw the graph in two parts. = + < = 3 The graph of =. We could have sketched this graph b first of all sketching the graph of = and then reflecting the negative part in the -ais. We will use this fact to sketch graphs of this tpe in Chapter.. Eercises. a. State the domain and range of f() = 9. b. Sketch the graph of = 9.. Given ψ() = ψ( + h) ψ() +5,find, in simplest form, h. h 3. Sketch the following functions stating the domain and range of each: a. =

12 Mathematics Learning Centre, Universit of Sdne 9 b. = c. = d. =.. a. Find the perpendicular distance from (, ) to the line + + k = b. If the line + + k =cuts the circle + =in two distinct points, find the restrictions on k. 5. Sketch the following, showing their important features. a. = ( ) b. =. 6. Eplain the meanings of function, domain and range. Discuss whether or not = 3 is a function. 7. Sketch the following relations, showing all intercepts and features. State which ones are functions giving their domain and range. a. = b. = c. = 3 d. =, e. =. 8. If A() = ++,,prove that A(p) =A( ) for all p. p 9. Write down the values of which are not in the domain of the following functions: a. f() = b. g() =. If φ() =log ( ), find in simplest form: a. φ(3) + φ() + φ(5) b. φ(3) + φ() + φ(5) + + φ(n). a. If = + and =(z ), find when z =3. b. Given L() = +and M() =, find i L(M()) ii M(L())

13 Mathematics Learning Centre, Universit of Sdne. Using the sketches, find the value(s) of the constants in the given equations: 5 = a b (,) = a b + (,) 3. a. Define a, the absolute value of a, where a is real. b. Sketch the relation + =.. Given that S(n) = n, find an epression for S(n ). n+ Hence show that S(n) S(n ) = (n )(n+).

14 Mathematics Learning Centre, Universit of Sdne More about functions In this Chapter we will look at the effects of stretching, shifting and reflecting the basic functions, =, = 3, =, =, = a, + = r. We will introduce the concepts of even and odd functions, increasing and decreasing functions and will solve equations using graphs.. Modifing functions b shifting.. Vertical shift We can draw the graph of = f() +k from the graph of = f() asthe addition of the constant k produces a vertical shift. That is, adding a constant to a function moves the graph up k units if k>ordownk units if k<. For eample, we can sketch the function = 3 from our knowledge of = b shifting the graph of = down b 3units. That is, if f() = then f() 3= 3. 3 = (,) = 3 3 (, ) We can also write = f() 3as +3=f(), so replacing b +3in = f() also shifts the graph down b 3 units... Horizontal shift We can draw the graph of = f( a) ifweknow the graph of = f() asplacing the constant a inside the brackets produces a horizontal shift. Ifwereplace b a inside the function then the graph will shift to the left b a units if a< and to the right b a units if a>.

15 Mathematics Learning Centre, Universit of Sdne For eample we can sketch the graph of = from our knowledge of = this graph to the right b units. That is, if f() = then f( ) =. 3 b shifting (,) (,3) (, ) (, ) 3 3 Note that the function = to (, 3). is not defined at =. The point (, ) has been shifted. Modifing functions b stretching We can sketch the graph of a function = bf() (b>) if we know the graph of = f() as multipling b the constant b will have the effect of stretching the graph in the - direction b a factor of b. That is, multipling f() bb will change all of the -values proportionall. For eample, we can sketch = from our knowledge of = as follows: (,) (,) The graph of =. The graph of =. Note, all the - values have been multiplied b, but the -values are unchanged. We can sketch the graph of = from our knowledge of = as follows:

16 Mathematics Learning Centre, Universit of Sdne 3 (,) (,/) The graph of =. The graph of =. Note, all the -values have been multiplied b, but the -values are unchanged..3 Modifing functions b reflections.3. Reflection in the -ais We can sketch the function = f() ifweknow the graph of = f(), as a minus sign in front of f() has the effect of reflecting the whole graph in the -ais. (Think of the -ais as a mirror.) For eample, we can sketch = from our knowledge of =. The graph of =. The graph of =. It is the reflection of = in the -ais..3. Reflection in the -ais We can sketch the graph of = f( ) ifweknow the graph of = f() asthe graph of = f( ) isthe reflection of = f() inthe-ais.

17 Mathematics Learning Centre, Universit of Sdne For eample, we can sketch =3 from our knowledge of =3. The graph of =3. The graph of =3.Itisthe reflection of =3 in the -ais.. Other effects We can sketch the graph of = f() if we know the graph of = f() asthe effect of the absolute value is to reflect all of the negative values of f() in the -ais. For eample, we can sketch the graph of = 3 from our knowledge of the graph of = 3. 3 The graph of = 3. The graph of = 3. The negative values of = 3have been reflected in the -ais..5 Combining effects We can use all the above techniques to graph more comple functions. For eample, we can sketch the graph of = (+) from the graph of = provided we can analse the combined effects of the modifications. Replacing b +(or ( )) moves the

18 Mathematics Learning Centre, Universit of Sdne 5 graph to the left b unit. The effect of the sign in front of the brackets turns the graph up side down. The effect of adding moves the graph up units. We can illustrate these effects in the following diagrams. The graph of =. The graph of =( +). The graph of = has been shifted unit to the left. The graph of = ( +). The graph of =( +) has been reflected in the -ais. The graph of = (+). The graph of = ( +) has been shifted up b units. Similarl, we can sketch the graph of ( h) +( k) = r from the graph of + = r. Replacing b h shifts the graph sidewas h units. Replacing b k shifts the graph up or down k units. (We remarked before that = f()+k could be written as k = f().) For eample, we can use the graph of the circle of radius 3, + =9,tosketch the graph of ( ) +( +) =9.

19 Mathematics Learning Centre, Universit of Sdne 6 The graph of + =9. This is a circle centre (, ), radius 3. (, ) (, ) (, ) (5, ) (, 7) The graph of ( ) +( +) =9. This is a circle centre (, ), radius 3. Replacing b has the effect of shifting the graph of + =9two units to the right. Replacing b +shifts it down units..6 Graphing b addition of ordinates We can sketch the graph of functions such as = + b drawing the graphs of both = and = on the same aes then adding the corresponding -values.

20 Mathematics Learning Centre, Universit of Sdne 7 = = 6 6 The graph of = +. At each point of the -values of = and = have been added. This allows us to sketch the graph of = +. This technique for sketching graphs is ver useful for sketching the graph of the sum of two trigonometric functions..7 Using graphs to solve equations We can solve equations of the form f() =k b sketching = f() and the horizontal line = k on the same aes. The solution to the equation f() =k is found b determining the -values of an points of intersection of the two graphs.

21 Mathematics Learning Centre, Universit of Sdne 8 For eample, to solve 3 =wesketch = 3 and =on the same aes. (,) (5,) 6 The -values of the points of intersection are and 5. Therefore 3 =when = or =5. Eample The graph of = f() issketched below. 3 (3, ) For what values of k does the equation f() =k have. solution. solutions 3. 3 solutions? Solution If we draw a horizontal line = k across the graph = f(), it will intersect once when k>ork<, twice when k =ork = and three times when <k<. Therefore the equation f() = k will have

22 Mathematics Learning Centre, Universit of Sdne 9. solution if k>ork<. solutions if k =ork = 3. 3 solutions if <k<..8 Eercises. Sketch the following: a. = b. = 3 c. = d. =( +). Sketch the following: a. = b. = c. = d. = Sketch the following: a. = 3 b. = 3 c. =3 ( ) 3. Sketch the following: a. = b. = c. = 5. Sketch the following: a. + =6 b. +( +) =6 c. ( ) +( 3) =6 6. Sketch the following: a. = 9 b. = 9 ( ) c. = Show that = +. Hence sketch the graph of =. 8. Sketch = Graph the following relations in the given interval: a. = + +for [Hint: Sketch b adding ordinates] b. = + for 3 c. = + for d. =for 3.. Sketch the function f() =.

23 Mathematics Learning Centre, Universit of Sdne. Given = f() assketched below, sketch a. =f() b. = f() c. = f( ) d. = f()+ e. = f( 3) f. = f( +) g. =3 f( 3) h. = f() 3.5. B sketching graphs solve the following equations: a. = b. = c. 3 = d. = 3. Solve =3. a. algebraicall b. geometricall.. The parabolas = ( ) and = ( 3) intersect at a point P. Find the coordinates of P. 5. Sketch the circle + +5=. [Hint: Complete the squares.] Find the values of k, sothat the line = k intersects the circle in two distinct points. 6. Solve =,using a graph Find all real numbers for which = Given that Q(p) =p p, find possible values of n if Q(n) =. 9. Solve =. a. algebraicall b. geometricall.

24 Mathematics Learning Centre, Universit of Sdne.9 Even and odd functions Definition: A function, = f(), is even if f() =f( ) for all in the domain of f. Geometricall, an even function is smmetrical about the -ais (it has line smmetr). The function f() = is an even function as f( ) =( ) = = f() for all values of. Weillustrate this on the following graph. The graph of =. Definition: A function, = f(), is odd if f( ) = f() for all in the domain of f. Geometricall, an odd function is smmetrical about the origin (it has rotational smmetr). The function f() = is an odd function as f( ) = = f() for all values of. This is illustrated on the following graph. The graph of =.

25 Mathematics Learning Centre, Universit of Sdne Eample Decide whether the following functions are even, odd or neither.. f() =3. g() = 3. f() = 3. Solution. f( ) =3( ) =3 =f() The function f() =3 iseven.. g( ) = ( ) = = = g() Therefore, the function g is odd. 3. f( ) =( ) 3 ( ) = 3 This function is neither even (since 3 3 ) nor odd (since 3 ( 3 )). Eample Sketched below is part of the graph of = f(). Complete the graph if = f() is. odd. even.

26 Mathematics Learning Centre, Universit of Sdne 3 Solution = f() isanodd function. = f() isaneven function.. Increasing and decreasing functions Here we will introduce the concepts of increasing and decreasing functions. In Chapter 5 we will relate these concepts to the derivative of a function. Definition: A function is increasing on an interval I, iffor all a and b in I such that a<b, f(a) <f(b). The function = is an eample of a function that is increasing over its domain. The function = is increasing for all real >.

27 Mathematics Learning Centre, Universit of Sdne f(b) The graph of =. This function is increasing for all real. f(a) The graph of =. This function is increasing on the interval >. a b Notice that when a function is increasing it has a positive slope. Definition: A graph is decreasing on an interval I, iffor all a and b in I such that a<b, f(a) >f(b). The function = is decreasing over its domain. The function = is decreasing on the interval <. f(a) f(b) a b The graph of =. This function is decreasing for all real. The graph of =. This function is decreasing on the interval <. Notice that if a function is decreasing then it has negative slope.. Eercises. Given the graph below of = f(): a. State the domain and range. b. Where is the graph

28 Mathematics Learning Centre, Universit of Sdne 5 i ii increasing? decreasing? c. if k is a constant, find the values of k such that f() =k has i no solutions ii solution iii solutions iv 3 solutions v solutions. d. Is = f() even, odd or neither?. Complete the following functions if the are defined to be (a) even (b) odd. = f() = g() 3. Determine whether the following functions are odd, even or neither. a. = + b. = c. = d. = 3 +3 e. = f. = g. = + h. = 3 +3 i. = + j. = + +. Given = f() isevenand = g() isodd, prove a. if h() =f() g() then h() isodd b. if h() =(g()) then h() iseven

29 Mathematics Learning Centre, Universit of Sdne 6 c. if h() = f(), g(),then h() isodd g() d. if h() =f() (g()) then h() iseven. 5. Consider the set of all odd functions which are defined at =. Can ou prove that for ever odd function in this set f() =? If not, give a counter-eample.

30 Mathematics Learning Centre, Universit of Sdne 7 3 Piecewise functions and solving inequalities In this Chapter we will discuss functions that are defined piecewise (sometimes called piecemeal functions) and look at solving inequalities using both algebraic and graphical techniques. 3. Piecewise functions 3.. Restricting the domain In Chapter we saw how functions could be defined on a subinterval of their natural domain. This is frequentl called restricting the domain of the function. In this Chapter we will etend this idea to define functions piecewise. Sketch the graph of = for. The graph of = for. Sketch the graph of = for <. The graph of = for <.

31 Mathematics Learning Centre, Universit of Sdne 8 We can now put these pieces together to define a function of the form for f() = for < We sa that this function is defined piecewise. First note that it is a function; each value of in the domain is assigned eactl one value of. This is eas to see if we graph the function and use the vertical line test. We graph this function b graphing each piece of it in turn. The graph shows that f defined in this wa is a function. The two pieces of = f() meet so f is a continuous function. The absolute value function for f() = for < is another eample of a piecewise function. Eample Sketch the function + for f() = for <

32 Mathematics Learning Centre, Universit of Sdne 9 Solution This function is not continuous at =as the two branches of the graph do not meet. Notice that we have put an open square (or circle) around the point (, ) and a solid square (or circle) around the point (, ). This is to make it absolutel clear that f() = and not. When defining a function piecewise, we must be etremel careful to assign to each eactl one value of. 3. Eercises. For the function for f() = for < evaluate a. f( ) + f() b. f(a ). For the function given in, solve f() =. 3. Below is the graph of = g(). Write down the rules which define g() given that its pieces are hperbolic, circular and linear.

33 Mathematics Learning Centre, Universit of Sdne a. Sketch the graph of = f() if for f() = for > b. State the range of f. c. Solve i f() = ii f() = 3. d. Find k if f() =k has i ii iii solutions. 5. Sketch the graph of = f() if for f() = + for < 6. Sketch the graph of = g() if + for < g() = for < for

34 Mathematics Learning Centre, Universit of Sdne 3 7. McMaths burgers are to modernise their logo as shown below Write down a piecewise function that represents this function using (a) (b) 3 (c) pieces (i.e. rules that define the function). 8. a. The following piecewise function is of the form a + b for < f() = c + d for > (,) Determine the values of a, b, c and d. b. Complete the graph so that f() isanodd function defined for all real,. c. Write down the equations that now define f(),.

35 Mathematics Learning Centre, Universit of Sdne Inequalities We can solve inequalities using both algebraic and graphical methods. Sometimes it is easier to use an algebraic method and sometimes a graphical one. For the following eamples we will use both, as this allows us to make the connections between the algebra and the graphs. Algebraic method Graphical method. Solve 3. This is a ( Unit) linear inequalit. Remember to reverse the inequalit sign when multipling or dividing b a negative number. 3 3 (,) When is the line =3 above or on the horizontal line =? From the graph, we see that this is true for.. Solve +3<. This is a ( Unit) quadratic inequalit. Factorise and use a number line. +3 < ( 3)( ) < The critical values are and 3, which divide the number line into three intervals. We take points in each interval to determine the sign of the inequalit; eg use =, =and =as test values. positive negative positive 3 Let = When does the parabola have negative -values? OR When is the parabola under the -ais? From the graph, we see that this happens when <<3. Thus, the solution is <<3.

36 Mathematics Learning Centre, Universit of Sdne Solve. This is a 3 Unit inequalit. There is a variable in the denominator. Remember that a denominator can never be zero, so in this case. First multipl b the square of the denominator ( ), ( )( 5) Mark the critical values on the number line and test =, =.5 and =6. positive neg positive Therefore, <or 5. Let =. 6 (5,) = is not defined for =. It is a hperbola with vertical asmptote at =. Tosolve our inequalit we need to find the values of for which the hperbola lies on or under the line =. (5, ) is the point of intersection. So, from the graph we see that when <or 5.. Solve 3 <. Consider 3=, Multipl b we get. 3 = 3 = ( 5)( +) = Therefore, the critical values are, and 5 which divide the number line into four intervals. We can use = 3, =, =and =6as test values in the inequalit. The points = 3 and = satisf the inequalit, so the solution is < or<<5. (Notice that we had to include as one of our critical values.) Sketch = 3 and then =. Note that second of these functions is not defined for = (, 5) 6 (, 3) (5,) For what values of does the line lie under the hperbola? From the graph, we see that this happens when < or <<5.

37 Mathematics Learning Centre, Universit of Sdne 3 Eample Sketch the graph of = 6. Hence, where possible, a. Solve i 6 = ii 6 > iii 6 = +3 iv 6 <+3 v 6 = 3 b. Determine the values of k for which 6 = + k has eactl two solutions. Solution 6 for 3 f() = 6 = ( 6) for <3 = 6. = (9,) 5. (,) -. = (.5,3) = a. i Mark in the graph of =. Itisparallel to one arm of the absolute value graph. It has one point of intersection with = 6 = +6(<3) at =.5. ii When is the absolute value graph above the line =? From the graph, when <.5.

38 Mathematics Learning Centre, Universit of Sdne 35 iii = +3intersects = 6 twice. To solve 6 = +3,take 6 = 6= +3when 3. This gives us the solution =9. Then take 6 = +6= +3when <3 which gives us the solution =. iv When is the absolute value graph below the line = +3? From the graph, <<9. v = 3intersects the absolute value graph at =3onl. b. k represents the -intercept of the line = + k. When k = 3, there is one point of intersection. (See (a) (v) above). For k> 3, lines of the form = + k will have two points of intersection. Hence 6 = + k will have two solutions for k> Eercises. Solve a. p b. p+3 c. 7 9 >. a. Sketch the graph of =( 3). b. Hence solve ( 3). 3. a. Find the points of intersection of the graphs =5 and =. b. On the same set of aes, sketch the graphs of =5 and =. c. Using part (ii), or otherwise, write down all the values of for which 5 >. a. Sketch the graph of =. b. Solve <. c. Suppose <a<band consider the points A(a, a ) and B(b, b )onthe graph of =. Find the coordinates of the midpoint M of the segment AB. Eplain wh a + b > a+b 5. a. Sketch the graphs of = and = 5 on the same diagram. b. Solve 5 >. c. For what values of m does m = 5 have eactl i two solutions ii no solutions 6. Solve

39 Mathematics Learning Centre, Universit of Sdne 36 Polnomials Man of the functions we have been using so far have been polnomials. In this Chapter we will stud them in more detail. Definition A real polnomial, P (), of degree n is an epression of the form P () =p n n + p n n + p n n + + p + p + p where p n,p, p,, p n are real and n is an integer. All polnomials are defined for all real and are continuous functions. We are familiar with the quadratic polnomial, Q() =a + b + c where a. This polnomial has degree. The function f() = + is not a polnomial as it has a power which is not an integer and so does not satisf the definition.. Graphs of polnomials and their zeros.. Behaviour of polnomials when is large One piece of information that can be a great help when sketching a polnomial is the wa it behaves for values of when is large. That is, values of which are large in magnitude. The term of the polnomial with the highest power of is called the leading or dominant term. For eample, in the polnomial P () = 6 3, the term 6 is the dominant term. When is large, the dominant term determines how the graph behaves as it is so much larger in magnitude than all the other terms. How the graph behaves for large depends on the power and coefficient of the dominant term. There are four possibilities which we summarise in the following diagrams:. Dominant term with even power and positive coefficient, eg =.. Dominant term with even power and negative coefficient, eg Q() =.

40 Mathematics Learning Centre, Universit of Sdne Dominant term with odd power and positive coefficient, eg = 3.. Dominant term with odd power and negative coefficient, eg Q() = 3. This gives us a good start to graphing polnomials. All we need do now is work out what happens in the middle. In Chapter 5 we will use calculus methods to do this. Here we will use our knowledge of the roots of polnomials to help complete the picture... Polnomial equations and their roots If, for a polnomial P (), P (k) =then we can sa. = k is a root of the equation P () =.. = k is a zero of P (). 3. k is an -intercept of the graph of P ()...3 Zeros of the quadratic polnomial The quadratic polnomial equation Q() =a + b + c =has two roots that ma be:. real (rational or irrational) and distinct,. real (rational or irrational) and equal, 3. comple (not real). We will illustrate all of these cases with eamples, and will show the relationship between the nature and number of zeros of Q() and the -intercepts (if an) on the graph.. Let Q() = +3. We find the zeros of Q() bsolving the equation Q() =. +3 = ( )( 3) = Therefore = or 3. The roots are rational (hence real) and distinct

41 Mathematics Learning Centre, Universit of Sdne 38. Let Q() = 3. Solving the equation Q() = we get, 3 = = ± 6 + Therefore = ± 7. The roots are irrational (hence real) and distinct Let Q() = +. Solving the equation Q() = we get, + = ( ) = Therefore =. The roots are rational (hence real) and equal. Q() = has a repeated or double root at =. 3 Notice that the graph turns at the double root =.. Let Q() = +5. Solving the equation Q() = we get, +5 = = ± 6 Therefore = ±. There are no real roots. In this case the roots are comple. 3 Notice that the graph does not intersect the -ais. That is Q() > for all real. Therefore Q is positive definite.

42 Mathematics Learning Centre, Universit of Sdne 39 We have given above four eamples of quadratic polnomials to illustrate the relationship between the zeros of the polnomials and their graphs. In particular we saw that: i if the quadratic polnomial has two real distinct zeros, then the graph of the polnomial cuts the -ais at two distinct points; ii iii if the quadratic polnomial has a real double (or repeated) zero, then the graph sits on the -ais; if the quadratic polnomial has no real zeros, then the graph does not intersect the -ais at all. So far, we have onl considered quadratic polnomials where the coefficient of the term is positive which gives us a graph which is concave up. If we consider polnomials Q() =a + b + c where a< then we will have a graph which is concave down. For eample, the graph of Q() = ( +)isthe reflection in the -ais of the graph of Q() = +. (See Chapter.) 3 3 The graph of Q() = +. The graph of Q() = ( + )... Zeros of cubic polnomials A real cubic polnomial has an equation of the form P () =a 3 + b + c + d where a,a, b, c and d are real. It has 3 zeros which ma be: i 3real distinct zeros; ii iii 3 real zeros, all of which are equal (3 equal zeros); 3 real zeros, of which are equal; iv real zero and comple zeros. We will illustrate these cases with the following eamples:

43 Mathematics Learning Centre, Universit of Sdne. Let Q() = Solving the equation Q() = we get: = 3( )( +) = Therefore = or or The roots are real (in fact rational) and distinct.. Let Q() = 3. Solving Q() =weget that 3 =. We can write this as ( ) 3 =. So, this equation has three equal real roots at =. 3. Let Q() = 3. Solving the equation Q() = we get, 3 = ( ) = Therefore = or. The roots are real with a double root at =and a single root at =. The graph turns at the double root.. Let Q() = 3 +. Solving the equation Q() = we get, 3 + = ( +) = Therefore =. There is one real root at =. + =does not have an real solutions. The graph intersects the -ais once onl.

44 Mathematics Learning Centre, Universit of Sdne Again, in the above eamples we have looked onl at cubic polnomials where the coefficient of the 3 term is positive. If we consider the polnomial P () = 3 then the graph of this polnomial is the reflection of the graph of P () = 3 in the -ais. The graph of Q() = 3. The graph of Q() = 3.. Polnomials of higher degree We will write down a few rules that we can use when we have a polnomial of degree 3. If P () isareal polnomial of degree n then:. P () =has at most n real roots;. if P () =has a repeated root with an even power then the graph of P () turns at this repeated root; 3. if P () =has a repeated root with an odd power then the graph of P () has a horizontal point of inflection at this repeated root. For eample,. tells us that if we have a quartic polnomial equation f() =. Then we know that f() =has real roots. We can illustrate. b the sketching f() =( ) ( +). Notice how the graph sits on the -ais at =. The graph of f() =( + )( ).

45 Mathematics Learning Centre, Universit of Sdne We illustrate 3. b sketching the graph of f() =( ) 3. Notice the horizontal point of inflection at =. 3 The graph of f() =( ) 3..3 Eercises. Sketch the graphs of the following polnomials if = P () is: a. ( + )( 3) b. ( + )(3 ) c. ( +) ( 3) d. ( + )( +5). The graphs of the following quartic polnomials are sketched below. Match the graph with the polnomial. a. = b. = c. = +d. = e. =( ) f. =( +) i ii iii iv v vi

46 Mathematics Learning Centre, Universit of Sdne 3 3. Sketch the graphs of the following quartic polnomials if = C() is: a. ( )( + )( +3) b. ( )( + )(3 ) c. ( )( 3) d. ( +) ( 3) e. ( +) 3 ( 3) f. ( +) 3 (3 ) g. ( + )( +5) h. ( +5).. B sketching the appropriate polnomial, solve: a. < b. ( + )( 3)(5 ) > c. ( +) (5 ) > d. ( +) 3 (5 ). 5. For what values of k will P () for all real if P () = + k? 6. The diagrams show the graph of = P () where P () =a( b)( c) d. In each case determine possible values for a, b, c and d. a. b. c. (, 8) d. e. f. (,8) 5 (3, 9) (, 8) 7. The graph of the polnomial = f() isgiven below. It has a local maimum and minimum as marked. Use the graph to answer the following questions. a. State the roots of f() =. b. What is the value of the repeated root. c. For what values of k does the equation f() =k have eactl 3 solutions.

47 Mathematics Learning Centre, Universit of Sdne d. Solve the inequalit f() <. e. What is the least possible degree of f()? f. State the value of the constant of f(). g. For what values of k is f()+k for all real. (.78,3.3) (.8,9.9) The graph of the polnomial = f(). Factorising polnomials So far for the most part, we have looked at polnomials which were alread factorised. In this section we will look at methods which will help us factorise polnomials with degree >... Dividing polnomials Suppose we have two polnomials P () and A(), with the degree of P () the degree of A(), and P () isdivided b A(). Then P () A() = Q()+R() A(), where Q() isapolnomial called the quotient and R() isapolnomial called the remainder, with the degree of R() < degree of A(). We can rewrite this as P () =A() Q()+R(). For eample: If P () = and A() =, then P () can be divided b A() as follows:

48 Mathematics Learning Centre, Universit of Sdne 5 The quotient is + +and the remainder is. We have = This can be written as 3 + 3=( )( + +)+. Note that the degree of the polnomial is... The Remainder Theorem If the polnomial f() isdivided b ( a) then the remainder is f(a). Proof: Following the above, we can write f() =A() Q()+R(), where A() =( a). Since the degree of A() is,the degree of R() iszero. That is, R() =r where r is a constant. f() = ( a)q()+r where r is a constant. f(a) = Q(a)+r = r So, if f() isdivided b ( a) then the remainder is f(a). Eample Find the remainder when P () = isdivided b a. +,b.. Solution a. Using the Remainder Theorem: Remainder = P ( ) = 3 ( ) 3 = 7 b. Remainder = P ( ) = 3( ) ( )3 + 3( ) = = 6

49 Mathematics Learning Centre, Universit of Sdne 6 Eample When the polnomial f() isdivided b, the remainder is What is the remainder when f() isdivided b ( )? Solution Write f() =( ) q()+(5 + 6). Then Remainder = f() = q()+6 = 6 A consequence of the Remainder Theorem is the Factor Theorem which we state below...3 The Factor Theorem If = a is a zero of f(), that is f(a) =,then ( a) isafactor of f() and f() ma be written as f() =( a)q() for some polnomial q(). Also, if ( a) and ( b) are factors of f() then ( a)( b) isafactor of f() and for some polnomial Q(). f() =( a)( b) Q() Another useful fact about zeros of polnomials is given below for a polnomial of degree 3. If a (real) polnomial P () =a 3 + b + c + d, where a,a, b, c and d are real, has eactl 3 real zeros α, β and γ, then P () =a( α)( β)( γ) () Furthermore, b epanding the right hand side of () and equating coefficients we get: i ii iii α + β + γ = b a ; αβ + αγ + βγ = c a ; αβγ = d a.

50 Mathematics Learning Centre, Universit of Sdne 7 This result can be etended for polnomials of degree n. We will give the partial result for n =. If P () =a + b 3 + c + d + e is a polnomial of degree with real coefficents, and P () has four real zeros α, β, γ and δ, then P () =a( α)( β)( γ)( δ) and epanding and equating as above gives αβγδ = e a. If a =and the equation P () =has a root which is an integer, then that integer must be afactor of the constant term. This gives us a place to start when looking for factors of a polnomial. That is, we look at all the factors of the constant term to see which ones (if an) are roots of the equation P () =. Eample Let f() = a. Factorise f(). b. Sketch the graph of = f(). c. Solve f(). Solution a. Consider the factors of the constant term,. We check to see if ± and ± are solutions of the equation f() =b substitution. Since f() =, we know that ( ) is a factor of f(). We use long division to determine the quotient So, f() = ( )( ) = ( )( )( +)

51 Mathematics Learning Centre, Universit of Sdne 8 b. 3 The graph of f() = c. f() when or. Eample Show that ( ) and ( 3) are factors of P () = , and hence solve =. Solution P () = = and P (3) = = so ( ) and ( 3) are both factors of P () and ( )( 3) = 5 +6is also a factor of P (). Long division of P () b 5 +6gives a quotient of ( + 5). So, P () = =( )( 3)( +5). Solving P () =weget ( )( 3)( +5)=. That is, =or =3or = 5. Instead of using long division we could have used the facts that i the polnomial cannot have more than three real zeros; ii the product of the zeros must be equal to 3. Let α be the unknown root. Then 3 α = 3, so that α = 5. Therefore the solution of P () = = is =or =3or = 5.

52 Mathematics Learning Centre, Universit of Sdne 9.5 Eercises. When the polnomial P () isdivided b ( a)( b) the quotient is Q() and the remainder is R(). a. Eplain wh R() isofthe form m + c where m and c are constants. b. When a polnomial is divided b ( ) and ( 3), the remainders are and 9 respectivel. Find the remainder when the polnomial is divided b c. When P () isdivided b ( a) the remainder is a. Also, P (b) =b. Find R() when P () isdivided b ( a)( b).. a. Divide the polnomial f() = bg() = Hence write f() = g()q() + r() where q() and r() are polnomials. b. Show that f() and g() have no common zeros. (Hint: Assume that α is a common zero and show b contradiction that α does not eist.) 3. For the following polnomials, i factorise ii solve P () = iii sketch the graph of = P (). a. P () = 3 8 b. P () = 3 6 c. P () = d. P () = e. P () =

53 Mathematics Learning Centre, Universit of Sdne 5 5 Solutions to eercises. Solutions. a. The domain of f() = 9 is all real where 3 3. The range is all real such that 3. b. The graph of f() = 9.. ψ( + h) ψ() h = ( + h) +5 ( +5) h = +h + h +5 5 h = h +h h = h + 3. a. b. 3 5 The graph of =. The domain is all real and the range is all real. The graph of =. Its domain is all real and range all real.

54 Mathematics Learning Centre, Universit of Sdne 5 c. 6 8 d. The graph of =. The domain is all real and the range is all real. The graph of =. The domain is all real, and the range is all real.. a. The perpendicular distance d from (, ) to + + k =isd = k. b. For the line ++k =to cut the circle in two distinct points d<. ie k < or <k<. 5. a. b. The graph of =( ). The graph of =. 6. = 3 is not a function.

55 Mathematics Learning Centre, Universit of Sdne 5 7. a. b. The graph of =. This is a function with the domain: all real such that and range: all real such that. The graph of =. This is not the graph of a function. c. d. The graph of = 3. This is a function with the domain: all real and range: all real. The graph of =. This is the graph of a function which is not defined at =. Its domain is all real,and range is = ±. e. The graph of =. This is not the graph of a function.

56 Mathematics Learning Centre, Universit of Sdne A( p ) = ( p ) ++ ( p ) = p ++ p = p ++p = A(p) 9. a. The values of in the interval << are not in the domain of the function. b. =and = are not in the domain of the function.. a. φ(3) + φ() + φ(5) = log(.5) b. φ(3) + φ() + φ(5) + + φ(n) =log( n). a. =3when z =3. b. i L(M()) =( )+ ii M(L()) = +. a. a =,b =so the equations is =. b. a =5,b =so the equation is = b. The graph of + =.. S(n ) = n n Hence S(n) S(n ) = n n + n n = n(n ) (n + )(n ) (n )(n +) = n n (n n ) (n )(n +) = (n )(n +)

57 Mathematics Learning Centre, Universit of Sdne 5.8 Solutions. a. b. The graph of =. The graph of = 3. c. d. The graph of =. The graph of =( +).. a. b. The graph of =. The graph of =.

58 Mathematics Learning Centre, Universit of Sdne 55 c. d. The graph of =. The graph of = a. b. The graph of = 3. The graph of = 3. c. The graph of =3 ( ) 3.. a. b. The graph of =. The graph of =.

59 Mathematics Learning Centre, Universit of Sdne 56 c. The graph of =. 5. a. b. (, ) The graph of + = 6. The graph of +( +) = 6. 6 c. 6 (,3) The graph of ( ) +( 3) = 6.

60 Mathematics Learning Centre, Universit of Sdne a. 3 b. 3 c. 3 3 The graph of = (,) The graph of = 9 ( ). 3 The graph of = ( ) += = The graph of =. 8. The graph of = +.

61 Mathematics Learning Centre, Universit of Sdne a. b. The graph of = + + for. The graph of = + for 3. c. d. The graph of = + for. The graph of =for 3.. The graph of f() =.

62 Mathematics Learning Centre, Universit of Sdne 59. a. b The graph of =f(). The graph of = f(). c. d The graph of = f( ). The graph of = f()+. e. f. 3.5 The graph of = f( 3). The graph of = f( +).

63 Mathematics Learning Centre, Universit of Sdne 6 g. h. 6 3 The graph of = f(). The graph of =3 f( 3).. a. b. c. = and =are solutions of the equation =. d. =is a solution of =. =and =are solutions of the equation 3 =. =is a solution of =. 3. a. For, = =3. Therefore =5is a solution of the inequalit. (Note that =5is indeed.) For <, = ( ) = + =3. Therefore = isasolution. (Note that = is<.)

64 Mathematics Learning Centre, Universit of Sdne 6 b. The points of intersection are (, 3) and (5, 3). Therefore the solutions of =3are = and =5.. The parabolas intersect at (, ). 5. (,7) = k intersects the circle at two distinct points when <k< (,) 5 5 The point of intersection is (, ). Therefore the solution of 5 =is =.

65 Mathematics Learning Centre, Universit of Sdne 6 7. The point of intersection is (, ). Therefore the solution of = + is =. 8. n = orn =. 9. a. For, = =when =, but this does not satisf the condition of soisnot a solution. For <, = + = when =. = is < soisasolution. 3 3 Therefore, = is a solution of =. 3 b. (/3,8/3) 6 The graph of = and = intersect at the point (, 8 ). So the solution 3 3 of = is =. 3. Solutions. a. The domain is all real, and the range is all real. b. i <<or> ii < or<< c. i k< ii There is no value of k for which f() =k has eactl one solution. iii k =ork>

66 Mathematics Learning Centre, Universit of Sdne 63 iv k = v <k< d. = f() iseven. a. b. = f() iseven. = f() isodd. a. b. = g() iseven. = g() isodd. 3. a. even b. even c. neither d. odd e. odd f. even g. even h. neither i. even j. even. a. h( ) = f( ) g( ) = f() g() = f() g() = h() Therefore h is odd. b. h( ) = (g( )) = ( (g()) = (g()) = h() Therefore h is even.

67 Mathematics Learning Centre, Universit of Sdne 6 c. h( ) = f( ) g( ) = f() g() = f() g() = h() Therefore h is odd. d. h( ) = f( ) (g( )) = f() ( g()) = f() (g()) = h() Therefore h is even. 5. If f is defined at = f() = f( ) (since = ) = f() (since f is odd) f() = (adding f() to both sides) Therefore f() =. 3. Solutions. a. f( ) + f() = ( ( ))+( () )=+( 3) =. b. f(a )= (a ) = a since a.. You can see from the graph below that there is one solution to f() =, and that this solution is at =.

68 Mathematics Learning Centre, Universit of Sdne g() = for < + for for >. a. The domain of f is all real. b. The range of f is all real >. c. i f() =when = or =. ii f() = 3 when =. d. i f() =k has no solutions when k. ii f() =k has solution when <k< ork>. iii f() =k has solutions when k. 5. Note that f() =. 6. The domain of g is all real,.

69 Mathematics Learning Centre, Universit of Sdne 66 The range of g is all real <or. 7. Note that there ma be more than one correct solution. a. Defining f as +6 for 3 for 3 << f() = for 3 +6 for >3 gives a function describing the McMaths burgers logo using pieces. b. Defining f as +6 for 3 f() = for 3 <<3 +6 for 3 gives a function describing the McMaths burgers logo using 3 pieces. c. Defining f as 3 +3 for f() = 3 3 for > gives a function describing the McMaths burgers logo using pieces. 8. a. Here a =,b =, c =and d =. So, for < f() = for > b. Defining f to be an odd function for all real,,weget

70 Mathematics Learning Centre, Universit of Sdne 67 c. We can define f as follows + for < for < f() = for < for > 3. Solutions. a. b. 3 <p c. < or 3 <<3or>. a. The graph of =( 3) is given below 5 5 b. From the graph we see that ( 3) when 3.

71 Mathematics Learning Centre, Universit of Sdne a. The graphs =5 and = intersect at the points (, ) and (, ). b. The graphs of =5 and = 5 5 c. The inequalit is satisfied for < or <<.. a. The graph of =. 5 b. < when <. c. The midpoint M of the segment AB has coordinates ( a+b, a + b ). Since the function = is concave up, the -coordinate of M is greater than f( a+b). So, a + b > a+b B 5 M A

72 Mathematics Learning Centre, Universit of Sdne a. 5 (.5,.5) 5 5 b. 5 >for all <.5. c. i m = 5 has eactl two solutions when <m<. ii m = 5 has no solutions when <m< Solutions. a. b The graph of P () =( + )( 3). The graph of P () =( + )(3 ). c. d The graph of P () =( +) ( 3). The graph of P () =( + )( + 5).. a. iv. b. v. c. i. d. iii. e. ii. f. vi.

73 Mathematics Learning Centre, Universit of Sdne 7 3. a. b. 5 5 The graph of P () =( )( + )( +3). The graph of P () =( )( + )(3 ). c. d. 5 5 The graph of P () = ( )( 3). The graph of P () =( +) ( 3). e. f. The graph of P () =( +) 3 ( 3). The graph of P () =( +) 3 (3 ). g. h. The graph of P () =( + )( + 5). The graph of P () = ( + 5).

74 Mathematics Learning Centre, Universit of Sdne 7. a < when <<6. b. 6 c. ( + )( 3)(5 ) > when < or3<<5. 6 ( +) (5 ) > when < k for all real when k = a. P () =( ) b. P () = ( ) c. P () = ( ) d. P () = 3 ( ) 3 e. P () = ( ) f. P () = (+)( ) 8 7. a. The roots of f() =are =, =and =.

75 Mathematics Learning Centre, Universit of Sdne 7 b. =is the repeated root. c. The equation f() =k has eactl 3 solutions when k =ork =3.3. d. f() < when <<. e. The least possible degree of the polnomial f() is. f. Since f() =, the constant in the polnomial is. g. f()+k for all real when k Solutions. a. Since A() =( a)( b) isapolnomial of degree, the remainder R() must be apolnomial of degree <. So, R() isapolnomial of degree. That is, R() =m + c where m and c are constants. Note that if m =the remainder is a constant. b. Let P () =( 5 +6)Q()+(m + c) =( )( 3)Q()+(m + c). Then P () = ()( )Q()+(m + c) = m + c = and P (3) = ()()Q(3) + (3m + c) = 3m + c = 9 Solving simultaneousl we get that m =5and c = 6. So, the remainder is R() =5 6. c. Let P () =( a)( b)q()+(m + c). Then P (a) = ()(a b)q(a)+(ma + c) = am + c = a and P (b) = (b a)()q(b)+(mb + c) = bm + c = b Solving simultaneousl we get that m = a + b and c = ab provided a b. So, R() =(a + b) ab.. a =( +5 + )( +3 ) b. Let α be a common zero of f() and g(). That is, f(α) =andg(α) =.

76 Mathematics Learning Centre, Universit of Sdne 73 Then since f() = g()q()+ r() wehave f(α) = g(α)q(α)+r(α) = ()q(α)+ r(α) since g(α) = = r(α) = since f(α) = But, from part b. r() = for all values of, sowehave acontradiction. Therefore, f() and g() donot have a common zero. This is an eample of a proof b contradiction. 3. a. i P () = 3 8=( + )( + )( ) ii =, = and =are solutions of P () =. iii 6 The graph of P () = 3 8. b. i P () = 3 6 = ( +) ( 5). ii = and =5are solutions of P () =. = isadouble root. iii 6 The graph of P () = 3 6. c. i P () = 3 + 8=(+)( + ) = (+)( ( + 5))( ( 5)) ii =, = + 5 and = 5 are solutions of P () =.

77 Mathematics Learning Centre, Universit of Sdne 7 iii The graph of P () = The zeros are =, = + 5 and = 5. d. i P () = 3 + 6=( )( + + 3). + +3=has no real solutions. ii =is the onl real solution of P () =. iii The graph of P () = There is onl one real zero at =. e. i P () = =( + )( 3)( ). ii =, = and =3are solutions of P () =. iii The graph of P () =