5Higher-degree polnomials 5. Kick off with CAS 5.2 Quartic polnomials 5.3 Families of polnomials 5.4 Numerical approimations to roots of polnomial equations 5.5 Review
5. Kick off with CAS Quartic transformations Using CAS technolog, sketch the following quartic functions. a = 4 = 4 c = 3 4 d = 2 4 e = 2 5 4 2 Using CAS technolog, enter = a 4 into the function entr line and use a slider to change the values of a. 3 Complete the following sentences. a When sketching a quartic function, a negative sign in front of the 4 term the graph of = 4. When sketching a quartic function, = a 4, for values of a < and a >, the graph of = 4 ecomes. c When sketching a quartic function, = a 4, for values of < a <, a, the graph of = 4 ecomes. 4 Using CAS technolog, sketch the following functions. a = 4 = ( + ) 4 c = ( 2) 4 d = 4 e = 4 + 2 f = 3 4 5 Using CAS technolog, enter = ( h) 4 into the function entr line and use a slider to change the value of h. 6 Using CAS technolog, enter = 4 + k into the function entr line and use a slider to change the value of k. 7 Complete the following sentences. a When sketching a quadratic function, = ( h) 4, the graph of = 4 is. When sketching a quadratic function, = 4 + k, the graph of = 4 is. 8 Use CAS technolog and our answers to questions 7 to determine the equation that could model the shape of the Gatewa Arch in St Louis. If the technolog permits, upload a photo of the ridge to make this easier. Please refer to the Resources ta in the Prelims section of our ebookplus for a comprehensive step--step guide on how to use our CAS technolog.
Units & 2 AOS Topic 4 Concept Quartic polnomials Concept summar Pratice questions Interactivit Graph plotter: Polnomials of higher degree int-2569 5.2 WorKed example think Quartic polnomials A quartic polnomial is a polnomial of degree 4 and is of the form P() = a 4 + 3 + c 2 + d + e, where a and a,, c, d, e R. graphs of quartic polnomials of the form = a( h) 4 + c The simplest quartic polnomial = 4 graph has the equation = 4. As = 2 oth negative and positive numers raised to an even power, in this case 4, will e positive, the longterm ehaviour of the graph of = 4 (, ) (, ) must e that as (, ) or as, then. The graph of = 4 is similar to that of the paraola = 2. Both graphs are concave up with a minimum turning point at (, ) and oth contain the points (, ) and (, ). However, for the intervals where < and >, the graph of = 4 lies aove the paraola. This is ecause 4 > 2 for these intervals. Likewise, the graph of = 4 lies elow that of the paraola for the intervals < < and < <, since 4 < 2 for these intervals. Despite these differences, the two graphs are of sufficient similarit to enale us to otain the ke features of graphs of quartic polnomials of the form = a( h) 4 + k in much the same manner as for quadratics of the form = a( h) 2 + k. Under a dilation of a units, a horizontal translation of h units and a vertical translation of k units, the graph of = 4 is transformed to that of = a( h) 2 + k. The graph of = a( h) 4 + k has the following features. A turning point with coordinates (h, k). If a >, the turning point is a minimum and if a < it is a maimum. Ais of smmetr with equation = h. Zero, one or two -intercepts. These are otained as the solution to the equation a( h) 4 + k =. Sketch the graph of: a = 4 ( + 3)4 4 = (3 ) 4 7 WritE a State the coordinates and tpe of turning point. a = 4 ( + 3)4 4 Turning point is ( 3, 4). As a =, a >, so the turning point is a 4 minimum. 224 maths QuesT mathematical methods VCe units and 2
2 Calculate the -intercept. -intercept: let = 3 Determine whether there will e an -intercepts. 4 Calculate the -intercepts. Note: ± is needed in taking the fourth root of each side. 5 Sketch the graph. Epress the equation in the form = a( h) 4 + k. = 4 (3)4 4 = 8 6 4 4 = 65 4 -intercept:, 65 4 As the -coordinate of the minimum turning point is negative, the concave up graph must pass through the -ais. -intercepts: let = ( + 4 3)4 4 = ( + 4 3)4 = 4 ( + 3) 4 = 6 Take the fourth root of oth sides. ( + 3) = ± 4 6 + 3 = ±2 = 5 or = -intercepts: ( 5, ) and (, ) ( 5, ) = (3 ) 4 7 = 3 3 ( 3, 4) 4 7 (, ), 65 ( 4 ) = 8 4 7 3 2 State the coordinates of the turning point The graph has a maimum turning point at and its tpe., 7. 3 3 Calculate the -intercept. -intercept: let = in the original form = (3 ) 4 7 = ( ) 4 7 = () 7 = 8 -intercept: (, 8) 4 Determine whether there will e an As the -coordinate of the maimum turning -intercepts. point is negative, the concave down graph will not pass through the -ais. Topic 5 Higher-degree polnomials 225
5 Sketch the graph. The graph is smmetric aout its ais of smmetr, = 3. Quartic polnomials which can e epressed as the product of linear factors Not all quartic polnomials have linear factors. However, the graphs of those which can e epressed as the product of linear factors can e readil sketched analsing these factors. A quartic polnomial ma have up to 4 linear factors since it is of fourth degree. The possile cominations of these linear factors are: four distinct linear factors = ( a)( )( c)( d) one repeated linear factor = ( a) 2 ( )( c) two repeated linear factors = ( a) 2 ( ) 2 one factor of multiplicit 3 = ( a) 3 ( ) one factor of multiplicit 4 = ( a) 4. This case in which the graph has a minimum turning point at (a, ) has alread een considered. Given the long-term ehaviour of a quartic polnomial where as ± for a positive coefficient of the term in 4, the sign diagrams and accompaning shape of the graphs must e of the form shown in the diagrams. + = ( a) ( ) ( c) ( d) a a c c d d + (, 8) = ( a) 2 ( ) ( c) a a, 7 ( 3 ) c c = ( a) 2 ( ) 2 = ( a) 3 ( ) + a + a a a 226 Maths Quest MATHEMATICAL METHODS VCE Units and 2
For a negative coefficient of 4, as ±, so the sign diagrams and graphs are inverted. The single factor identifies an -intercept where the graph cuts the ais; a repeated factor identifies an -intercept which is a turning point; and the factor of multiplicit 3 identifies an -intercept which is a stationar point of inflection. WorKed example 2 Sketch the graph of = ( + 2)(2 ) 3. think WritE Calculate the -intercepts. = ( + 2)(2 ) 3 -intercepts: let = ( + 2)(2 ) 3 = + 2 = or (2 ) 3 = = 2 or = 2 -intercepts: ( 2, ) and (2, ) 2 Interpret the nature of the graph at each Due to the multiplicit of each factor, at = 2 the -intercept. graph cuts the -ais and at = 2 it saddle-cuts the -ais. The point (2, ) is a stationar point of inflection. 3 Calculate the -intercept. -intercept: let = = (2)(2) 3 4 Determine the sign of the coefficient of the leading term and identif the long-term ehaviour of the graph. = 6 -intercept: (, 6) 5 Sketch the graph. Leading term is ()( ) 3 = 4. The coefficient of the leading term is negative, so as ± then. This means the sketch of the graph must start and finish elow the -ais. (, 6) ( 2, ) (2, ) equations and inequations Factorisation techniques ma enale a quartic polnomial P() given in its general form to e rewritten as the product of linear factors. Its graph = P() can then e readil sketched from this form and the equation P() = solved using the Null Factor Law. With the aid of a sign diagram, or a graph, an inequation such as P() can e solved. The factor theorem ma e one method emploed if no simpler method can e found. Once one zero and its corresponding factor are otained using the factor theorem, division of the quartic polnomial this factor would produce a cuic quotient Topic 5 HigHer-degree polnomials 227
which in turn could e factorised with further use of the factor theorem. Alternativel, if two zeros can e found from trial and error, then division of the quartic the product of their corresponding factors would produce a quadratic quotient. WorKed example 3 Factorise P() = 3 4 5 3 5 2 + 5 + 2 and hence solve the inequation 3 4 5 3 5 2 + 5 + 2. think Use the factor theorem to otain one linear factor of the polnomial. 2 Use the factor theorem to otain a second linear factor. 3 State a quadratic factor of the polnomial. 4 Divide the known quadratic factor into the polnomial. WritE/drAW P() = 3 4 5 3 5 2 + 5 + 2 P() = 3 5 5 + 5 + 2 = ( ) is a factor. P( ) = 3 + 5 5 5 + 2 = ( + ) is a factor. Hence, ( )( + ) = 2 is a quadratic factor of P(). 3 4 5 3 5 2 + 5 + 2 = ( 2 ) a 2 + + c = 2 3 2 + 2 Equating coefficients of 3 : 5 = 3 4 5 3 5 2 + 5 + 2 = 2 3 2 5 2 5 Completel factorise the polnomial. P() = ( 2 )(3 2 5 2) = ( 2 )(3 + )( 2) = ( )( + )(3 + )( 2) 6 State the zeros of the polnomial. The zeros of the polnomial are: =, =, = 3, = 2 7 Draw the sign diagram. The leading term has a positive coefficient so the sign diagram is: + 2 3 8 Sketch the graph. (, 3 ) (, 2) (, ) (, ) (2, ) 9 State the solution to the inequation. P() 3 or 2 228 maths QuesT mathematical methods VCe units and 2
Eercise 5.2 PRactise Work without CAS Consolidate Appl the most appropriate mathematical processes and tools Quartic polnomials WE Sketch the following graphs. a = ( 2) 4 = (2 + ) 4 2 A graph with the equation = a( ) 4 + c has a maimum turning point at ( 2, 4) and cuts the -ais at =. a Determine its equation. Sketch the graph and so determine : a( ) 4 + c >. 3 WE2 Sketch the graph of = ( + 2) 2 (2 ) 2. 4 Give a suitale equation for the graph of the quartic polnomial shown. ( 3, 3) ( 4, ) (, ) (2, ) (5, ) 5 WE3 Factorise P() = 4 + 5 3 6 2 32 + 32 and hence solve the inequation 4 + 5 3 6 2 32 + 32 >. 6 Solve the equation ( + 2) 4 3( + 2) 2 48 =. 7 a On the same set of aes, sketch the graphs of = 4, = 2 4 and = 2 4. Lael the points for which =, and. On the same set of aes, sketch the graphs of = 4, = 4, = 2 4 and = ( 2) 4. Lael the points for which =, and. c On the same set of aes, sketch the graphs of = 4, = ( + ) 4 and = ( ) 4. Lael the points for which =, and. d On the same set of aes, sketch the graphs of = 4, = 4 + 2 and = 4. Lael the points for which =, and. 8 Sketch the following graphs, identifing the coordinates of the turning point and an point of intersection with the coordinate aes. a = ( ) 4 6 = 9 ( + 3)4 + 2 c = 25.4( + 5) 4 d = 6( 2) 4 + e = 8 (5 3)4 2 f = 2 7 3 9 Determine a posssile equation for each of the following. a A quartic graph with the same shape as = 2 3 4 ut whose turning point has the coordinates ( 9, ). The curve with the equation = a( + ) 4 + c (, ) which has a minimum turning point at ( 3, 8) and passes through the point ( 4, 2). c A curve has the equation = (a + ) 4 (, ) ( 9, ) where a > and <. The points (, 6) and (2, 256) lie on the graph. d The graph shown has the equation = a( h) 4 + k. 4 Topic 5 Higher-degree polnomials 229
The curve with equation = a 4 + k passes through the points (, ) and a Determine the values of a and k. State the coordinates of the turning point and its nature. c Give the equation of the ais of smmetr. d Sketch the curve. The graph of = a( + ) 4 + c passes through the points ( 2, 3) and (4, 3). 2, 3 8. a State the equation of its ais of smmetr. Given the greatest -value the graph reaches is, state the coordinates of the turning point of the graph. c Determine the equation of the graph. d Calculate the coordinates of the point of intersection with the -ais. e Calculate the eact value(s) of an intercepts the graph makes with the -ais. f Sketch the graph. 2 Sketch the following quartic polnomials without attempting to locate an turning points that do not lie on the coordinate aes. a = ( + 8)( + 3)( 4)( ) = ( + 3)( 2)(2 5)(3 ) c = 2( + 7)( ) 2 (2 5) d = 2 3 2 (4 5) 2 e = 3( + ) 3 (4 ) f = (3 + )(3 ) 3 3 For each of the following quartic graphs, form a possile equation. a c (, 5) ( 6, ) (4, ) ( 5, ) ( 3, ) ( 3, 54) d ( 6, ) (, ) (3, 75) ( 2, ) (, ) (4, ) (, 54) (.5, ) (.8, ) 4 Solve the following equations and inequations. a ( + 3)(2 )(4 )(2 ) < 9 4 49 2 = c 3 4 + 2 3 + 28 2 d 3 4 + 2 3 + 2 2 7 = e 4 + 3 8 = 8 f 2(2 ) 4 8( 2) 3 23 Maths Quest MATHEMATICAL METHODS VCE Units and 2
Master 5.3 Units & 2 AOS Topic 4 Concept 2 Families of polnomials Concept summar Pratice questions 5 a Factorise 4 + 8 2 8. Sketch the graph of = 4 + 8 2 8. c Use the graph to otain : 4 + 8 2 8 >. d State the solution set for : 4 8 2 + 8 >. 6 The graphs of = 4 and = 2 3 intersect at the origin and at a point P. a Calculate the coordinates of the point P. The paraola = a 2 and the straight line = m pass through the origin and the point P. Determine the values of a and m. c Using the values otained for a and m in part, sketch the graphs of = 4, = 2 3, = a 2 and = m on the same set of aes. d i At what points would the graphs of = n 3 and = 4 intersect? ii If each of the four curves = 4, = n 3, = a 2 and = m intersect at the same two points, epress a and m in terms of n. 7 Sketch the graph of = 4 3 2 2 4 + 4, locating turning points and intersections with the coordinate aes. Epress coordinates to 2 decimal places where appropriate. 8 a Sketch the graph of = 4 7 8, locating an turning points and intersections with the coordinate aes. Hence epress 4 7 8 as the product of a linear and a cuic polnomial with rational coefficients. Families of polnomials Of the polnomials, the simplest is the linear polnomial. In some was it is the eception, ecause its graph is a straight line, whereas the graphs of all other polnomials are curves. Nevertheless, the graphs of linear and all other polnomials of odd degree do displa some similarities. Likewise, the graphs of polnomials of even degree also displa similarities with each other. In considering families of polnomials it is therefore helpful to separate them into two categories: those with an odd degree and those with an even degree. Graphs of = n, where n N and n is odd Consider the shapes of the graphs of some odd-degree polnomials with simple equations. Once the asic shape is estalished, we can deduce the effect of transformations on these graphs. Comparison of the graphs of =, = 3 and = 5 shows a numer of similarities. = 5 = 3 = Each graph ehiits the same long-term (, ) ehaviour that as ±, ±. Each graph passes through the points (, ), (, ), (, ). With the eception of =, the other two graphs have a stationar point of inflection at (, ) and essentiall similar shapes. (, ) (, ) The larger the power, the narrower the graph for > and for <. Thus, if n is an odd positive integer, n, the graph of = n will have a stationar point of inflection at (, ) and essentiall resemle the shape of = 3. Topic 5 Higher-degree polnomials 23
WorKed example 4 Under a sequence of transformations where the graph of = n, n N \ is dilated factor a from the -ais and translated horizontall h units right and verticall k units up, the equation of the transformed graph takes the form = a( h) n + k. The ke features of the graphs of the famil of odd-degree polnomials, with the equation = a( h) n + k, n N \ {}, and n is odd, are as follows: There is a stationar point of inflection at (h, k). If a >, then as ±, ±. If a <, then as ±,. There is one -intercept which is calculated solving a( h) n + k =. There is one -intercept which is calculated sustituting =. Sketch the graph of = ( + 6 2)5 7. think WritE State the coordinates of the stationar = point of inflection. 6 ( + 2)5 7 Stationar point of inflection at ( 2, 7) 2 Calculate the -intercept. -intercept: let = = 6 (2)5 7 = 32 6 7 = 5 -intercept: (, 5) 3 Calculate the -intercept. -intercept: let = 6 ( + 2)5 7 = 6 ( + 2)5 = 7 ( + 2) 5 = 2 Take the fifth root of oth sides: + 2 = 5 2 = 5 2 2 -intercept: approimatel (.6, ) 4 Sketch the graph. = ( + 2) 5 7 6 ( 5 2 2, ) ( 2, 7) (, 5) 232 maths QuesT mathematical methods VCe units and 2
WorKed example 5 graphs of = n, where n N and n is even Comparison of the graphs of = 2, = 4 and = 6 shows a numer of similarities. Each graph ehiits the same long-term ehaviour that as ±,. Each graph passes through = 6 = 4 = 2 the points (, ), (, ), (, ). All graphs have a minimum (, ) (, ) turning point at (, ) and essentiall similar shapes. The larger the power, the narrower the graph for > and for <. (, ) Thus, if n is an even positive integer, the graph of = n will have a minimum turning point at (, ) and essentiall resemle the shape of = 2. Under a sequence of transformations, the equation takes the form = a( h) n + k. If n is an even positive integer, the ke features of the graphs of the famil of even-degree polnomials with the equation = a( h) n + k are as follows: There is a turning point, or verte, at (h, k). For a >, the turning point is a minimum; for a < it is a maimum. For a >, as ±,. For a <, as ±,. The ais of smmetr has the equation = h. There ma e, or 2 -intercepts. There is one -intercept. Sketch the graph of = ( 2) 6. think WritE State the tpe of turning point and its = ( 2) 6 coordinates. As a >, minimum turning point at (2, ) 2 Calculate the -intercept. -intercept: let = = ( 2) 6 = 63 -intercept: (, 63) 3 Calculate the -intercepts if the eist. -intercepts: let = Note: ± is needed in taking an even root of ( 2) 6 = each side. ( 2) 6 = 2 = ± 6 2 = ± =, = 3 -intercepts: (, ) and (3, ) Topic 5 HigHer-degree polnomials 233
4 Sketch the graph. Families of polnomials which can e epressed as the product of linear factors The factors of a polnomial determine the -intercepts of the graph. From the multiplicit of each linear factor, the ehaviour of a graph at its -intercepts can e determined. For eample, if ou look at the equation = ( + 3)( 2) 2 ( 5), the graph of this quartic polnomial is predicted to: cut the -ais at = 3 touch the -ais at = 2 cut the -ais at = 5. Its graph confirms this prediction. ( 3, ) (2, ) (, 6) (5, ) This interpretation can e etended to an polnomial epressed in factorised form. Effect of multiplicit of zeros and linear factors The graph of = ( a)( ) 2 ( c) 3 ( d) would: cut the -ais at = a (single zero from factor of multiplicit ) touch the -ais at = (doule zero from factor of multiplicit 2) saddle-cut the -ais at = c (triple zero from factor of multiplicit 3) cut the -ais at = d (single zero from factor of multiplicit ). At a touch -intercept, there is a turning point; at a saddle-cut -intercept there is a stationar point of inflection. This polnomial has degree 7 since 2 3 = 7. As the coefficient of the leading term is positive, the graph follows the long-term ehaviour of an odd-degree polnomial. It must initiall start elow the -ais and it must end aove the -ais, since as ±, ±. A possile graph of = ( a)( ) 2 ( c) 3 ( d) is shown in the diagram. (, ) (3, ) (2, ) a c d = ( a)( ) 2 ( c) 3 ( d) 234 Maths Quest MATHEMATICAL METHODS VCE Units and 2
Sign diagrams or graphs can e drawn using the nature of the zeros to solve inequations. On a graph, there is a significant difference in shape etween the wa the graph crosses the -ais at a saddle-cut -intercept compared with a cut -intercept. However, on a sign diagram, a saddle cut is not treated differentl to a cut since the sign diagram is simpl showing the graph crosses through the -ais at these points. The sign diagram for the graph drawn for = ( a)( ) 2 ( c) 3 ( d) is: + a c d WorKed example 6 Sketch the graph of = ( + 2) 2 (2 ) 3. think WritE Calculate the -intercepts. = ( + 2) 2 (2 ) 3 -intercepts: let = ( + 2) 2 (2 ) 3 = + 2 = or 2 = = 2 (touch), = 2 (saddle cut) -intercepts: ( 2, ) and (2, ) 2 State the nature of the graph at each -intercept. 3 Calculate the -intercept. -intercept: let = = (2) 2 (2) 3 Due to the multiplicit of each factor, at ( 2, ) the graph has a turning point and at (2, ) there is a stationar point of inflection. = 32 -intercept: (, 32) 4 Identif the degree of the polnomial. Leading term is () 2 ( ) 3 = 5, so the polnomial is of degree 5. 5 Sketch the graph. As the coefficient of the leading term is negative, the graph starts aove the -ais. (, 32) ( 2, ) (2, ) Topic 5 HigHer-degree polnomials 235
WorKed example 7 other families of polnomials Various sets of polnomials which share a common feature or features ma e considered a famil. Often these families are descried a common equation which contains one or more constants that can e varied in value. Such a varing constant is called a parameter. An eample is the set of linear polnomials, each with a gradient of 3 ut with a differing -intercept. This is the famil of parallel lines defined the equation = 3 + c, c R, some memers of which are shown in the diagram. This set of lines is generated allowing the parameter c to take the values 2,,,, 2 and 3. As could e anticipated, these values of c are the -intercepts of each line. 4 3 2 3 2 2 3 2 3 4 = 3 + c, c = 2,,,, 2, 3 Consider the famil of cuic polnomials defined the equation = 3 + m 2 where the parameter m is a real non-zero constant. a Calculate the -intercepts and epress them in terms of m, where appropriate. Draw a sketch of the shape of the curve for positive and negative values of m and comment on the ehaviour of the graph at the origin in each case. c Determine the equation of the memer of the famil that contains the point (7, 49). think WritE a Calculate the -intercepts. a = 3 + m 2 -intercepts: let = 3 + m 2 = 2 ( + m) = = or = m -intercepts: (, ) and ( m, ) Descrie the ehaviour of the Due to the multiplicit of the factor there is a turning point curve at each -intercept. at (, ). At ( m, ) the graph cuts the -ais. 2 Sketch the shape of the If m < then ( m, ) lies to the right of the origin. graph, keeping in mind If m > then ( m, ) lies to the left of the origin. whether the parameter is positive or negative. = 3 + m 2, m > ( m, ) (m, ) (, ) = 3 + m 2, m < 236 maths QuesT mathematical methods VCe units and 2
3 Comment on the ehaviour of the graph at the origin. c Use the given point to determine the value of the parameter. 2 State the equation of the required curve. Eercise 5.3 PRactise Work without CAS Families of polnomials WE4 Sketch the graph of = 32 ( ) 5. 2 On the same set of aes sketch the graphs of = 7 and =, laelling the points of intersection with their coordinates. Hence state : 7. 3 WE5 Sketch the graph of = (2 + ) 6 6. 4 A graph with equation = a( ) 8 + c has a minimum turning point at (, 2) and passes through the origin. a Determine its equation. State the equation of its ais of smmetr. c What is its other -intercept? 5 WE6 Sketch the graph of = (2 + ) 3 (2 ) 2. 6 The graph of a polnomial is shown. The graph has a maimum turning point at the origin if m is negative. If m is positive there is a minimum turning point at the origin. c = 3 + m 2 Sustitute the point (7, 49). 49 = 7 3 + m 7 2 49 = 49(7 + m) = 7 + m m = 6 If m = 6 then the memer of the famil which passes through the point (7, 49) has the equation = 3 6. a State the degree of the polnomial. Given the point (5, 24.3) lies on the graph, form its equation. 7 WE7 Consider the famil of quartic polnomials defined the equation = 4 m 3 where the parameter m is a real non-zero constant. a Calculate the -intercepts and epress them in terms of m where appropriate. Draw a sketch of the shape of the curve for positive and negative values of m and comment on the ehaviour of the graph at the origin in each case. c Determine the equation of the memer of the famil that contains the point (, 6). 8 Consider the famil of quadratic polnomials defined = a( 3) 2 + 5 4a, a R \. (5, 24.3) ( 4, ) (, ) (2, ) (4, ) a Show that ever memer of this famil passes through the point (, 5). For what value(s) of a will the turning point of the paraola lie on the -ais? c For what value(s) of a will the paraola have no -intercepts? Topic 5 Higher-degree polnomials 237
Consolidate Appl the most appropriate mathematical processes and tools 9 a On the same set of aes, sketch the graphs of = 6, = 6, = and =. On the same set of aes, sketch the graphs of = 5, = 2 5 and = 2 5. a i On the same set of aes, sketch the graphs of = and =. ii Hence state the solution set to : >. i Sketch the graphs of = 6 and = 7. ii Hence or otherwise, solve the equation 6 = 7. c i On the same set of aes sketch the graphs of = ( + ) 4 + and = ( + ) 5 +. ii Hence state the coordinates of the points of intersection of the two graphs. a State the coordinates of the turning point and specif its nature for each of the following: i = 6 ( 4) + 3 ii = 6 3 25 2 5 State the coordinates of the stationar point of inflection for each of the following: i = 5 + 27 ii = 6 (2 + ) 7 54 2 Sketch the shape of the following, identifing an intercepts with the coordinate aes. a = ( + 3) 2 ( + )( 2) 2 = 4 ( + 2)3 (8 ) c = 3 ( 2 ) d = 2 ( ) 2 ( 3) 2 3 a The curve elonging to the famil of polnomials for which = a( + ) 5 + c has a stationar point of inflection at (, 7) and passes through the point ( 2, 2). Determine the equation of the curve. The graph of a monic polnomial of degree 4 has a turning point at ( 2, ) and one of its other -intercepts occurs at (4, ). Its -intercept is (, 48). Determine a posssile equation for the polnomial and identif the coordinates of its other -intercept. c Give a possile equation for the graph of the polnomial shown. d The curve has the equation = (a + ) 4. ( 5, ) (, ) (, ) (3, ) When epanded, the coefficients of ( 3, 76.8) the terms in and 2 are equal and the coefficient of 3 is 536. Determine the equation of this curve. 4 a Solve the following equations: i 9 3 36 5 = ii 6 2 3 + = iii 5 + 4 3 2 2 2 = Solve the following inequations: i ( + 4) 3 ( 3) 2 < ii 8 (9 )5 3 iii ( 2) 4 8 238 Maths Quest MATHEMATICAL METHODS VCE Units and 2
5 a i Using a parameter, form the equation of the famil of lines which pass through the point (2, 3). ii Use the equation from part i to find which line in this famil also passes through the origin. i What is the point which is common to the famil of paraolas defined the equation = a 2 +, a? ii Epress the equation of those paraolas, in the form = a 2 +,, which pass through the point (2, 6) in terms of the one parameter, a, onl. c Calculate the -intercepts of the graphs of the famil of polnomials defined the equation = m 2 4, m > and draw a sketch of the shape of the graph. 6 Consider the two families of polnomials for which = k and = 2 + +, where k and are real constants. a Descrie a feature of each famil that is shared all memers of that famil. Sketch the graphs of = 2 + + for = 7, and 7. c For what values of k will memers of the famil = k intersect = 2 + 7 +? i Once onl ii Twice iii Never d If k = 7, then for what values of will = k intersect = 2 + +? i Once onl ii Twice iii Never 7 Consider the curves for which = ( a) 3 ( + a) 2, where a is a positive real constant. a Identif the intercepts with the aes in terms of a and comment on their nature. Draw a sketch of the shape of the curve. c In how man places will the line = intersect these curves? a d For what value(s) of a will the line = intersect the curve at the point 2, a 2? 8 Consider the famil of cuic polnomials for which = a 3 + (3 2a) 2 + (3a + ) 4 2a where a R \. a Show that the point (, ) is common to all this famil. For the memer of the famil which passes through the origin, form its equation and sketch its graph. c A memer of the famil passes through the point (, ). Show that its graph has eactl one -intercept. d B calculating the coordinates of the point of intersection of the polnomials with equations = a 3 + (3 2a) 2 + (3a + ) 4 2a and = (a ) 3 2a 2 + (3a 2) 2a 5, show that for all values of a the point of intesection will alwas lie on a vertical line. 9 Use CAS technolog to sketch the famil of quartic polnomials for which = 4 + a 2 + 4, a R, for a = 6, 4, 2,, 2, 4, 6, and determine the values of a for which the polnomial equation 4 + a 2 + 4 = will have: a four roots two roots c no real roots. 2 A paraola has an ais of smmetr with equation =. If the paraola intersects the graph of = 8 ( + 2)6 2 at oth the turning point and the -intercept of this curve, determine the equation of the paraola and sketch each curve on the same set of aes. Master Topic 5 Higher-degree polnomials 239
5.4 Units & 2 AOS Topic 4 Concept 3 Numerical approimations to roots of polnomial equations Concept summar Pratice questions Numerical approimations to roots of polnomial equations For an polnomial P() the values of the -ais intercepts of the graph of = P() are the roots of the polnomial equation P() =. These roots can alwas e otained if the polnomial is linear or quadratic, or if the polnomial can e epressed as a product of linear factors. However, there are man polnomial equations that cannot e solved an algeraic method. In such cases, if an approimate value of a root can e estimated, then this value can e improved upon a method of iteration. An iterative procedure is one which is repeated using the values otained from one stage to calculate the value of the net stage and so on. Eistence of roots For a polnomial P(), if P(a) and P() are of opposite signs, then there is at least one value of (a, ) where P() =. For eample, in the diagram shown, P(a) < and = P() P() >. The graph cuts the -ais at a point for which a < <. P() > a This means that the equation P() = has a root P(a) < which lies in the interval (a, ). This gives an estimate of the root. Often the values of a and are integers and these ma e found through trial and error. Alternativel, if the polnomial graph has een sketched, it ma e possile to otain their values from the graph. Ideall, the values of a and are not too far apart in order to avoid, if possile, there eing more than one -intercept of the graph, or one root of the polnomial equation, that lies etween them. The method of isection Either of the values of a and for which P(a) and P() are of opposite sign provides an estimate for one of the roots of the equation P() =. The method of isection is a procedure for improving this estimate halving the interval in which the root is known to lie. Let c e the midpoint of the interval [a, ] so c = (a + ). 2 The value = c ecomes an estimate of the root. B testing the sign of P(c) it can e determined whether the root lies in (a, c] or in [c, ). In the diagram shown, P(a) < and P(c) < = P() so the root does not lie etween a and c. It lies etween c and since P(c) < and P() >. The midpoint d of the interval [c, ] can then e a c d calculated. The value of d ma e an acceptale approimation to the root. If not, the accurac of the approimation can e further improved testing which of [c, d] and [d, ] contains the root and then halving that interval and so on. The use of some form of technolog helps consideral with the calculations as it can take man iterations to achieve an estimate that has a high degree of accurac. An other roots of the polnomial equation ma e estimated the same method once an interval in which each root lies has een estalished. 24 Maths Quest MATHEMATICAL METHODS VCE Units and 2
WorKed example 8 think a Calculate the value of the polnomial at each of the given values. Consider the cuic polnomial P() = 3 3 2 + 7 4. a Show the equation 3 3 2 + 7 4 = has a root which lies etween = and =. State a first estimate of the root. c Carr out two iterations of the method of isection hand to otain two further estimates of this root. d Continue the iteration using a calculator until the error in using this estimate as the root of the equation is less than.5. WritE a P() = 3 3 2 + 7 4 P() = 4 P() = 3 + 7 4 = 2 Interpret the values otained. As P() < and P() >, there is a value of which lies etween = and = where P() =. Hence the equation 3 3 2 + 7 4 = has a root which lies etween = and =. B comparing the values calculated select the one which is closer to the root. c Calculate the midpoint of the first interval. P() = 4 and P() =, so the root is closer to = than to =. A first estimate of the root of the equation is =. c The midpoint of the interval etween = and = is = 2. 2 State the second estimate. =.5 is a second estimate. 3 Determine in which half interval the root lies. 4 Calculate the midpoint of the second interval. P 2 = 8 3 4 + 7 2 4 = 9 8 < As P() < and P() >, the root lies etween = and =. 2 = 3 4 =.75 5 State the third estimate. =.75 is a third estimate. d Continue the calculations using a calculator. 2 State the estimate of the root. Note: The value is still an estimate, not the eact value. The midpoint of the interval etween = and = is: 2 = 2 2 + d P(.75) =.5625 As.5 < P(.75) <.5, this estimate provides a solution to the equation with an error that s less than.5. Hence =.75 is a good estimate of the root of the equation which lies etween = and =. Topic 5 HigHer-degree polnomials 24
Using the intersections of two graphs to estimate solutions to equations Consider the quadratic equation 2 + 2 6 =. Although it can e solved algeraicall to give = ± 7, we shall use it to illustrate another non-algeraic method for solving equations. If the equation is rearranged to 2 = 2 + 6, then an solutions to the equation are the -coordinates of an points of intersection of the paraola = 2 and the straight line = 2 + 6. Both of these polnomial graphs are relativel simple graphs to draw. The line can e drawn accuratel using its intercepts with the coordinate aes, and the paraola can e drawn with reasonale accurac plotting some points that lie on it. The diagram of their graphs shows there are two points of intersection and hence that the equation 2 + 2 6 = has two roots. = 2 + 6 8 5 2 9 6 3 = 2 5 4 3 2 2 3 4 5 3 6 9 Estimates of the roots can e read from the graph. One root is approimatel = 3.6 and the other is approimatel =.6. (This agrees with the actual solutions which, to 3 decimal places, are = 3.646 and =.646). Alternativel, we can confidentl sa that one root lies in the interval [ 4, 3] and the other in the interval [, 2] and appling the method of isection the roots could e otained to a greater accurac than that of the estimates that were read from the graph. To use the graphical method to solve the polnomial equation H() = : Rearrange the equation into the form P() = Q() where each of the polnomials P() and Q() have graphs that can e drawn quite simpl and accuratel. Sketch the graphs of = P() and = Q() with care. The numer of intersections determines the numer of solutions to the equation H() =. The -coordinates of the points of intersection are the solutions to the equation. Estimate these -coordinates reading off the graph. Alternativel, an interval in which the -coordinates lie can e determined from the graph and the method of isection applied to improve the approimation. 242 Maths Quest MATHEMATICAL METHODS VCE Units and 2
WorKed example 9 Use a graphical method to estimate an solutions to the equation 4 2 2 =. think Rearrange the equation so that it is epressed in terms of two familiar polnomials. 2 State the equations of the two polnomial graphs to e drawn. 3 Determine an information which will assist ou to sketch each graph with some accurac. 4 Carefull sketch each graph on the same set of aes. 5 State the numer of solutions to the original equation given. WritE 4 2 2 = 4 = 2 + 2 The solutions to the equation are the -coordinates of the points of intersection of the graphs of = 4 and = 2 + 2. The straight line = 2 + 2 has a -intercept at (, 2) and an -intercept at ( 6, ). The quartic graph = 4 has a minimum turning point at (, ) and contains the points (±, ) and (±2, 6). = 4 = 2 + 2 2 8 6 (2, 6) 4 2 8 6 4 2 (, ) 9 8 7 6 5 4 3 2 2 3 4 5 6 7 8 9 2 4 6 8 As there are two points of intersection, the equation 4 2 2 = has two solutions. 6 Use the graph to otain the solutions. From the graph it is clear that one point of intersection is at (2, 6), so = 2 is an eact solution of the equation. An estimate of the -coordinate of the other point of intersection is approimatel.7, so =.7 is an approimate solution to the equation. estimating coordinates of turning points If the linear factors of a polnomial are known, sketching the graph of the polnomial is a relativel eas task to undertake. Turning points, other than those which lie on the -ais, have largel een ignored, or, at est, allowed to occur where our pen and empath for the polnomial have placed them. Promises of rectifing this later when calculus is studied have een made. While this remains the case, we will address this unfinished aspect of our graph-sketching considering a numerical method of sstematic trial and error to locate the approimate position of a turning point. Topic 5 HigHer-degree polnomials 243
For an polnomial with zeros at = a and =, its graph will have at least one turning point etween = a and =. To illustrate, consider the graph of = ( + 2)( )( 4). The factors show there are three -intercepts: one at = 2, a second one at = and a third at = 4. There would e a turning point etween = 2 and =, and a second turning point etween = and = 4. The first turning point must e a maimum and the second one must e a minimum in order to satisf the long-term ehaviour requirements of a positive cuic polnomial. The interval in which the -coordinate of the maimum turning point lies can e narrowed using a tale of values. 2.5.5.5 6.875.25 8 4.375 As a first approimation, the maimum turning point lies near the point (.5,.25). Zooming in further around =.5 gives greater accurac..75.5.25.39625.25 9.2989 An improved estimate is that the maimum turning point lies near the point (.75,.39). The process could continue zooming in around =.75 if greater accurac is desired. An approimate position of the minimum turning point could e estimated the same numerical method of sstematic trial and error. The shape of this positive cuic with a -intercept at (, 8) could then e sketched. (.75,.39) ( 2, ) (, 8) (, ) = ( + 2)( )( 4) (4, ) An alternative approach For an polnomial P(), if P(a) = P() then its graph will have at least one turning point etween = a and =. This means for the cuic polnomial shown in the previous diagram, the maimum turning point must lie etween the -values for which = 8 (the -intercept value). 244 Maths Quest MATHEMATICAL METHODS VCE Units and 2
WorKed example Sustitute = 8 into = ( + 2)( )( 4): 3 3 2 6 + 8 = 8 3 3 2 6 = ( 2 3 6) = = or 2 3 6 = As the cuic graph must have a maimum turning point, the quadratic equation must have a solution. Solving it would give the negative solution as =.37. Rather than test values etween = 2 and = as we have previousl, the starting interval for testing values could e narrowed to etween =.37 and =. The positive solution = 4.37 = ( + 2)( )( 4) indicates that the minimum (.37, 8) turning point lies etween (, 8) (4.37, 8) = and = 4.37. In this = 8 case the interval etween ( 2, ) the two positive -intercepts provides a narrower and therefore etter interval to zoom into. (, ) (4, ) a State an interval in which the -coordinate of the minimum turning point on the graph of = ( 2)( + 3) must lie. Use a numerical method to zoom in on this interval and hence estimate the position of the minimum turning point of the graph, with the -coordinate correct to decimal place. think a State the values of the -intercepts in increasing order. 2 Determine the pair of -intercepts etween which the required turning point lies. Construct a tale of values which zooms in on the interval containing the required turning point. 2 State an estimate of the position of the turning point. 3 Zoom in further to otain a second estimate. WritE a = ( 2)( + 3) The -intercepts occur when =, = 2, 3. In increasing order the are = 3, =, = 2. The graph is a positive cuic so the first turning point is a maimum and the second is a minimum. The minimum turning point must lie etween = and = 2. Values of the polnomial calculated over the interval [, 2] are taulated..5.5 2 2.625 4 3.375 The turning point is near (, 4). Zooming in around = gives the tale of values:.9..2 3.86 4 4.59 4.32 4 State the approimate position. The minimum turning point is approimatel (., 4.59). Topic 5 HigHer-degree polnomials 245
Eercise 5.4 PRactise Consolidate Appl the most appropriate mathematical processes and tools Numerical approimations to roots of polnomial equations WE8 Consider the cuic polnomial P() = 3 + 3 2 7 4. a Show the equation 3 + 3 2 7 4 = has a root which lies etween = and = 2. State a first estimate of the root. c Carr out two iterations of the method of isection hand to otain two further estimates of this root. d Continue the iteration using a calculator until the error in using this estimate as the root of the equation is less than.5. 2 The graph of = 4 2 2 has two -intercepts. a Construct a tale of values for this polnomial rule for = 3, 2,,,, 2, 3. Hence state an eact solution to the equation 4 2 2 =. c State an interval within which the other root of the equation lies and use the method of isection to otain an estimate of this root correct to decimal place. 3 WE9 Use a graphical method to estimate an solutions to the equation 4 + 3 4 =. 4 Use a graphical method to estimate an solutions to the equation 3 6 + 4 =. 5 WE a State an interval in which the -coordinate of the maimum turning point on the graph of = ( + 2)( 3) must lie. Use a numerical method to zoom in on this interval and hence estimate the position of the maimum turning point of the graph with the -coordinate correct to decimal place. 6 Consider the cuic polnomial = 2 3 2 5 + 9. a State the -intercept. What other points on the graph have the same -coordinate as the -intercept? c Between which two -values does the maimum turning point lie? d Use a numerical method to zoom in on this interval and hence estimate the position of the maimum turning point of the graph, with the -coordinate correct to decimal place. 7 For each of the following polnomials, show that there is a zero of each in the interval [a, ]. a P() = 2 2 +, a =, = 2 P() = 2 3 + 8 + 3, a = 2, = c P() = 4 + 9 3 2 +, a = 2, = d P() = 5 4 3 + 2, a =, = 8 The following polnomial equations are formed using the polnomials in question 7. Use the method of isection to otain two narrower intervals in which the root lies and hence give an estimate of the root which lies in the interval [a, ]. a 2 2 + =, a =, = 2 2 3 + 8 + 3 =, a = 2, = c 4 + 9 3 2 + =, a = 2, = d 5 4 3 + 2 =, a =, = 246 Maths Quest MATHEMATICAL METHODS VCE Units and 2
9 The quadratic equation 5 2 26 + 24 = has a root in the interval for which 2. a Use the method of isection to otain this root correct to decimal place. What is the other root of this equation? c Comment on the efficienc of the method of isection. Consider the polnomial defined the rule = 4 3. a Complete the tale of values for the polnomial. 2 2 Hence, state an interval in which 4 3 lies. c Use the method of isection to show that 4 3 =.32 to 2 decimal places. Consider the polnomial equation P() = 3 + 5 2 =. a Determine an interval [a, ], a, Z in which there is a root of this equation. Use the method of isection to otain this root with an error less than.. c State the equations of two graphs, the intersections of which would give the numer of solutions to the equation. d Sketch the two graphs and hence state the numer of solutions to the equation P() = 3 + 5 2 =. Does the diagram support the answer otained in part? 2 The diagram shows that the line = 3 2 is a tangent to the curve = 3 at a point A and that the line intersects the curve again at a point B. B A = 3 = 3 2 a Form the polnomial equation P() = for which the -coordinates of the points A and B are solutions. Descrie the numer and multiplicit of the linear factors of the polnomial specified in part a. c Use an algeraic method to calculate the eact roots of the polnomial equation specified in part a and hence state the coordinates of the points A and B. d Using the graph, state how man solutions there are to the equation 3 3 + =. 3 Use a numerical sstematic trial and error process to estimate the position of the following turning points. Epress the -coordinate correct to decimal place. a The maimum turning point of = ( + 4)( 2)( 6) The minimum turning point of = (2 + 5)(2 + ) c The maimum and minimum turning points of = 2 4 Topic 5 Higher-degree polnomials 247
4 For the following polnomials, P() = d. Solve the equations P() = d and hence state intervals in which the turning points of the graphs of = P() lie. a P() = 3 3 2 4 + 9 P() = 3 2 + 8 c P() = 2 3 + 2 8 + d P() = 3 + 2 + 7 5 The weekl profit, in tens of dollars, from the sale of containers of whe protein sold a health food usiness is given = 3 + 7 2 3 4,. 3 6 (, 4) The graph of the profit is shown in the diagram. a Show that the usiness first started to make a profit when the numer of containers sold was etween and 2. Use the method of isection to construct two further intervals for the value of required for the usiness to first start making a profit. c Hence, use a numerical sstematic trial-and-error method to calculate the numer of containers that need to e sold for a profit to e made. d Use the graph to state an interval in which the greatest profit lies. e Use a numerical sstematic trial and error process to estimate the numer of containers that need to e sold for greatest profit, and state the greatest profit to the nearest dollar. f As the containers are large, storage costs can lower profits. State an estimate from the graph of the numer of containers eond which no profit is made and improve upon this value with a method of our choice. 6 a For a polnomial P() it is found that the product P( )P( 2 ) <. Eplain what conclusion can e made aout an roots of the equation P() =. Consider the polnomial P() = 6 3 2 4 5. i Show that P(2)P(3) <. ii Use the method of isection to otain a root of the equation 6 3 2 4 5 =. iii Show that there is onl one root to the equation 6 3 2 4 5 =. iv Determine the numer of points of intersection of the graph of = 6 3 2 4 5 and the graph of = 5. What information does this provide aout the graph of = 6 3 2 4 5? v Use the information gathered to draw a sketch of = 6 3 2 4 5. 7 Consider the cuic polnomial = 3 4 and the famil of lines = a. a Sketch = 3 4 and the famil of lines = a. What is the largest integer value of a for which the equation 3 + a + 4 = has three roots? c If a >, how man solutions are there to 3 + a + 4 =? d Determine the root(s) of the equation 3 + + 4 = using the method of isection. Epress the answer correct to 2 decimal places. Master 248 Maths Quest MATHEMATICAL METHODS VCE Units and 2
8 A rectangular sheet of cardoard measures 8 cm 4 cm. Four equal squares of side length cm are cut from the corners and the sides are folded to form an open rectangular o in which to place some clothing. a Epress the volume of the o in terms of. State an interval within which lies the value of for which the volume is greatest. c Use the spreadsheet option on a CAS calculator to sstematicall test values in order to determine, to 3 decimal places, the side length of the square needed for the volume of the o to e greatest. Topic 5 Higher-degree polnomials 249
ONLINE ONLY 5.5 Review www.jacplus.com.au the Maths Quest review is availale in a customisale format for ou to demonstrate our knowledge of this topic. the review contains: short-answer questions providing ou with the opportunit to demonstrate the skills ou have developed to efficientl answer questions without the use of CAS technolog Multiple-choice questions providing ou with the opportunit to practise answering questions using CAS technolog ONLINE ONLY Activities to access ebookplus activities, log on to www.jacplus.com.au Interactivities A comprehensive set of relevant interactivities to ring difficult mathematical concepts to life can e found in the Resources section of our ebookplus. Etended-response questions providing ou with the opportunit to practise eam-stle questions. A summar of the ke points covered in this topic is also availale as a digital document. REVIEW QUESTIONS Download the Review questions document from the links found in the Resources section of our ebookplus. studon is an interactive and highl visual online tool that helps ou to clearl identif strengths and weaknesses prior to our eams. You can then confidentl target areas of greatest need, enaling ou to achieve our est results. Units & 2 Higher-degree polnomials Sit topic test 25 maths QuesT mathematical methods VCe units and 2
5 Answers Eercise 5.2 a (, 5) = ( 2) 4 (, ) (3, ) (2, ) Minimum turning point (2, ); -intercept (, 5); -intercepts (, ), (3, ) (.5, ) (, ) = (2 + ) 4 -intercept and maimum turning point, ; 2 -intercept (, ) 2 a = 4 ( + 2)4 + 4 3 ( 4, ) = ( + 2) 4 + 4 4 ( 2, 4) -intercepts ( 4, ), (, ); (, ) : 4 ( + 2)4 + 4 > = : 4 < < (, 6) = ( + 2) 2 (2 ) 2 5 P() = ( )( 2)( + 4) 2, < 4, or 4 < < or > 2 6 = 6 or = 2 7 a c d (, 2) = 4 = 2 4 = 4 2 (, 2) (, ) (, ) 2 (, ) ( ), 2 ( ), (, 6) (, 6) (, ) (, 2) (, ) (, 6) (, ) (, ) (, ) = ( 2) 4 (, ) (, ) = 2 4 (, ) (, ) = 4 = 4 = 4 (, ) (, 2) (, ) (, 6) = ( ) 4 (, ) = ( + ) 4 = 4 + 2 = 4 (, 2) (, 3) (, 3) ( 2, ) (2, ) -intercepts ( 2, ) and (2, ) are turning points; -intercept (, 6) 4 = ( + 4)( 2)( 5) 4 (, ) (, ) (, ) (, 2) (, 2) (, ) = 4 Topic 5 Higher-degree polnomials 25
8 Turning point -intercept -intercepts 9 a = 2 ( + 3 9)4 a (, 6) minimum ( 3, 2) minimum (, 5) (, 2) (, ), (3, ) none = 6( + 3) 4 8 c = (3 2) 4 d = ( + ) 4 + c ( 5, 25) maimum (, ) (, ), (, ) a a = 2 3, k = 3 d (2, ) maimum (, 7) none e 3 65, 2 minimum, 5 8,, (, ) 5 Minimum turning point, 3 c =,, 5, PROOFS 7 7 3 (, ) (, 2 ) 8 (, ) (3, ) (, 3 ) a = (, ) c = 7 ( 8 )4 + (, 5) (, 6) d, 83 8 e 4 8,, + 4 8 (, 2), 7 7 PAGE f ( 3, 2) (, ) (, 83 8 ) 4 8 4 (, + 8, 7 ( 7 ( 5, 25) 2 a -intercepts ( 8, ), ( 3, ), (4, ), (, ); (, ) (, ) -intercept (, 96) ONLINE (, 96) ( 8, ) ( 3, ) (4, ) (, ) (, 7) 65 -intercepts ( 3, ), (2, ), 5,,, ; 2 3 ( 8 ), -intercept (, 9) (, 5 ) (, ) (, 9) 3 (, 5 2) 5 (, (, 3 ) 2 ) 2 (, ( 4, ) 7 ) (2, ) 65 ( 8), 5 (, 7 ) f 2 65, maimum, 7 8 a c d (2, ) e f (, 7 ) d ( ( 252 Maths Quest MATHEMATICAL METHODS VCE Units and 2
c -intercepts ( 7, ), (2.5, ); turning point (, ); -intercept (, 7) e =, = 2 f 3 or 2 5 a ( 3) 2 ( + 3) 2 ( 3, ) (3, ) (, 7) (, 7) (2.5, ) (, ) d -intercepts and turning points (, ), 5, 4 5 4, (, ) ( ) e -intercept and stationar point of inflection (, ); -intercept (4, ); -intercept (, 2) (, ) (, 2) (4, ) f -intercept and stationar point of inflection, ; 3 -intercept, ; -intercept, 3 (, 3 ) ( 3 ), (,, ) 3 a = ( + 6)( + 5)( + 3)( 4) 72 = ( 4)( + 2) 2 c = 2 3 3 ( + 6) d = 3 8 (2 + 3)2 (5 4) 2 4 a < 3 or.5 < < 4 or > 5.5 c (, 8) -intercepts and maimum turning points ( 3, ), (3, ); -intercept and minimum turning point (, 8) : 4 + 8 2 8 > = d R \ 3, 3 6 a (2, 6) 7 8 a a = 4, m = 8 c = 4 = 3 = 4 2 ( 2, 6) d i (, ) and n, n 4 ii a = n 2, m = n 3 P(2, 6) (.43, ) ( 2.7, ) (.84, ) = 2 (4., ) -intercepts are ( 2.7, ), (.84, ), (.43, ), (4., ); minimum turning points ( 2, 2), (2.92, 62.9); maimum turning point (.7, 4.34) (, ) (2.9, ) = 7 3, =, = 7 3 c 7 5 5 or = d =, = 3, =, = 7 Minimum turning point (.2, 4.33); -intercepts (, ), (2.9, ) 4 7 8 = ( + )( 3 2 + 8) Topic 5 Higher-degree polnomials 253