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# Col Alg for Math,Science,Engin MTH 111C

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This 18 page Class Notes was uploaded by August Feeney on Monday October 19, 2015. The Class Notes belongs to MTH 111C at Portland Community College taught by Staff in Fall. Since its upload, it has received 46 views. For similar materials see /class/224641/mth-111c-portland-community-college in Math at Portland Community College.

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Date Created: 10/19/15

Haberman King MTH lllc Section III Graph Transformations Module 4 Combinations of Transformations EXAMPLE A function y mx is defined by the graph below Match each transformation of m given in a 7 d with one of the graphs in z 7 v1 below a y max 17 y 2mm c y mx 2 d y mx 2 it iii iv SOLUTION Since y m2x is a horizontal stretch of y mx by a factor of Le a compression the graph it must represent the function in a Since y 2mx is a vertical stretch of ymx by a factor of 2 the graph 1 must represent the function in b Since y mx 2 is a shift up 2 units of y mx the graph of iii must represent the function in c Since y mx 2 is a shift left 2 units of y mx the graph of iv must represent the function in d When transformations are combined the order that you perform the transformations matters In practice there are usually many different reasonable sequences of transformations that work As long as you provide a correct sequence you need not worry about the order you choose But in general the order that you list the transformations matters so if you mix up their order you won t obtain the proper graph In the generalization below we suggest an order to use each time you are confronted with a transformation problem It isn t the only possible generalization but we think it is the most useful SUMMARY OF GRAPH TRANSFORMATIONS Suppose that f and g are functions such that gx afbx h k and a b h k e R In orderto transform the graph of the function f into the graph of g 1 horizontally stretchcompress the graph of f by a factor of m and if b lt 0 reflect it about the yaxis 2quot shift the graph horizontally h units shift right if h is positive and left if h is negative 3quot vertically stretchcompress the graph by a factor of lal and if a lt 0 reflect it about the xaxis 4 shift the graph vertically k units shift up if k is positive and down if k is negative The order in which these transformations are performed matters EXAMPLE Describe a sequence of transformations that warp the graph of rx x3 into the graph of tx x 13 SOLUTION First let s rewrite tx in terms rx and adjust its form roe x 1f r x 1 r x 1 So to warp the graph of y rx into the graph of tx 1 Reflect the graph of rx x3 about the yaxis 2quot Shift left one unit to obtain the graph of y tx Below we ve graphed rx x3 and tx x 13 39139 v The graphs of rx x3 and tx x 13 EXAMPLE Describe a sequence of transformations that would warp the graph of y fx into the graph of hx 5fx 2 7 SOLUTION First write hx in terms of fx and adjust it s form hx 5fx 2 7 5fx 8 7 So to warp the graph of y fx into the graph of y hx 1 Horizontally stretch the graph by a factor of 4 2quot Shift right 8 units 339 Vertically stretch by a factor of 5 4 Shift up 7 units EXAMPLE Describe a sequence of transformations that warps the graph of rx x3 into the graph of ux 32x 73 4 SOLUTION First write ux in terms of rx and adjust its form ux 32x 73 4 3r2x 7 4 3r2x 4 We can now see that a sequence of transformations that warps the graph of y rx into the graph of y ux is 1 Stretch the graph horizontally by a factor of ie compress the graph 2quot Shift right units 339 Stretch vertically by a factor of 3 and reflect about the x axis 4 Shift up 4 units EXAMPLE The function yfx is graphed below Sketch a graph of sx fx3 SOLUTION To sketch the graph of sx fx 3 we need to vertically compress the graph of y fx by a factor of ie compress and reflect the graph about the xaxis Then we need to shift the graph up 3 units Let s just think about what happens to a few key points Reflecting about the x axis and stretching vertically by a factor of will not affect the points 6 0 or 4 0 So after shifting up 3 units these points will end up at 6 3 and 4 3 The point 4 4 becomes 4 2 after we stretch it by a factor of then 4 2 after reflecting about the xaxis Finally shifting up 3 units brings takes 4 2 to 4 1 The point 2 2 will end up at the point 2 4 and the point 6 4 will end up at the point 6 5 Verify these for yourself Now use these key points to sketch sx fx 3 The graph of sx fx 3 is given below The graphs uf yx arm 3 EXAMPLE The mhmuh yx s graphed be uvv sketch a graph uf 1x7f1 SOLUTION Let s start by rewriting tx in terms offx trying to mimic the form given in the Generalization of Graph Transformations above To sketch the graph of tx f 1 3 we need to reflect y fx about the x axis then stretch horizontally by a factor of 2 Next shift left 2 units and then shift down 3 units The points 6 0 and 4 0 are not affected by the reflection about the xaxis However stretching horizontally by a factor of 2 takes the points 6 0 and 4 0 to the points 12 0 and 8 0 respectively Shifting left 2 units and then shifting down 3 units takes the points 12 0 and 8 0 to the points 14 3 and 6 3 respectively The point 4 4 moves to 4 4 upon reflection about the xaxis then to 8 4 upon stretching horizontally by a factor of 2 Finally shifting left 2 units and then shifting down 3 units takes the point 8 4 to the point 10 7 The point 2 2 will end up at the point 2 1 and the point 6 4 will end up at the point 10 1 You should verify these for yourself Now we can use the points we just found to sketch tx f 1 3 see the graph below The graphs of yfx and tx f 1 3 EXAMPLE Describe a sequence of transformations that warp the graph of fx x2 into the graph of gx x2 6x 1 SOLUTION In order to determine how to transform the graph of y fx into the graph of y gx we need to write of fx in terms of gx We can do this by completingthesquare See 55 and pp 227229 in the text for a review of this procedure gxx2 6x1 2 add and subtract 6 9 to create a perfect square trinomial x2 6x9 91 x2 6x9 8 x 32 8 fx 3 8 So to transform the graph of y fx into the graph of gx fx 3 8 first shift y fx right 3 units then down 8 units see the graph below You should recognize y gx as a parabola opening upwards with vertex 3 8 The graphs of fx x2 and gx fx 3 8 EXAMPLE Complete the square to determine a sequence of transformations that warp the graph fx x2 into the graph of hx 2x2 3x 5 SOLUTION Mx2x2 3x5 z 3 factor out the coefficient of the squared 2x x 5 term from the first two terms x x 2 z 2 2 3 9 i add and subtract i i 9 2 16 16 5 2 to create a perfect square trinomial 2x 5 2x2 2f x So to transform the graph of y fx into the graph of y hx 1 Shift the graph right of a unit 2quot Vertically stretch by a factor of 2 339 Shift up units Haberman King MTH 1110 Section IV Power Polynomial and Rational Functions Module 3 Graphing Polynomial Functions In the last module we looked at the longterm behavior of polynomials After studying that module you should be able to recognize that polynomials like fx x4 3x3 63x2 27x 486 and gx x4 12x3 27x2 270x 648 have similar longterm behavior Since they are both 4 degree polynomials with a positive leading coefficient we know that their graphs must have arrows pointing up at the extreme left and rightsides ie the outputs of both functions increases without bound as the inputs increase without bound and as the inputs decrease without bound See Figure 1 below Jquot 75000 50000 25000 X 20 10 10 20 Figure 1 y fx and y gx Although the functions fx x4 3x3 63x2 27x 486 and gx x4 12x3 27x2 270x 648 have similar longrun behavior they are not identical functions Let s study the shortrun behavior of their graphs to see how these functions differ The shortrun behavior of the graph of a function concerns graphical features that occur when the input values aren t very large It s hard to specify what not largequot means since it will be different for each function but we ll for particular graphical features rather than look within a particular interval so we don t need to worry about being more specific Clearly x 0 isn t a large xvalue so the yintercept will be part of the shortrun behavior of a polynomial function s graph Notice that the ycoordinate of the yintercept ofa polynomial function is its the constant term The yintercept of fx x4 3x3 63x2 27x 486 is 0486 The yintercept of gx x4 12x3 27x2 270x 648 is 0648 Zeros or roots are another important part of the shortrun behavior of the graph of a polynomial function To find the roots of a polynomial function we can write it in factored form fx x4 3x3 63x2 27x 486 x 3x 3x 6x 9 eh raemr er gwes Hse m a rerun smce wheh eaeh factur Equa s zeru the uutput fur the mheheh rs zere Tu nd the ruuts determme whreh numbers make the ram Equa m zere FACTOR 2 ROOT h urdertu graph f we eah mm ts rem a d yrmtercept see Frgure 2 be uvv y 797573 35912 7500 Fi ure2 2 can euhheet these pmnts 7 makmg sure that our graph has the correct longd39un 7 3x3 7 63x 27K 486 see Frgure 3 Nuvv we behaviorem nmp etethe graph e f x e x y 500 x 79 7M 5 12 7500 igure 3 The raph er y fx Nuvv et sfactur g m determme rtsruuts W x 712x 7 27x2 270K 648 x32x7 0712 FACTOR 2 ROOT shee 1he mm x e3 15 assumated W11h a sguareg pr duub1e factur 11 15 men named a double rook ur a rook ofmultiplicity two 1h urder 1p graph g We eah pm 11 rpm and wmereepp see Rama 4 The prahge pmnt at x e 3 15 a root ormumprcrzyzwo y 1000 500 x 797573 3591215 7500 71000 igurea W we eah EDHHEEI 1hese pumts 7 making sure 11m our graph has the correct longd39un a a 12x 7 h pr W x e 27x2 270K 648 see Rama 5 1 beha Nuuee that the why Way 1p ED men the pawl W11hu111 usmg aggmuha1 rum ur WEEIrrECL 1e1th run behavmr 151 have the graph puuhee unhe guume rpm at x 73 Try 11yuurse1r1 1000 50 x 79 as 73 3 9 15 7500 71000 Figure5 Th graph pr y go Q EXAMPLE Wmte ah a1gepra1e ru1erpr1he pp1yhprh1a1 mhehph p grapheg Rama 6 Nate 1ha11he graph passes1hrpugh 1he pu1nt7318 50y A 25 X 75 47372 1 2 4 725 F gures The gra h pr y pa SOLUTION Since there are roots at x 4 x 2 x 0 and x 3 we know that p has form px kxx 3x 4x 2 where k is a constant To find k we can use the point 3 18 3 18 3 p 3 18 k 3 3 3 3 4 3 2 3 18 k 3611 3 18 18k 3 k 1 Therefore px xx 3x 4x 2 EXAMPLE Write an algebraic rule for the polynomial function w graphed in Figure 7 Note that the yintercept of w is 012 50 y 25 x 4 3 2 1 1 2 4 25 SO F gure 7 The graph of y wx SOLUTION Since there are roots at x 3 x 1 x 1 and x 3 and since the graph bounces off the xaxis at x 1 this is a root of multiplicity two Le a doubleroot Thus we know that w has form wx kx 3x 12x 1x 3 where k is a constant To find k we can use the fact that the yintercept of w is 012 012 2 f0 12 kO 30 120 10 3 2 12 k312 1 3 3 12 9k 3 k Thus wx x 3x 12x 1x 3 EXAMPLE Whte ah a1geprate rme turthe pp1yhprhta1 mhettph h grapheg W thure 8 hate that the y nterceptuf h 1 013 25y o 1 1o 7 7 7 473727175 1 4 5 710 Figure 8 The gr ph nyhx SOLUT ON Nuttee that the graph at y 10 due hut have rupts at mtEgEr vames But there are same htee pmnts uh the graph at y 10 a1phg the huthhta1 hhe y 1t We treat thtshhe as the XVaX1S 1e Amagmethatthe graph at h 1ssh1 ed dEIWn 5 umts then the graph meg have rupts x x 1 ah x 74 s tu nd a rmetpr h we eah ere a mhettph that hasthese three rants and then shttt 11 up 5 WM 10 k x e 4x 71mm 5 Tphhg k We eah use the y ntercept 013 013 2 10 13 k 0 7 4x0 71x0 4 5 131c 74e145 j j 816k j Therefure ah a1geprate rme tpr h 1 hx x e 4x 71 4 5 Properties of Polynomial Functions I The graph of a polynomial is a smooth unbroken curve By smooth we mean that the graph does not have any sharp corners as turning points I The graph of a polynomial always exhibits the characteristic that as lxl gets very large lyl gets very large I If p is a polynomial of degree n then the polynomial equation px0 has at most n distinct solutions that is p has at most n zeros This is equivalent to saying that the graph of y px crosses the xaxis at most n times Thus a polynomial of degree five can have at most five xintercepts I The graph of a polynomial function of degree n can have at most n l turning points see Key Point below For example the graph of a polynomial of degree five can have at most four turning points In particular the graph of a quadratic 2quotd degree polynomial function always has exactly one turning point its vertex amp Key Point A turning point on a graph is a point where the function changes from increasing to decreasing or vice versa These points are the relative maximums or relative minimums of the function An example of a turning point is the vertex of a parabola EXAMPLE What is the minimum possible degree of the polynomial function in Figure 9 4y 4 V2 1234 4 Figure 9 SOLUTION The polynomial function graphed in Figure 9 has four zeros and five turning points The properties of polynomials tell us that a polynomial function with four zeros must have a degree of at least four These properties also tell us that if a polynomial has degree n then it can have at most n l turning points In other words the degree of a polynomial must be at least one more than the number of turning points Since this graph has four turning points the degree of the polynomial must be at least six Keep in mind that although a 6th degree polynomial may have as many as six real zeros it need not have that many The graph in Figure 9 only has four real zeros Concavity There are four graphs in Figure 10 below The first two graphs a and b have different shapes but they are both increasing on the interval 2 1 The second two graphs c and d also have different shapes but they are both decreasing on the interval 2 1 4 Jquot 4 y 2 x x 2 2 2 2 2 2 a b 4 J 4 J quot N 2 A x 2 2 2 2 2 2 c c Figure 10 As the first graph a rises it bends or curves upward but as the second graph b rises it bends or curves downward Similarly the third graph c is falling and curved upward whereas the fourth graph d is falling and curved downward A graph the curves upward like the graphs a and c is called concave up a graph that curves downward like the graphs b and d is called concave down It might help to remember thata parabola that opens upward is concave up and a parabola that opens downward is concave down The graph in Figure 11 is always increasing concave down on the interval 00 1 and concave up on the interval 1 00 At the point 1 1 the concavity changes from concave up to concave down Such a point is called a point of inflection 4y 2 2 igure11 An inflection point occurs at 1 1 The puTyhurmaT y x3x2 7 2x ts grapheg W thure 12 the graph appears tn have an th eetmh puht at x a 7 meat Neg EXAMPLE Use the graph m thure 13 tpthg a the apprpxtrhate TnteNa sWherE the graph at ts hereasthg b the apprpxtrhate TnteNa sWherE the graph at ts decreasmg the apprpxtrhate TnteNa sWherE the graph at ts cuncave up st the apprpxtrhate TnteNa sWherE the graph at ts cuncave duvvn e the number at h eetph pphts an the graph at SOLUTTON a The apprpxtrhate mtENa S where the graph at ts hereasthg are 7min and o 4 5 b The appruxwmate ntENa S where the graph uf Ts decreasmg are 74 50 and 4 5 w c It appears that the graph of f is concave up on the interval 2 2 D It appears that the graph of f is concave down on the intervals 00 2 and 2 00 e As we move from left to right the graph changes from concave down to concave up and then back to concave down Each change occurs at an inflection point so the graph of f has two inflection points EXAMPLE Suppose p is a polynomial function that satisfies the following conditions The graph of p has exactly three turning points 25 5 0 2 25 5 and the graph of p has exactly two inflection points 1 0 and 1 0 Sketch a graph of p based upon this information How many real zeros does p have SOLUTION Let s start by plotting the given points see Figure 14 Figure14 There is only one way to connect the points to create a polynomial function without adding turning points or inflection points see Figure 15 You should verify this for yourself 1 Figure 15 The graph of y px Since it has four xintercepts p has four real zeros

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