10.5 Graphing Quadratic Functions


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1 0.5 Grphing Qudrtic Functions Now tht we cn solve qudrtic equtions, we wnt to lern how to grph the function ssocited with the qudrtic eqution. We cll this the qudrtic function. Grphs of Qudrtic Functions The grph of f x x x c The completed squre form f x x h k vertex of the prol occurs t the point symmetry. is clled prol (Ushped curve). is clled stndrd form of the function. The h, k nd the verticl line h x is clled the xis of There re two possile grphs of the qudrtic function. For 0 : For 0: vertex Opens up xis of symmetry vertex xis of symmetry Opens down From our sic grphs we cn see the following:. If is negtive, f x x x c vlue of y k.. If is positive, f x x x c y k. opens down nd the prol hs mximum opens up nd the prol hs minimum vlue of All qudrtic functions hve mximum or minimum nd it lwys occurs t the vertex. For this reson, the vertex is very importnt piece of informtion. Also notice tht the vlue of the mximum or minimum is the y vlue of the vertex. The x vlue of the vertex just indictes the vlue t which tht mximum or minimum occurs. We lso note tht the grph is symmetry round the xis of symmetry. Tht is, the grph is mirror imge cross the line tht is the xis of symmetry. The xis of symmetry for qudrtic functions lwys goes through the vertex nd is lwys the x vlue of the vertex. Lstly efore we egin grphing qudrtic functions, we need to recll the following informtion. Finding Intercepts To find the xintercepts, let y = 0 nd solve for x. To find the yintercept, let x = 0 nd solve for y.
2 Now we re redy to grph our qudrtic functions. Exmple : Find the vertex, x nd yintercepts, xis of symmetry, mximum or minimum vlues nd grph the following.. x 5 y. x 6x 5 y c. f x x 8x Solution:. First we notice tht the eqution is lredy in stndrd form. Therefore, using the formul ove we cn simply red off the vertex. Notice the signs of h nd k. Stndrd form is set up so tht we must switch the sign of the vlue with the x nd keep the sign of the vlue on the outside. So for our eqution x 5 5,. y we get the vertex is Also, since the vlue out in front is, we know tht the prol must open down. Therefore, the vertex must represent mximum. So the mximum vlue is the y vlue of the vertex. Thus we hve mximum vlue of. Also, the xis of symmetry lwys goes through the x vlue of the vertex. Thus the xis of symmetry is x 5. So we still need the intercepts nd grph of the function. xintercepts: (Set y=0) yintercepts: (Set x=0) Extrcting roots is esiest here x 5 0 x 5 x 5 x 6, x 5 y 0 5 x So now to grph we simply need to plot ll of this given informtion nd connect with smooth Ushped curve. We get. This time the eqution is not given to us in stndrd form. So we must egin y putting it in stndrd form. To do this we need use completing the squre. Recll, for completing the squre, the leding coefficient hd to e + nd we dd to oth sides of the eqution. The only thing tht is different this time is insted of dding the sme vlue to oth sides of the eqution, we will dd the completed squre vlue in nd sutrct tht sme vlue from
3 the sme side. This wy the eqution will sty lnced nd we will hve everything on one side nd y on the other side, just s stndrd form requires. We proceed s follows y x x 6x 5 6x 5 x 6x x 5 9 x So now tht the eqution is in stndrd form we cn red off the vertex, xis of symmetry nd the minimum vlue (since the prol is opening up ecuse = which is positive) Vertex:, Axis of symmetry: x Minimum vlue is Now we simply need the intercepts. We hve our choice of which eqution to use to find the intercepts. For the xintercepts we will use the eqution in stndrd form since we cn esily extrct the roots to solve. For the yintercept, we will use the eqution tht ws given to us since we cn see tht if we let x = 0 the eqution is prcticlly trivil. xintercepts: yintercepts: x 0 x x x x 6.7, 0.7 y We use deciml equivlents for the xintercepts since we wnt to use them for grphing. It is clerly esier to plot 0.7 thn to plot. Now we put it ll together nd grph dding from the sme side to lnce the eqution sutrcting c. Agin we must strt y getting the eqution to stndrd form. Agin, rememer tht the leding coefficient must e +. So we will fctor the out of the first two terms only (since those re the only terms we wnt to complete the squre on) nd then dd our completed squre piece. We proceed s follows
4 f x x x 8x x x x sutrcting x The  is here since wht we relly dded in is x 5 prenthesis So now tht the eqution is in stndrd form we cn red off the vertex, xis of symmetry nd the mximum vlue (since the prol is opening down ecuse = ) Vertex:, 5 Axis of symmetry: x Mximum vlue is 5 Now we find the intercepts. Like efore, for the xintercepts we will use the eqution in stndrd form since we cn esily extrct the roots nd for the yintercept, we will use the eqution tht ws given to us since if we let x = 0 the eqution very esy. xintercepts: yintercepts: x x 5 x 0 x 5 5 x 5 x., 0. 5 Now we put it ll together nd grph Fctor out  to get the leding coefficient + dding y 0 times the leding coefficient since it ws in front of the 8 0 There is nother wy to find the vertex which is little esier thn getting the function into stndrd form. Lets get the qudrtic function f x x x c We proceed s follows into stndrd form.
5 f x x x x c x c Fctor out to get the leding coefficient + dding x x c sutrcting x c from the sme side to lnce the eqution while ccounting for the leding coefficient So the vertex hs n x vlue of the following. The y vlue of the vertex is relly f x. So, we get Finding the Vertex of Qudrtic Function The vertex of f x x x c is given y the formul. f, We cn use either this formul method or the completing the squre method shown ove to find the vertex of prol. Exmple : Grph. Find the vertex, x nd yintercepts, nd ny mximum or minimum vlues.. y x 0x 0. h x x x Solution:. Lets use the formul x to solve this prolem. First, we find the vertex. So we notice tht nd 0 x. Therefore the vertex is t 5 0 nd f f So the vertex is 5, 5. Since the prol opens down, this vlue represents mximum. So the mximum vlue is 5. Next we find the intercepts s usul. We will use the qudrtic formul for the xintercepts since in this cse it is the most efficient method to use. xintercepts: 0 x x 0 5 0x , 0.9 yintercept: y
6 Now we simply plot ll of our informtion nd finish the grph.. Agin we use the formul x to solve this prolem. First, we find the vertex. In this nd cse x. Therefore the vertex is t nd h h So the vertex is,. Since the prol opens up, this vlue represents minimum. So the minimum vlue is. Next we find the intercepts s usul. xintercepts: 0 x x x i yintercept: y 0 0 Since the xintercepts re complex numers, the grph must hve no xintercepts. We could hve lso seen this y noticing tht the minimum vlue is. Since the grph cn get no smller thn it cnnot intercept the xxis. Now plot ll of our informtion nd finish the grph we cn use little symmetry to ssist us in getting enough informtion to get good grph. We cn use the point symmetric to the yintercept. We get
7 So in generl, we use the following procedure for grphing qudrtic function ( prol). Sketching Prol. Determine the vertex y completing the squre or using the formul x.. Find the x nd yintercepts.. Plot the vertex nd intercepts.. Use the fct tht the prol opens up if 0 nd down if 0 nd symmetry to complete the grph. 0.5 Exercises Find the vertex, x nd yintercepts, xis of symmetry nd mximum or minimum vlues of the following.. y x. y x. y x. y x 5. x 7. x y 6. y x y 8. y x 9. f x x 0. f x x. y x. h x x. y x 8. h x x 6 5. h x 6 x 6. y x 7. y x 8. y x 9. y x 0. 7 y x 6 Grph y completing the squre. Find the vertex, x nd yintercepts, nd ny mximum or minimum vlues.. h x x x. h x x x 5. f x x 6x 7. f x x 0x 5. g x x 8x 6. g x x x 5 y 8. y x x 9. y x 8x 7 7. x x 0. x 6x 9 y. y x x. y x 6x y. y x x 0 5. y x x. x 8x 9 6. y x x 7. f x x x 8. f x x x 9. g x x x 0. g x x x Grph y using the formul minimum vlues. x. Find the vertex, x nd yintercepts, nd ny mximum or. h x x x 5. h x x x. f x x 0x. f x x 6x 7 5. g x x x 5 6. g x x 8x y 8. y x x 9. y x 6x 9 7. x x 50. x 8x 7 y 5. y x 6x 5. y x x 5. x x 0 y 5. y x 8x y x x 56. x x y 57. f x x x 58. f x x x
8 59. g x x x 60. g x x x Grph. Find the vertex, x nd yintercepts, nd ny mximum or minimum vlues. 6. x y 6. x x y x 7x 6. x 5x 8 y 6. y 65. f x x x g x x x 67. x x x 70. h 68. x 9x y y x x 7. y x 6x y 8x x 7. y x x 7. f x x x 7. f x x x 75. g x x x 5 x 6x 76. x x x 79. y x x 80. y x x x x g 77. y y
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