1105G4sf11 - MACl 105 Quadratic Functions Names K E B Group...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MACl 105 Quadratic Functions Names: K E B Group work 4 f (x) = ax? + bx + c (a 9’: 0) September 27', 2011 l. The simplest formula for a parabola is f(x) = x2. Graph this function. 2. Draw the graph of f(x) = ~x2. ‘K 00 :Il 2:111 3. What does the Sign of the leading-coefficient tell about the shape of the graph? CL>o Mews U 0.40 mews KN f(x) = x2 g(x) = 2x2 11(x) = 0.53:2 k(x) = —x2 j(x) = —2x2 m(x) : —{l.5x2 5. Where is the vertex of the parabola if the formula is y == axz , a at 0 ? (O ,Q‘} What is the effect of making the leading coefficient larger in absolute value (either more positive or more negative)? He. lacl owls larger, 4!». pemJoolat gals narrower (owlpwlschngLr) 6. In the formula f(x) = a):2 + bx + c , HO) = __C.__ . On the graph, this point represents the m. Evaluating, C(o) will gm. 4L1. y-inlemepl- «for we fibula». Jvlwl- Mus and. b b b 7. The parabola f(x) = 3x2 + bx + c has its axis of syimnetiy at the line it = —~2— vertex at the point (_.2__ ,f (“z—J] a a a Consider: f(x) =12 Mix—12 and 7 4?? f(1'()= 3x2 + bx+(c) Xe “fig," y “ _ . axz+lox=o I xfax H3) :0 30-0 cumbzo ax=~b K: -h- 0.. I wk ; “5.. a. : a0... XE 'tx ~l’t -. (sumac—e) 3“) )) :4 the: _ Ls _ y,_ WWW" < 53») QC :6.) 8. Find the vertices of the following parabolas. a)f{X)=2X2+8 (Jul/q,“ la: a, m. LCM CL: 3 K: '% = 0 WIM is. «Be (01% he 4w; y—iw‘emefi. C: 3 QCO) '3? b)g(x)=2x2+8x bum €50,4th b) (—93-?) - .. e- _._ at": x: “q ‘ 3 flfifikfi’fi‘“ 36:): D,x(»<+lt):o at 2-. . _ O - 3- co “5(4)“ “'5' ordain C L‘ “'N“ ’4 ”3.4) pm L64): 2(x-tl)(x+35=0 d": X'avl ‘X3’3 c)h(x)=2x2+8x+6 wt 5mpk .1? h 053 K'-'- -_‘5‘_= _3_ is 6 buM‘S 52% H higlxor 4km». Czé. Mam—a. teams at: 3. 10. Both of the above functions above are of the form y = (x "- h)2 . For f we have 11 = 2 , and for g we have h = ~2. Explain the effect on the graph ofchanging the formula fi'om y = x2 to y = (x — hf . Use the above graphs as examples. Ckmtng “NI. permuflk Raw. )1: x2 (in y: (xnm‘ will MMM 63mph Matthieu), [Ll bwai‘l'fi- I: h>o JAN. amps wfumme. 4-0 “M- F‘LQM’ Gawa- 4|» express?“ w?" [mm “M. 89mm up (Ff/«Md!» tics»: (m? m we. :9 Mo Mamet» tam mm. M stun. Let”? mil 4%». expreSStth tutu law. the MM efi Metals acts): (met wm m —a. e [ta-tap“ teams [my] 11. On the following grids, graph the formulas f (X) = x2 + 4 and g(x) = x2 — 4 . 12. Both of the above functions above are of the form y = x2 + R. For f we have R = 4 , and for g we have R = —4 . Explain the effect on the graph of changing the formula from y = x2 to y = x2 + k . Use the above graphs as examples. Chm-M9 4L3. PDVMAQO. (Pram y: x?— +, y: Kit-K will NM 414. finial. wlln‘cuflt/ [kl bum-H. TIT-F K70 Mu. maple mm; up (4:29. ampk .F “5&3. 'I-F kCO 41a gapk mm} claw. (9&1.- gmp‘as diam). 13. On the following grids, graph the formulas f(x) = (x — 2)2 + 4 and f(x) == (x + 2)2 -— 4. Label the vertices and y-intereepts with appropriate coordinates. (F66 = X '2_ [{X 4' 3') (4- (K5 1 x1. 4' LIX) 1t“- ‘km 6"13; ”M3 J“- WVF Th lam (rt-n)“ mm; 'l’Ln. WW 3 +0 4‘“- V‘ESM' TN- “! News 2 lo 'HM. Le-Ft. m -H MMS ll»; 3 M Merl-bx H up. 90 44M. Ulrika: WNW LI dawn. so “u. WW ‘u .3: (my Make. 15 a>%(-1,-Lh. £(o)=o E4. The form of a quadratic function f(X) = a(x — h)2 + k is calied vertex form. The vertex of the corresponding parabola is the point (h,k) . On page 1 'we discussed quadratic equations given in form f(x) = axz + bx + c ,which is called standard form. Some equations can be considered to be in either form. For example, in standard form the quadratic function f(x) = 2)::2 + 3 could be thought of as f(x) = 22:2 + 0 - X + 3 or in vertex form as f(X) = 2(X — (Dz + 3 . Either way, the vertex and y-intercept is at the point (0, 3;). Use whichever form is appropriate and any appropriate algebra to find the vertices and vertical intercepts of the following parabolas. Vertex (x,y) y—intercept y=? a)f(x)=5(x+3)2+1i} Ho): S(%)a+to =65 a) (‘3 IO 55 Vat-M Pam: ("3,10) (Opens up) b) g(x)=—4(x-—3)2-2 flab-«43": = ‘39 b) (3."3.) ”38' Vetting; Fwy“, : (3J’3) (opens 46%) c) s(t)=2t2—7 (Open; up} QM Am '0: 0 so W‘i-Mc': y- \néeveep'i' a)j(1-)=—2(r+7r +o 3(o)=->.(n’=-as d) (”9:02 4i? Write/x gm : (-0.0) (Opens dam) e)11(t)=160t+15-—16t2 e) (.5 Lug) L = — .. is .. léo m X‘- 513." 33'3“"5 |o= loo c; [5 i" 5): ins Mot-ls (Opus) “9) 15. Just as with a line, if enough information is know about the graph of a parabola the corresponding equation can be determined. Use the graph below and vertex form to determine the equation of the following parabola. See example 4 on p. 181 of your text for a similar example. a 2 mam-is $00: c.0040 +i< or- My Veda; (Dy-19) g :(x): 0‘(X*a)a..|s egjfitewi Firm MM. graph we. see. ~ilawi' :(o)=-S'. J, -S=..- deaf-to rca- a 6(3“ =39x- le+to-ISI —$= Lia-ls maxim»: Si Lin—514’ I HS H5 u¢=toa>a=%=—M 60*“) ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern