1105G4sf11

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

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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*“) ...
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