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Unformatted text preview: MACl 105 Names: K8 3 Group work 5 October 6, 2011 Graphs of Polynomials The exponent of a factor is called the multiplicity of that factor. In each of the given formulas the multiplicity of the factor
(X + 4) is 1, while the multiplicity of the factor (x ~— 2) starts at 1 and increases to 4. 1. Draw the following graphs. Use [~5, 5] for [X11].lEl ,Xmax] and an appropriate scale of your own choosing for the y axis so all the extremes are clearly shown. Put the values you use for ymin and yum at the ends of each vertical axis. Locate and
label all x— and y-intercepts. State the degree of each polynomial and multiplicity of the factor (x —- 2) . Use your calculator’s max/min feature to label the local extremes with coordinates (x, y) , using one—decimal—place accuracy as necessary. a) y = (x + 4)(x —2)‘ degree: 3 b) y = (x + 4)(x -— 2)2 degree: 3
the: multiplicity of (x-Z): l multiplicity of (ii-2): Q
‘1: (H)(‘\)=-8 l ‘ , w
y That. is a quhG,
heme, o. parabola. Tl». Mia. J: (4,4) X=Oi
v: (HM-13‘- l6 (Alla) C’ l) " a
c) y =(x + 4)(x — 2)3 degree: H a) y = (x + 4)(x — 2)‘1 degree: 5
multiplicity of (Ii-2): multiplicity of (ii—2): LI
x=ot 00 ' -|»OO° How does a graph behave at an x—intercept that comes item a factor with an even multiplicity? What if the multiplicity is odd?
Think about y—values changing signs and explain your answers to these two questions. Use the above graphs as examples. ‘I-F' 4kg. mull-ipllci-l-y is cum (b,d) the. ample elm Ml- crocs Mu. x-Mis «Jr All»: lnkmept,l§wi l? “M. Mulllplicih l5 (Meoq (4,6) graph A“; cwsg.
Thus lAAppmb bums—c. exam paw-us eliminvlt Sign changes-Ii? n is am,
0:40" mu eley [meme a. el-ilau' s‘rle e-P Km. Tswi it a is eat 4%». lg a» sign chm-«ea mepwtl' wﬁ. lbw. graph cuescss He. wands. If the degree of a polynomial, the horizontal intercepts of its graph, and the coordinates of one other point are known, we can
derive a formula for this polynomial. For example, the graph shownﬁhas degree 3 and passes through the labeled points. We are seeing three x-intercepts, which matches the degree. Using these
intercepts we can derive the three factors. Therefore the formula must be of the form y = a(x + 4)(X + 2)(X —- 2) for a suitable value for leading
coefﬁcient a. Substituting the coordinates of the other point into the formula, we can solve
for the leading coefﬁcient. )7 = a(x + 4)(x + 2X: — 2) 72 = a(4 +4)(4 + 2)(4 m2)
72 = a(3)(6)(2) 72 = 963
a = 7—2 = 0.75
96 So the formula becomes 3’ = 0.75(x + 4)(X + 2)(X - 2) . This is called
the factored form. If the multiplication is performed, this formula would appear instandardform as y = 0.75):3 + 3X2 — 3x — 12 . 2. Derive a formula in factored form for the following graph. Assume the multiplicities of all factors are l, and that all the key
behavior of the graph is shown. Use your caleulator’s max/min feature to label the local extremes with appropriate coordinates (Ly) , using one-decimal~place accuracy as necessary.
:26 - - - v= a» (K+H)(x~ soon) so l°¢°9~ “MG TLL Pong-l (OJ-9°) is M ilnl. a“ 48 (34.116) Va..le l-o solve t-‘a- M."- _5 _ - s -30: a.(.o+u)(o-aD(0-lb
' ‘ ' ‘ —%o=ct('sa) -_g-9I$—-'
0».- 3a '35 (-t h, surfs)
(Goad). Mln ‘ Factored form: 3’ a .. .5 X + Ll .. .. r ‘ 3
‘7: — 1.5x + 5x" + H0 K ~80
If the factored form was multiplied out to standard form, what would the constant term be? You do not have to multiply the
factors together to answer this question.
Ans: " 8 0 TLQ, CW$+QM+ lax-w. ‘15 ll“. y—in-lcrcepi-l ...
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