activity 34
DESCRIPTION
Activity 34. Review (Sections 3.6+3.7+4.1+4.2+4.3). Problems 3 and 5:. Sketch the graph of the function by transforming an appropriate function of the form y = x n . Indicate all x- and y-intercepts on each graph. No x – intercepts. y – intercepts is (0,8). y = x 4. y = 2x 4. y = 2x 4 + 8. - PowerPoint PPT PresentationTRANSCRIPT
ACTIVITY 34
Review (Sections 3.6+3.7+4.1+4.2+4.3)
Problems 3 and 5:Sketch the graph of the function by transforming an appropriate function of the form y = xn. Indicate all x- and y-intercepts on each graph.
3)2()( xxP3xy 32 xy 32 xyintercept-y
0x 3)20(0 P 8
8,0
intercept-x0y
3)2(0 x
3)2(0 x)2(0 x 2x
0,2
82)( 4 xxP
y = 2x4
y = x4
y = 2x4 + 8
No x – interceptsy – intercepts is (0,8)
Problems 11, 13, and 15:Match the polynomial function with the below graphs
)4()( 2 xxxP xxxxR 45)( 35 34 2)( xxxT
xxxxR 45)( 35 )4()( 2 xxxP34 2)( xxxT
Problems 21:Sketch the graph of the polynomial function. Make sure your graph shows all intercepts and exhibits the proper end behavior.
2323)( xxxxPszero' / intercepts-x
3x2x
32
x
intercept-y
0x 2032030)0( P 223 12
The leading term is 3x3
Problem 31:Factor the polynomial P(x) = −x3 + x2 + 12x and use the factored form to find the zeros. Then sketch the graph.
xxxxP 12)( 23 122 xxx 34 xxx
0 x 04 x 03 x0x 4x 3x
Zeros
Problem 3:Let P(x) = x4 − x3 + 4x + 2 and Q(x) = x2 + 3. (a)Divide P(x) by Q(x). (b)Express P(x) in the form P(x) = D(x) · Q(x) + R(x).
x4 − x3 + 4x + 2x2 + 3
x2
x4 + 3x2 –x4 – 3x2
− x3 – 3x2 + 4x + 2
– x
– x3 – 3x + x3 + 3x
– 3x2 + 7x + 2
)(xQ
)(xR
)()()()( xRxDxQxP 1173324 2234 xxxxxxx
– 3
– 3x2 – 9 +3x2 + 9
7x + 11
Problem 9:Find the quotient and remainder using long division for the expression
2236
2
3
xxxx
x3 + 6x + 3x2 – 2x+2
x
x3 – 2x2 + 2x –x3 + 2x2 – 2x
2x2 + 4x + 3
+ 2
2x2 – 4x + 4– 2x2 + 4x – 4
8x – 1
)(xQ
)(xR
)()()()( xRxDxQxP
)()()()(
)()(
xDxRxDxQ
xDxP
)()(
)()()(
xDxR
xDxDxQ
)()()(
xDxRxQ
22
1822236
22
3
xxxx
xxxx
Problem 21:Find the quotient and remainder using synthetic division for the expression
3283
xxx 13 0 8
1
3
3
9
1)(xQ )(xR
13)( 2 xxxQ
23
1
1)( xR
Problem 35:Use synthetic division and the Remainder Theorem to evaluate P(c) if P(x) = 5x4 + 30x3 − 40x2 + 36x + 14 and c = −7.
57 30 40
5
35
5
35
5)(xQ )(xR
)()( cRcP
363571
)7(P
14497
483
483
Problem 43:Use the Factor Theorem to show that x − 1 is a factor of P(x) = x3 − 3x2 + 3x − 1.
11 3 3
11
2
2
1
1
10
Showing that x = 1 is a zero.Therefore, (x – 1) is a factor
Problem 53:Find a polynomial of degree 3 that has zeros 1, −2, and 3, and in which the coefficient of x2 is 3.
1x 2x 3xa
3222 xxxxa 322 xxxa 6233 223 xxxxxa 652 23 xxxa
aaxaxax 652 23
32 a
23
a
1x 2x 3x23
Problem 3:List all possible rational zeros given by the Rational Zeros Theorem (but don’t check to see which actually are zeros).
8432)( 235 xxxxR
qp ,1
,1 2
,1possible rational zeros:
2 of divisors8 of divisors
,21
,2 ,4
,2 ,4 8
8
Problems 13 and 23:Find all rational zeros of the polynomial.
23)( 3 xxxP
qp ,1
1
,1possible rational zeros:
1 of divisors2 of divisors
2
2
11 0 3
111
12
2
2
4
11 0 3
111
12
2
20
233 xx 22 xx )1(x 22 xx 1x
233 xx 22 xx 1x
22 xx 12 xx
112233 xxxxxConsequently, the zero’s are x = 2 and x = – 1
8676)( 234 xxxxxP
qp
1 of divisors8 of divisors
,1
1
,2 ,4 8
possible rational zeros: ,1 ,2 ,4 8
8676)( 234 xxxxxP possible rational zeros: ,1 ,2 ,4 8
11 6 7
1
17
7
14
6
148
8
80
8676 234 xxxx 8147 23 xxx 1x
So we need only factor 8147 23 xxxpossible rational zeros: ,1 ,2 ,4 8possible rational zeros: ,1 ,2 ,4 8
8147 23 xxx
11 7 14
116
6
8
8
8
0 8676 234 xxxx 8147 23 xxx 1x
1)1(862 xxxx
11862 xxxxSo we need only factor 862 xx 14 xx
possible rational zeros: ,1 ,2 ,4 8
11248676 234 xxxxxxxx
Consequently, the roots are x = -4, x = -2, x = -1, and x = 1