Algebra Ii 2nd Semester Study Guide

Algebra II 2nd Semester Study Guide italicized =may be incorrect ; bold = not done Pg. 2/key concept

Pg. 55/3 [decide whether the given number is a solution of the equation] |b-1| = 14; -13 |-13-1| = 14 |-14| = 14 14 = 14  Pg. 55/4 [decide whether the given number is a solution of the equation] |d+6| = 10; -4 |-4+6| = 10 |2| = 10 2 ≠ 10 Pg. 55/5 [decide whether the given number is a solution of the equation] |32-6f| = 20; -2 |32-6(-2)| = 20 |32-(-12)| = 20 |32+12| = 20 |44| = 20 44 ≠ 20 Pg. 93/29 [find the x- and y-intercepts of the line with the given equation] -6x + 8y = -36 x-intercept y-intercept -6x + 8(0) = -36 -6(0) + 8y = -36 -6x = -36 8y = -36 x=6 y = -9/2 or -4.5 Pg. 145/15 [the variables x and y vary directly. Write an equation that relates x and y. then find x when y=6] x = -2, y = -1 y = ½x y = ax 6=½x -1 = -2a 12 = x ½=a Pg. 164/31 [solve the system using any algebraic method] 4x – 10y = 18 -2x + 5y = -9 4x – 10y = 18 2[-2x + 5y] = -9 4x – 10y = 18 -4x + 10y = -18 0; infinitely many solutions Pg. 227/20 [use the given matrices to evaluate the expression, if possible. If not possible, state the reason]

DE Not possible; both are dimensions of 3 x 2 Pg. 323/15 [solve the equation] (x+2)2 – 12 = 36

(x+2)2 = 48 (x+2) = √48 x = -2 ± 2√12 Pg. 323/18 [write the expression as a complex number in standard form] 3 + I . 2 + 3i 6+9i+2i+3i2 = 3 + 11i = 2-3i 2 + 3i 4-9i2 13 Pg. 323/20 [solve the equation by completing the square] x2 – 10x – 7 = 0 x2 – 10x =0 x2 – 10x + 25 = 7 + 25 (x-5) 2 = 32 x – 5 = √32 x = 5 ± 4√2 Pg. 342/29 [Describe the end behavior of the graph of the polynomial function by completing these statements: f(x)  ? as x-∞ and f(x)  ? as x+∞] f(x) = -x6 + 4x3 – 3x f(x)  -∞ as x-∞ f(x)  -∞ as x+∞ Pg. 366/29 [given the polynomial function f and a zero of f, find the other zeros] f(x) = x3 – 2x2 – 21x – 18 -3 1

-2 -21 -18 x2 – 5x – 6 = 0 -3 15 18 (x-6)(x+1) = 0 1 -5 -6 0 Pg. 390/5 [graph the function] g(x) = ⅓(x-5)(x+2)(x-3)

(x-6) = 0 x=6

(x+1) = 0 x = -1

Pg. 397/3 [write the cubic function whose graph is shown]

y = 0.5x3 – 2x2 + 0.5x + 3 Pg. 397/7 [write a cubic function whose graph passes through the points] (-2,0), (-1,0), (0,-8), (2,0) y = 2x3 + 2x2 – 8x - 8 Pg. 407/12 [perform the indicated operation] (3x3 – 14x2 +16x – 22) ÷ (x - 4) 10

3x2 - 2x + 8 + x-4 x-4|3x3 – 14x2 +16x - 22 -) 3x3 – 12x2 -2x2 + 16x -)-2x2 + 8x 8x – 22 -)8x – 32 10 Pg. 407/13 [perform the indicated operation] (6x4 + 7x2 + 4x – 17) ÷ (3x2 -3x + 2) Pg. 407/17 [find all real zeros of the function] f(x) = x3 + x2 – 22x - 40

Pg. 417/9 [rewrite the expression using rational exponent notation] (3√10)7 (n√a)m = am/n  (3√10)7 = 107/3 Pg. 424/31 [write the expression in simplest form] 3 15 √9  3√9 . 5√9 = √9 5 5 5 5 √27 √27 √9 √243 Pg. 432/29 [Let f(x) = 3x-1, g(x) = 2x-7, and h(x) = x-4/5. perform the indicated operation and state the domain] g(f(2)) =g(3(2)-1) =g(5) =2(5)-7 =3 Pg. 469/4 [evaluate the expression without using a calculator] 3 √27 =271/3 =3 . 3 . 3 =3 Pg. 469/7 [evaluate the expression without using a calculator] (3√-27)2 =(271/3)2 =32 =9 Pg. 469/29 [find the inverse of the function] f(x) = x3 + 5 x = f(x)3 +5 x – 5 = f(x)3 3 √x-5 = f(x)-1 Pg. 469/34 [solve the equation. check for extraneous solutions] √3x + 7 = 4 (√3x + 7)2 = 42 3x + 7 = 16 3x = 9 x=3 Pg. 495/6 [simplify the expression] (2e-2)-4 =2-4(e8) = 1 . e8 16

= e8 16 Pg. 543/6 [graph the function. state the domain and range] g(x) = (2/3)x + 2 Pg. 543/13 [graph the function. state the domain and range] y = log2x Pg. 543/18 [condense the expression] log 5 + log x – 2 log 3 =log5x - 2 log 3 =log5x 32 =log5x 9 Pg. 543/19 [use the change-of-base formula to evaluate the logarithm] log550 =log 50 log 8 ≈1.6989 0.9030 ≈1.8813 Pg. 543/22 [solve the equation. check for extraneous solutions] 72x = 30 log7 72x = log7 30

2x = log7 30 2x = log 30 log 7 2x ≈ 1.7478 x ≈ 0.8739 Pg. 543/24 [solve the equation. check for extraneous solutions] log4 x + log4 (x + 6) = 2 4 log4 x + 4 log4 (x + 6) = 42 x (x + 6) = 16 x2 + 6x = 16 x2 + 6x –16 = 0 (x + 8) (x – 2) = 0 x = -8 or 2 Pg. 562/29 [graph the function. state the domain and range] y=x+6 4x – 8

domain: all real numbers except 3, range: all real numbers except 1 Pg. 568/9 [identify the x-intercept(s) and vertical asymptote(s) of the graph of the function] f(x) = x2 + 9 . x2 – 2x – 15 x-intercept: vertical asymptotes: x2 + 9 = 0 x2 – 2x – 15 = 0 x2 = -9 (x – 5) (x+ 3) = 0 no solution x = 5 and x = -3 Pg. 569/15 [graph the function] y = 2x f x2 – 1

Pg. 607/5 [the variables x and y vary inversely. use the given values to write an equation relating x and y. then find y when x=4] x = -4, y = 7/2 y=

a x

y = −14 x

y=

−14 4

7/2 = −a4 y = −27 a = -14 Pg. 607/26 [solve the equation. check for extraneous solutions] x +1 13 1 x +6 + x = x +6 x (x+6) [

1 x+6

+

x +1 x

=

13 x +6

]

1 ( 3) + 6

x + (x+6)(x+1) = 13x

+

( 3) +1 ( 3)

= ( 313 ) +6

13 1 4 9 + 3 = 9 13 13 9 = 9 

1 ( 2 )+ 6

+

( 2 ) +1 ( 2)

=

3 13 1 8 + 2 = 8 13 13 8 = 8 

13 ( 2 )+ 6

x2 + 8x + 6 = 13x x2 –5x + 6 = 0 (x – 3) (x – 2) = 0 x = 3 and 2 Pg. 617/31 [write an equation for the perpendicular bisector of the line segment joining the two points] (3, 8), (7, 14) (

x1+ x 2 2

,

y1+ y 2 2

) = ( 3+27 , 8+214 ) = (5, 11)

m=

y 2 − y1 x 2 − x1

=

14 −8 7 −3

=

6 4

=

3 2

- m1 = - 31/ 2 = - 23 y – 11 = - 23 (x – 5) y – 11 = - 23 x +

10 3

y = - 23 x + 43 3 Pg. 623/21graph the equation. identify the focus, directrix, and axis of symmetry of the parabola] 5x2 + 12y = 0 Pg. 623/29 [write the standard form of the equation of the parabola with the given focus and vertex at (0, 0)] (0, -4) x2 = 4py x2 = 4(-4)y x2 = -16y Pg. 630/53 [write an equation of the line tangent to the given circle at the given point] x2 + y2 = 17; (1, 4) focus = (0, p) p = -4

slope m =

4 −0 1− 0

=4

- m1 =- 14

y – 4 = - 14 (x – 1) y – 4 = - 14 x +

1 4

y = - 14 x + 174 Pg. 673/3 [find the distance between the two points. then find the midpoint of the line segment joining the two points] (-1, -6), (1, 5) d= M(

( x 2 − x1)2 + ( y 2 − y1)2 = (1 − ( −1))2 + (5 − (−6))2 = 122 ≈ 11.045

x1+ x 2 2

,

y1+ y 2 2

1 ) = ( −1+ 2 ,

−6+5 2

) = (0, - 12 )

Pg. 673/11 [graph the equation] (x – 6)2 + (y + 1)2 = 36 Pg. 673/12 [graph the equation] (x + 4)2 = 6(y – 2) Pg. 673/25 [classify the conic section and write its equation in standard form] x2 + y2 + 8x + 12y + 3 = 0 A = 1, B = 0, C = 1 value of discriminant is: B2 – 4AC = 02 – 4(1)(1) = -4 < 0 ; conic is a circle (x2+ 8x) + (y2 + 12y) = -3 (x2+ 8x + 16) + (y2 + 12y +36) = -3 + 16 + 36 (x +4)2 + (y + 6)2 = 49 Pg. 673/27 [classify the conic section and write its equation in standard form] y2 –16y –12x + 40 = 0 A = 0, B = 0, C = 1 value of discriminant is: B2 – 4AC = 02 – 4(0)(1) = 0; conic is a parabola Pg. 717/key concept

Pg. 806/27 [write a rule for the nth term of the arithmetic sequence. then graph the first six terms of the sequence] 7 a10 = 30, d = 2 an = a1 + (n - 1) d 30 = a1 + (10 –1)

7 2

an = a1 + (n - 1) d an = - 32 + (n - 1) 72

- 32 = a1 an = -5 + 72 n Pg. 807/47 [find the sum of the arithmetic series] -1 + 4 + 9 + … + 34 Sn = n ( a1+2a 2 ) S8 = 8( −1+234 ) S8 = 132 Pg. 807/49 [write a rule for the sequence whose graph is shown]

an = -3 + 5n Pg. 843/4 [tell whether the sequence is arithmetic, geometric, or neither. explain] 4, 7, 12, 19, … neither; there is no common difference or ratio Pg. 843/5 [write the first six terms of the sequence] an = 6 – n2 a1 = 6 – (1)2 = 5 a2 = 6 – (2)2 = 2 a3 = 6 – (3)2 = -3 a4 = 6 – (4)2 = -10 a5 = 6 – (5)2 = -19 a6 = 6 – (6)2 = -30 Pg. 843/11 [write the next term of the sequence, and then write a rule for the nth term] 6 7 8 9 5 , 10 , 15 , 20 ,… 10 25

n ; 6+ 5n Pg. 843/16 [find the sum of the series] 19

∑ 2i + 5 i =1

a1 = 2(1) + 5 = 7 a5 = 2(5) + 5 = 5 =39 a2 = 2(2) + 5 = 9 a6 = 2(6) + 5 = 5 =41 a3 = 2(3) + 5 = 11 a7 = 2(7) + 5 2(19) + 5 =43 a4 = 2(4) + 5 = 13 a8 = 2(8) + 5

15

a9 = 2(9) + 5 =

23

a13 = 2(13) + 5 =31

a17 = 2(17) +

17

a10 = 2(10) + 5 =25

a14 = 2(14) + 5 =33

a18 = 2(18) +

=

19

a11 = 2(11) + 5 =27

a15 = 2(15) + 5 =35

a19 =

=

21

a12 = 2(12) + 5 =29

a16 = 2(16) + 5 =37

=475