Algebra II Chapter 9 Review SOLUTIONS

Here are the solutions to the Chapter 9 Review for Algebra 2. Please remember that you may use a SINGLE 3x5 index card - both sides. Come prepared.

Solutions:
1) vertical asymptote: x = -2; horizontal asymptote: y = 1; x-intercepts: -5 and 2; y-intercept: -2.5
2) (5x + 34)/((x + 5)(x - 3)(x + 4)); (x + 8)/(x - 2), x ≠ -1, x ≠ 5
3) 16 in
4) y = 0
5) d = sqrt(76^2 + x^2) + 42 - x
6) (4, 0) and (-4, 0)
7) (x/48)^2 + (y/18)^2 = 1
8) hole: x = 1; vertical asymptotes: x = -1, x =2
9) skip
10) vertical parabola with f = -1, (h, k) = (0, o)
11) (sqrt(40), 0) and (-sqrt(40), 0)
12) ((x + 2)/5)^2 + ((y + 2)/4)^2 = 1
13) focus: (3, 0) and directrix: x = -3
14) (x + 4)^2 = y + 3
15) a locus of points, the difference of whose distances to the foci is constant
16) C
17) vertices: (-4/3, 2) and (-8/3, 2); foci: -2 + sqrt(85)/3 and -2 - sqrt(85)/3; asymptotes have slope 9/2 or -9/2
18) C
19) Center at (0, 0) with rectangle 4 wide and 10 tall, horizontal hyperbola
20) vertex: (3, -9); focus: (7, -9); directrix: x = -1
21) Center at (4, 3) with rectangle 6 wide and 10 tall, horizontal hyperbola
22) vertical, with (h, k) = (5, -1) and k = -1
23) skip
24) skip
25) (5/3, 0) and (-5/3, 0)
26) B
27) (2x - 19)/(x - 5)
28) (x(x - 3)(x + 8))/((x + 4)(x - 4))
29) (x + 3)^2 - ((y - 1)^2/ 2) = 1
30) vertical asymptote: x = 6; slant asymptote: y = 3x + 2
31) (w^2 + 1)/((w - 1)(w - 2))
32) (-7x + 71)/((x - 9)(x + 1)(x - 1))
33) 2t/3s; 8m; -32a^2b/27
34) (y/2)^2 = x
35) a locus of points whose distance from the focus is equal to its distance to the directrix
36) B