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COLLEGE ALGEBRA II (MATH 010)
SPRING 2015
——PRACTICE FOR EXAM II(SOLUTIONS)——
1. Find the intercepts, asymptotes, and domain of each rational function.
(a) s(x) =
(b) r(x) =
(c) f (x) =
x−1
x2 −4
2x−1
x+1
x2 −x−6
x+1
Solution:
x−1
(a) s(x) = xx−1
2 −4 = (x−2)(x+2)
−1
x−int: x = 1, y−int: y = −4
= 14
Horiz. Asymp: y = 0, Vert. Asymp: x = ±2
y
5
−5
5
x
−5
Figure 1: Domain: {x : x 6= ±2}, Range: (−∞, ∞)
1
(b) r(x) = 2x−1
x+1
x−int: 2x − 1 = 0, x = 12 , y−int: y =
Horiz. Asymp: y = 12 = 2
Vert. Asymp: x + 1 = 0, x = −1
y
−1
1
= −1
5
−5
5
x
−5
Figure 2: Domain: {x : x 6= −1}, Range: {y : y 6= 2}
2
−x−6
(c) f (x) = x x+1
= (x−3)(x+2)
x+1
x−int: (x − 3)(x + 2) = 0, x = 3, and x = −2
y−int: y = −6
1 = −6
Horiz. Asymp: None, Vert. Asymp: x + 1 = 0, x = −1
Oblique/Slant: y = x − 2(Long Division)
y
5
−5
5
x
−5
Figure 3: Domain: {x : x 6= −1}, Range: (−∞, ∞)
2
2. Find the focus and directrix of the parabola
y = − 18 x2 , and sketch the graph.
Solution:
x2 = −8y
4p = −8, p = −2
Focus (0, −2) and Directrix y = 2.
4
3
2
1
0
−5 −4 −3 −2 −1 0
−1
1
2
3
4
5
−2
−3
−4
−5
3. Find the focus and directrix of the parabola. Sketch the
graph.
y 2 − 12x = 0 Solution:
y 2 = 12x, 4p = 12, p = 3
Focus (3, 0) and Directrix x = −3
7
6
5
4
3
2
1
0
−4
−3
−2
−1 0
−1
1
2
3
4
5
−2
−3
−4
−5
−6
−7
3
4. Find the foci, the vertices, and the lengths of the major
and minor axes of the ellipse, and sketch the graph.
x2 y 2
+
=1
16
9
Solution:
√
√
2
2
2
a = √ 16 = 4, b =
√ 9 = 3, and c = a − b
c = 7, Foci (± 7, 0) and Vertices (±4, 0).
Major Axis Length: 2a = 8 and Minor Axis Length: 2b = 6
4
3
2
1
0
−5 −4 −3 −2 −1 0
−1
1
2
3
4
5
−2
−3
−4
4
5. The vertices of an ellipse are (0, ±4), and the foci are (0, ±3).
Find its equation, and sketch the graph.
Solution:
a = 4, c = 3, and c2 = a2 − b2 or b2 = a2 − c2 = 7
2
y2
=1
Equation: x7 + 16
5
4
3
2
1
0
−4 −3 −2 −1 0
−1
1
2
3
4
−2
−3
−4
−5
6. Find the vertices, foci, and asymptotes of the hyperbola,
and sketch the graph.
x2 − 4y 2 = −4
Solution:
√
y2
x2
Rewrite
as
−
=
1,
a
=
1,
b
=
4 = 2, and c2 = a2 + b2
1
4
√
√
c = 5, Vertices (0, ±1), Foci (0, ± 5), y = ± ab x = ± 12 x
5
4
3
2
1
0
−5 −4 −3 −2 −1 0
−1
1
2
3
4
5
−2
−3
−4
−5
5
7. Find the equation of the hyperbola with vertices (±2, 0)
and foci (±3, 0). Sketch the graph.
Solution:
a =√
2, c = 3, and c2 = a2 + b2
2
2
b = 5, Equation x4 − y5 = 1
6
5
4
3
2
1
0
−5 −4 −3 −2 −1 0
1
2
3
4
5
−2
−3
−4
−5
−6
8. Complete the square to put each equation in standard form.
Identify the type of conic, and its components (vertices,
center, foci, etc.)
(a) 4x2 + 9y 2 − 16x − 36y + 16 = 0
(b) x2 + 8x + 8y = 0
(c) −9x2 + 16y 2 − 72x − 96y − 144 = 0
Solution:
(a) 4x2 + 9y 2 − 16x − 36y + 16 = 0
(4x2 − 16x) + (9y 2 − 36y) = −16
4(x2 − 4x) + 9(y 2 − 4y) = −16
4(x2 − 4x + 4) + 9(y 2 − 4y + 4) = −16 + 4(4) + 9(4)
4(xh − 2)2 + 9(y − 2)2 = 36
i
1
1
2
2
4(x
−
2)
+
9(y
−
2)
= 36
· 36
36
6
(x−2)2
9
2
+ (y−2)
=1
4
Ellipse: Center=
(2, 2), vertices= (±3, 0),
√
foci= (± 5, 0)
(b) x2 + 8x + 8y = 0
(x2 + 8x) = −8y
(x2 + 8x + 16) = −8y + 16
(x + 4)2 = −8(y − 2)
Parabola: Vertex= (−4, 2), focus= (−4, 0),
directrix: y = 4
(c) −9x2 + 16y 2 − 72x − 96y − 144 = 0
(−9x2 − 72x) + (16y 2 − 96y) = 144
−9(x2 + 8x) + 16(y 2 − 6y) = 144
−9(x2 + 8x + 16) + 16(y 2 − 6y + 9) = 144 − 9(16) + 16(9)
16(yh − 3)2 − 9(x + 4)2 = 144
i
1
1
2
2
· 144
16(y
−
3)
−
9(x
+
4)
= 144
144
2
2
(y−3)
− (x+4)
=1
9
16
Hyperbola: Center= (−4, 3),
Vertices= (−4, 3 ± 3), foci= (−4, 3 ± 5),
asymptotes: y − 3 = ± 34 (x + 4)
7
9. Solve each system.
(a)



x − y = 1
4x + 3y = 18
(b)



3x + 4y = 10
x − 4y = −2
Solution:
(a) Substitution:
x=1+y
4(1 + y) + 3y = 18
4 + 4y + 3y = 18
7y = 14
y=2
x=1+y =3
(3, 2)
(b) Elimination:
3x + 4y = 10
x − 4y = −2
4x
= 8
x
= 2
8
2 −
4y
−4y
y
= −2
= −4
=
1
(2, 1)