Algebra II / Andreoli, Graham, Hansen
Partial Answer Key to Practice Midterm Exam

Name: _____________________________

1.(a)

(b)

(c)

6y² – 4y – 66

(d)

w³ – 2w² – 11w + 12

2.

S = (– ¥ , 3/4] È [11/4, ¥ )
Graph consists of two rays: one has a solid circle on 0.75 and includes everything to the left, while the other has a solid circle on 2.75 and includes everything to the right.

3.

4/3 (w + y) – 2

4.

S = {1, 7/3}

5.

x = cdwy / (a – bdw)

6.

S = (23/4, ¥ ) = {x: x > 23/4}
Graph consists of a ray having an open circle on 23/4 and including everything to the right.

7.

S = {0, 5 + Ö2, 5 – Ö2}. There are no extraneous solutions here, but always check!

8.

9.

y = 2x – 21

11.

(–15/7, 8/7)
Note that the equation for C is irrelevant. All you need to do is find a corner of the feasible region (and in this problem there happens to be only one corner).

12.

{(–1/2, –3/4, 2)}
Write your answer either as an ordered triple or as x = –1/2, y = –3/4, z = 2. Using your calculator to compute rref([A]) where [A] is the 3 ´ 4 augmented matrix is a useful check, but work is required for full credit. The substitution method works fairly well here.

13.

14.

11/(x² – 4)

15.(a)

(b)

Be sure to sketch this graph (using calculator as a check afterward). From (a), you know the vertex is (–3/2, 15/2). From the original form of the problem, you know the y-intercept is (0, 3) since 3 is the constant term. Finally, you can use the quadratic formula to find the x-intercepts: (–3/2 + .5Ö(15), 0) and (–3/2 – .5Ö(15), 0)

(c)

x = –3/2

16.(a)

9/4 · x^3/2

(b)

5/4 · 5^1/3

17.(a)

{–2, (–7 + Ö 33)/4, (–7 – Ö 33)/4}

(b)

P(x) = (x + 2) (2x² + 7x + 2)

18.(a)

3/2 · |x| / y^3/4

(b)

3c⁴ · 3^1/3

19.

2 – 2Ö6 + 3i

20.

vertical asymptote: x = 4
horizontal asymptote: y = 0
removable discontinuity: (3, –1)
x-intercepts: none
y-intercept: (0, –1/4)

21.(a)

(b)

8z · (y/x)⁶

22.

by 15 m

23.(a)

height = y = –16t² + 106t + 10

(b)

6.718 seconds

(c)

106 ft/sec
This is faster than 60 mph, since 60 mph = 88 ft/sec (true fact).

24.(a)

Let M = mass (kg), n = # of round trips of sound in a second
M = 23,328,000/n³

(b)

68,012 kg

25.

13, 14, and 15