Algebra 2 Level 2
Final Exam Review 3
Problems 1-4: Solve the equation. Show all steps neatly. Keep rational solutions in exact
form, and round off irrational solutions to the nearest hundredth.
1. 2𝑥 − 7 + 𝑥 + 4 = 2(3𝑥 − 8)
3𝑥 − 3 = 6𝑥 − 16
−3𝑥 = −13
13
𝑥= 3
2. |5𝑥 − 2| = 6𝑥 + 10
5𝑥 − 2 = 6𝑥 + 10
−𝑥 = 12
𝑥 = −12
CHECK:
|5(−12) − 2| = 6(−12) + 10
|−62| = −62
False – extraneous solution
5𝑥 − 2 = −6𝑥 − 10
11𝑥 = −8
8
𝑥 = − 11
8
8
|5 (− 11) − 2| = 6 (− 11) + 10
40
22
48
|− 11 − 11| = − 11 +
62
62
110
11
|− 11| = 11
True
3. 3𝑥 2 + 𝑥 + 4 = 20
3𝑥 2 + 𝑥 − 16 = 0
𝑥=
−1±√1−4(3)(−16)
2(3)
−1±√193
𝑥=
6
𝑥 ≈ 2.15, 𝑥 ≈ −2.48
4. 25(0.82)𝑥 = 3.5
0.82𝑥 = 0.14
𝑥 = log 0.82 0.14
𝑥 ≈ 9.91
5. Solve by completing the square. Show all steps, and keep solutions in exact form.
𝑥 2 + 22𝑥 + 140 = 25
𝑥 2 + 22𝑥 + _____ = −115 + _____
𝑥 2 + 22𝑥 + 121 = −115 + 121
(𝑥 + 11)2 = 6
𝑥 + 11 = ±√6
𝑥 = −11 ± √6
6. Simplify the expression completely.
𝑥
12
𝑥
12
𝑥∙𝑥
12∙3
𝑥 2 −36 (𝑥+6)(𝑥−6)
− 𝑥 2 +6𝑥 = 3(𝑥+6) − 𝑥(𝑥+6) = 3𝑥(𝑥+6) − 3𝑥(𝑥+6) = 3𝑥(𝑥+6)
3𝑥+18
3𝑥(𝑥+6)
=
𝑥−6
3𝑥
7. Simplify the complex fraction completely.
3𝑥+3
−2
𝑥+1
4
7+
𝑥
𝑥(𝑥+1)
∙ 𝑥(𝑥+1) =
3𝑥 2 +3𝑥−2𝑥(𝑥+1)
7𝑥(𝑥+1)+4(𝑥+1)
𝑥 2 +𝑥
𝑥(𝑥+1)
𝑥
= (7𝑥+4)(𝑥+1) = (7𝑥+4)(𝑥+1) = 7𝑥+4
8. Solve the system of equations using the method of your choice.
𝑥
𝑥 + 2𝑦 − 3𝑧 = 1
1 2 −3 −1 1
13
3𝑥 − 2𝑦 + 𝑧 = 55
[𝑦] = [3 −2 1 ] ∙ [55] = [−9]
𝑧
𝑥 − 2𝑦 − 𝑧 = 33
1 −2 −1
33
−2
(13, −9, −2)
9. Expand the power. Show all work.
(𝑥 + 5)3
(𝑥 + 5)(𝑥 + 5)(𝑥 + 5)
(𝑥 + 5)(𝑥 2 + 10𝑥 + 25)
𝑥 3 + 10𝑥 2 + 25𝑥 + 5𝑥 2 + 50𝑥 + 125
𝑥 3 + 15𝑥 2 + 75𝑥 + 125
10. Factor completely.
𝑥 8 + 4𝑥 6 − 9𝑥 2 − 36
𝑥 6 (𝑥 2 + 4) − 9(𝑥 2 + 4)
(𝑥 2 + 4)(𝑥 6 − 9)
(𝑥 2 + 4)(𝑥 3 + 3)(𝑥 3 − 3)
11. State all zeros of the function.
𝑓(𝑥) = 3𝑥 6 + 9𝑥 4 − 12𝑥 2
𝑓(𝑥) = 3𝑥 2 (𝑥 4 + 3𝑥 2 − 4)
𝑓(𝑥) = 3𝑥 2 (𝑥 2 + 4)(𝑥 2 − 1)
𝑓(𝑥) = 3𝑥 2 (𝑥 2 + 4)(𝑥 + 1)(𝑥 − 1)
Zeros: 0, ±2𝑖, ±1
12. Write the equation of the inverse function.
𝑓(𝑥) = 𝑥 5 − 2
𝑥 = 𝑦5 − 2
𝑦5 = 𝑥 + 2
5
𝑦 = √𝑥 + 2
13. Tell whether the function represents growth or decay. Then state the rate of change.
𝑦 = 16(1.3)𝑥
This function represents 30% growth.
14. Refer to this equation: 𝑓(𝑥) = 𝑎|𝑥 − ℎ| + 𝑘
Choose values for 𝑎, ℎ, and 𝑘 which will cause the graph of the function to have a
maximum output which occurs in Quadrant II.
For the function to have a maximum, 𝑎 must be negative.
For the vertex to be in Q2, ℎ must be negative and 𝑘 must be positive.
15. Write the equation of the function
shown on the graph.
16. Use your calculator to
write the equation of a
best-fit line for the data.
Round off your slope and
y-intercept to the nearest
tenth.
𝑦 = 7.7𝑥 + 24.9
x
2
4
6
8
10
12
14
16
18
20
𝑦 = (𝑥 − 3)2 − 2
17. Compare and contrast the graphs of each pair of function types.
a. linear and absolute value
Absolute value graphs have two linear pieces which form a “V” shape.
b. quadratic and absolute value
Both types have both increasing and decreasing portions, but quadratic graphs are
parabolas while absolute value graphs form a “V”.
c. quadratic and exponential
Both of these are curved rather than straight, but exponential curves only go up or
down but not both.
d. exponential and logarithm
Both of these have asymptotes, but they are horizontal asymptotes on exponential
graphs and vertical asymptotes on logarithm graphs.
e. logarithm and square root
The parent graphs of both of these curve up and to the right, concave down. But
the logarithm has an asymptote on the left, while the square root has an endpoint
on the left.
y
40
56
71
87
102
117
133
148
164
179