Algebra 2 Final (Units 4–6)

#	Answer	Lesson Origin
1A) 1B) 1C)	$\begin{matrix} \log 6^{2 x - 5} = \log 25 & Exponents to common \log s \\ (2 x - 5) \log 6 = \log 25 & Power rule \\ \frac{(2 x - 5) \log 6}{\log 6} = \frac{\log 25}{\log 6} & Isolate x \\ 2 x - 5 = \frac{\log 25}{\log 6} \\ + 5 = + 5 \\ (\frac{1}{2}) 2 x = (\frac{\log 25}{\log 6} + 5) (\frac{1}{2}) \\ x = \frac{\log 25}{2 \log 6} + \frac{5}{2} \\ x = 3.3982 \end{matrix}$	40, 41, 43
2A)	5 possible roots	35
2B)		35
2C)	Relative minimum: $(0.49, - 2.35)$ Relative maximum: $(1.95, 6.03)$ Real zeros: $x = - 1, 0, 1, 3 (multiplicity 2)$	35
2D)	The graph f(x) increases across the intervals: $(- \infty, - 0.63) and (0.49, 1.95) and (3, + \infty)$ The graph f(x) decreases across the intervals: $(- 0.63, 0.49) and (1.95, 3)$	35
3)	D	42
4)	A	36
5)	A	31
6)	B	39
7)	D	32
8)	C	39, 40, 42, 44
9)	B	31
10)	A	33
11)	C	38
12)	D	39
13)	B	36
14)	C	33
15)	A	38
16)	C	42
17)	B	33
18)	D	40
19)	B	37
20)	C	39, 40
21)	B	43
22)	A	38, 44
23)	C	32
24)	D	6, 15, 34
25)	C	37
Unit 6 Solutions
26)	C	46
27)	D	47
28)	C	51, 52
29)	A	55
30)	C	48
31)	B	45
32)	C	53, 54
33)	D	56
34)	B	49
35)	C	50

Answer all parts of the open response problem.

Solve: $6^{2 x - 5} = 25$

Write your answer as a logarithm.
Using technology, approximate the value of the logarithm to four decimal places.
Name the properties you used to solve.

$\begin{matrix} \log 6^{2 x - 5} = \log 25 & Exponents to common logs \\ (2 x - 5) \log 6 = \log 25 & Power rule \\ \frac{(2 x - 5) \log 6}{\log 6} = \frac{\log 25}{\log 6} & Isolate x \\ 2 x - 5 = \frac{\log 25}{\log 6} \\ + 5 = + 5 \\ (\frac{1}{2}) 2 x = (\frac{\log 25}{\log 6} + 5) (\frac{1}{2}) \\ x = \frac{\log 25}{2 \log 6} + \frac{5}{2} \\ x = 3.3982 \end{matrix}$

Answer all parts of the open response problem.

Complete the problem using the polynomial function: $f (x) = x {(x - 3)}^{2} (x^{2} - 1)$

Determine the number of roots with the Fundamental Theorem of Algebra (FTA).

$FTA : 1 + 2 + 2 = 5$

According to the FTA, a total of 5 roots are possible.

Sketch a graph of the function using technology. Include roots and turning points rounded to the hundredth place.

Name the following points:

Relative minimum on the interval $[0, 2]$
$(0.49, - 2.35)$

Relative maximum on the interval $[0, 2]$
$(1.95, 6.03)$

Real zeros
$x = - 1, 0, 1, 3 (multiplicity 2)$

Name the intervals in which the function is increasing and decreasing.

The graph f(x) increases across the intervals:
$(- \infty, - 0.63) and (0.49, 1.95) and (3, + \infty)$

The graph f(x) decreases across the intervals:
$(- 0.63, 0.49) and (1.95, 3)$

Note

Remember that when using interval notation, you are representing the range of x-values.

Multiple Choice

Write $y = \ln 5$ as an exponential function with base e.

$e = \ln 5$
$e^{5} = y$
$e^{5} = \ln y$
$e^{y} = 5$

Note

This option incorrectly switches y for e. (e does not equal ln 5.)

B, C) The foundational property was not correctly applied.

Foundational property: $\ln (e^{x}) = x and e^{\ln x} = x \ln 5 = y e^{y} = 5$

The volume of a cylinder varies jointly with π, the radius squared, and the height. Solve for the height in terms of π when the radius is 2 units and the volume is 12 cubic units.

$\frac{3}{π}$ units
$\frac{6}{π}$ units
$3 π$ units
$48 π$ units

Note

This option is the result when r is not squared.
This option is the result when π is incorrectly moved to the numerator.
This option is the result if you solve for V by making h equal 12.

$V = 12, r = 2 \begin{matrix} V = π r^{2} h \\ 12 = π {(2)}^{2} h \\ \frac{12}{4 π} = \frac{4 π h}{4 π} \\ h = \frac{3}{π} \end{matrix}$

Given $g (x) = 3 x^{2} + 5 x$ , determine $g (x^{2} - 1) .$

$3 x^{4} - x^{2} - 2$
$8 x^{2} - 8$
$3 x^{4} + 5 x^{2} - 5$
$2 x^{2} - 2$

Note

This option occurs when the binomial is not squared for the first term.
This option occurs when the binomial $x^{2} - 1$ is not correctly squared.
This option occurs when the terms are incorrectly combined.

$\begin{array}{rcl} g (x^{2} - 1) & = & 3 {(x^{2} - 1)}^{2} + 5 (x^{2} - 1) \\ = & 3 (x^{4} - 2 x^{2} + 1) + 5 x^{2} - 5 \\ = & 3 x^{4} - 6 x^{2} + 3 + 5 x^{2} - 5 \\ = & 3 x^{4} - x^{2} - 2 \end{array}$

If $\log_{b} (\sqrt[3]{25}) = \frac{2}{3},$ find x for $\log_{b} 625 = x .$

$\frac{2}{3}$
4
5
25

Note

This option is the value of x in the first equation.
This option is the value of b.
This option ignores the value of b and takes the square root of 625.

$\begin{matrix} \log_{b} (\sqrt[3]{25}) = \frac{2}{3} & \log_{b} (625) = x \\ b^{\frac{2}{3}} = \sqrt[3]{25} & \log_{5} (625) = x \\ b^{\frac{2}{3}} = \sqrt[3]{5^{2}} & 5^{x} = 625 \\ b^{\frac{2}{3}} = 5^{\frac{2}{3}} & 5^{x} = 5^{4} \\ b = 5 & x = 4 \end{matrix}$

Select the functions that make the composition: $[j \circ k] (x) = \sqrt{2 x - 3}$

$j (x) = 2 x - 3 k (x) = \sqrt{x}$
$j (x) = \sqrt{2 x} k (x) = - \sqrt{3}$
$j (x) = \sqrt{2 x} k (x) = x - 3$
$j (x) = \sqrt{x} k (x) = 2 x - 3$

Note

This option is $[k \circ j] (x)$ .

B, C) Functions with unlike radicands cannot be combined.

$j (x) = \sqrt{x} k (x) = 2 x - 3 \begin{matrix} \begin{array}{rcl} [j \circ k] (x) & = & \sqrt{(2 x - 3)} \\ = & \sqrt{2 x - 3} \end{array} \end{matrix}$

Customer First Credit Union ran a promotion 20 years ago for a savings account that earned continuously compounded interest using the formula: $y = P e^{r t} .$ Jamie invested $8,000 and now has $28,000 in the account. What interest rate did Jamie earn? Round to the hundredth of a percent.

2.72%
0.06%
6.26%
6.3%

Note

This option is an approximate value of e.
This option did not convert the decimal to a percent.
This option is not rounded correctly.

$\begin{array}{rcl} y & = & P e^{r t} \\ 28000 & = & 8000 e^{r (20)} \\ 3.5 & = & e^{20 r} \\ \ln 3.5 & = & \ln e^{20 r} \\ \ln 3.5 & = & 20 r \\ r & = & \frac{\ln 3.5}{20} = 0.06264 \\ r & = & 6.26 % \end{array}$

Determine the domain for the combination of functions $(\frac{h}{g}) (x)$ where $g (x) = 4 x^{2} - 9$ and $h (x) = 2 x + 3 .$

${x | x \in ℝ, x \neq \frac{3}{2}}$
${x | x \in ℝ, x \neq \pm \frac{3}{2}}$
${x | x \in ℝ}$
${x | x \in ℝ, x \neq 0}$

Note

This domain is missing the value that simplifies out of the problem.
This option is the domain of each function before they are combined.
The denominator cannot equal zero, but zero is not an excluded value for this combination of functions.

$\begin{matrix} \frac{h (x)}{g (x)} = \frac{2 x + 3}{4 x^{2} - 9} & {Domain}_{g} : \{x | x \in ℝ\} \\ = \frac{2 x + 3}{(2 x + 3) (2 x - 3)} & {Domain}_{h} : \{x | x \in ℝ\} \\ = \frac{1}{2 x - 3} & {Domain}_{\frac{h}{g}} : \{x | x \in ℝ, x \neq \pm \frac{3}{2}\} \end{matrix}$

Sketch the zeros of the function: $f (x) = {(x - 1)}^{3} (x + 2) (x + 1)$

Note

This option is the function when $a = - 1 .$
This option ignores the cubed binomial and only has cross-throughs.
This option is the result when $x = 2$ (instead of $x = - 2) .$

$0 = {(x - 1)}^{3} (x + 2) (x + 1) x = 1 (snake) x = - 2 (cross through) x = - 1 (cross through)$

Solve: $49^{2 x - 5} = 343^{x + 2}$

7
11
16
no solution

Note

This option is the solution if the exponents of the bases were ignored.
This option is the solution if the distributive property is applied incorrectly on the left side of the equation.

$\begin{matrix} \begin{array}{rcl} 49 & = & 7^{2} 343 = 7^{3} \end{array} \end{matrix} \begin{matrix} 7^{2 (2 x - 5)} = 7^{3 (x + 2)} \\ 2 (2 x - 5) = 3 (x + 2) \\ 4 x - 10 = 3 x + 6 \\ x = 16 \end{matrix}$

Write the equation $2^{6} = 64$ in logarithmic form.

$\log_{2} 6 = 64$
$\log_{6} 64 = 2$
$\log_{64} 2 = 6$
$\log_{2} 64 = 6$

Note

A, B, C) These options incorrectly convert from an exponential equation to a logarithmic equation.

If the distance varies directly as time in hours and $d = 100$ and $t = 2$ , determine the time when the distance is 425 miles.

0.11 hours
8.5 hours
2.13 hours
850 hours

Note

This option is the result when k is divided by 425 (the fraction is flipped).
This option is the result if 100 and 2 were multiplied, rather than divided, to determine the value of k.
This option is the result if 2 was divided by 100 to determine the value of k.

$\begin{array}{rcl} \begin{matrix} d = k t \\ \frac{100}{2} = \frac{k (2)}{2} & \frac{425}{50} = \frac{50 t}{50} \\ k = 50 & t = 8.5 \end{matrix} \end{array}$

Which function is both a power and a polynomial function?

$y = \frac{x^{3} - 5}{6}$
$y = - \frac{1}{2} | x + 12 |$
$y = 3 x^{2}$
$y = x^{\frac{3}{2}}$

Power function: a monomial function with real number exponents

Polynomial function: a function with real number coefficients, excluding zero, and whole number exponents

Note

This option is not a power function because it is not a monomial.
This option is neither a power nor a polynomial function.
Polynomial functions must have whole-number exponents.

Solve: ${(\frac{27}{8})}^{x + 1} \leq {(\frac{16}{81})}^{2 x}$

$\begin{array}{rcl} x & \leq & - \frac{3}{11} \end{array}$
$\begin{array}{rcl} x & \geq & - \frac{3}{11} \end{array}$
$x \geq \frac{3}{5}$
no solution

Note

The inequality symbol should be is less than or equal to.
This option is the solution when 4, instead of $- 4,$ is used to solve.

$\frac{27}{8} = {(\frac{3}{2})}^{3} \frac{16}{81} = {(\frac{2}{3})}^{4} = {(\frac{3}{2})}^{- 4} \begin{matrix} {(\frac{3}{2})}^{3 (x + 1)} \leq {(\frac{3}{2})}^{- 4 (2 x)} \\ 3 (x + 1) \leq - 4 (2 x) \\ 3 x + 3 \leq - 8 x \\ 11 x \leq - 3 \\ x \leq - \frac{3}{11} \end{matrix}$

Solve $\ln (x - 1) - 2 \ln 5 = 3$ in terms of e.

$10 e^{3} + 1$
$75 e + 1$
$25 e^{3} + 1$
$e^{76}$

Note

This option multiplies five by two instead of squaring it.
This option multiplies 25 and 3, but an argument and a constant cannot be combined.
This option combines all of the numbers in the argument and the constant.

$\begin{array}{rcl} \ln (x - 1) - \ln 5^{2} & = & 3 \\ \ln (x - 1) - \ln 25 & = & 3 \\ \ln \frac{x - 1}{25} & = & 3 \\ e^{3} & = & \frac{x - 1}{25} \\ 25 e^{3} & = & x - 1 \\ x & = & 25 e^{3} + 1 \end{array}$

Name the end behavior for the polynomial function.

$x \to - \infty, f (x) \to + \infty, and x \to + \infty, f (x) \to - \infty$
$x \to - \infty, f (x) \to - \infty, and x \to + \infty, f (x) \to + \infty$
$x \to \pm \infty, f (x) \to + \infty$
$x \to \pm \infty, f (x) \to - \infty$

odd

Note

This option is the end behavior when the graph starts positive and ends negative.

C, D) These options represent the end behavior for an even function.

Completely expand the expression $\log \frac{3 \sqrt{x}}{z^{4}}$ with the logarithmic properties.

$\frac{1}{2} (\log 3 + \log x) - 4 \log z$
$\log 3 + 2 \log - 4 \log z$
$\log 3 + \frac{1}{2} \log - 4 \log z$
$\log 3 + \frac{1}{2} \log x - 4 \log z$

Note

This option would be the answer for the expression $\sqrt{3 x} .$
This option incorrectly used x-squared instead of the square root of x.
The variable x is missing from the second expression.

$\log \frac{3 x^{\frac{1}{2}}}{z^{4}} \log 3 x^{\frac{1}{2}} - \log z^{4} \log 3 + \log x^{\frac{1}{2}} - 4 \log z \log 3 + \frac{1}{2} \log - 4 \log z$

Describe the transformation from f(x) to g(x) when $f (x) = 6^{x}$ and $g (x) = - {(6)}^{(x + 2)} + 7 .$

g(x) is reflected across the y-axis, translated 7 units up, and 2 units left.
g(x) is reflected across the x-axis, translated 7 units up, and 2 units left.
g(x) is reflected across the y-axis, translated 7 units up, and 2 units right.
g(x) is reflected across the x-axis, translated 7 units up, and 2 units right.

	f(x)	g(x)
b	6	6
a	1	–1
h	0	–2
k	0	7

Note

A, C) The graph is not reflected over the y-axis.
C, D) The graph is translated left, not right, two units.

Solve: $\log_{2} x + \log_{2} (x - 4) = 5 \log_{2} 2$

$x = - 8, 4$
$x = - 4, 8$
$x = 8$
$x = 4$

Note

A, D) These options would be solutions for the equation $x^{2} + 4 x - 32 = 0 .$

A, B) It is not possible to have a negative solution because the argument of a log cannot be negative.

$\begin{array}{rcl} \log_{2} (x) (x - 4) & = & 5 (1) \\ \log_{2} (x^{2} - 4 x) & = & 5 \\ 2^{5} & = & x^{2} - 4 x \\ 32 & = & x^{2} - 4 x \\ 0 & = & x^{2} - 4 x - 32 \\ 0 & = & (x + 4) (x - 8) \\ x & = & - 4, 8 \end{array}$

The equation, $y = \log_{2} (x),$ is transformed to the given graph. Select the equation that represents the graph.

$y = \log_{2} (x) - 3$
$y = \log_{2} (x + 1) - 3$
$y = \log_{2} (x - 1) - 2$
$y = \log_{2} (x - 7)$

The given graph shifts the x-intercept of $y = \log_{2} (x)$ left one unit and down three units.

The graph has values of $h = - 1, k = - 3 .$

Note

This option does not demonstrate the horizontal shift on the graph.
This option would shift the graph right one and down two units.
This option would shift the graph right seven units.

Solve the problem using the formula: $A (t) = A_{0} {(\frac{1}{2})}^{\frac{t}{h}}$
A drink contains 200 milligrams (mg) of caffeine, and the half-life of caffeine, h, is 6 hours. How much caffeine, to the nearest whole milligram, would be in a person’s system after 7 hours?

89 mg
100 mg
110 mg
200 mg

$A_{0} = 200, h = 6, t = 7 A (t) = 200 {(\frac{1}{2})}^{\frac{7}{6}} A (t) = 89.0899$

Note

This option is half of the amount of caffeine and ignores the time.
This option is the solution when the exponent is $\frac{6}{7}$ .
This option is the initial amount of caffeine in the problem.

Find the composite function $[f \circ g] (x)$ when $f (x) = 4 x^{2}$ and $g (x) = \frac{1}{x + 1} .$

$\frac{4 x^{2}}{x + 1}, \{x | x \in ℝ, x \neq - 1\}$
$\frac{1}{4 x^{2} + 1}, \{x | x \in ℝ, x \neq \pm 0.5 i\}$
$\frac{4}{x^{2} + 2 x + 1}, \{x | x \in ℝ, x \neq - 1\}$
$\frac{4}{x^{2} + 2 x + 1}, \{x | x \in ℝ\}$

Note

This option is the combination $(f g) (x)$ .
This option is $[g \circ f] (x)$ .
This option is missing a domain restriction.

$\begin{matrix} [f \circ g] (x) = f [g (x)] \\ = 4 {(\frac{1}{x + 1})}^{2} & {Domain}_{f} : \{x | x \in ℝ\} \\ = 4 (\frac{1}{x^{2} + 2 x + 1}) & {Domain}_{g} : \{x | x \in ℝ, x \neq - 1\} \\ = \frac{4}{x^{2} + 2 x + 1} & {Domain}_{f \circ g} : \{x | x \in ℝ, x \neq - 1\} \end{matrix}$

Determine all roots of the function: $h (x) = x^{5} - 6 x^{3} + 6 x^{2} - 7 x + 6$

$x = \pm 1, \pm 2, \pm 3, \pm 6$
$x = - 3, 1, 2, i$
$x = - 3, 1, 2$
$x = - 3, 1, 2, \pm i$

Note

This option includes the possible rational roots.
Imaginary numbers must come in conjugate pairs.
This option does not include complex roots.

$RRT = \frac{\pm 1, \pm 2, \pm 3, \pm 6}{\pm 1}$

$x^{2} + 1 = 0 x^{2} = - 1 x = \pm i$

Select the graph of f(x) that transforms $w (x) = 2^{x}$ by reflecting the function, and shifting the function three units right and four units up.

$y = a {(b)}^{x - h} + k w (x) = 2^{x} a = - 1, h = 3, k = 4 w (x) = - {(2)}^{x - 3} + 4$

Note

The graph is not reflected across the x-axis.
This option is the transformation if h is –3.
This option is the transformation if k is –4.

Unit 6 Multiple Choice

What is the area in a standard normal distribution that falls between $z = - 1$ and $z = 2 ?$

1.00
0.95
0.82
0.68

$z = - 1 z ‐ table : 0.1587 z = 2 z ‐ table : 0.9772 0.9772 - 0.1587 = 0.8185$

Note

All data (100%) is not between the given z-values.
This option is the data within two standard deviations of the mean.
This option is the data within one standard deviation of the mean.

Which type of variable will provide categorical data?

Height in inches
Weight in pounds
Age in years
Brand of car

Categorical data: characteristics or categories

Note

A, B, C) These options are all quantitative data.

Calculate $n C r (8, 3)$ using Pascal’s Triangle.

Note

This option is the row of Pascal’s Triangle to be referenced, not the calculation.
This option is $n C r (8, 2) .$
This option is $n P r (8, 3) .$

8th row of Pascal’s Triangle: 1, 8, 28, 56, 70, 56, 28, 8, 1

$n C r (8, 3) = \frac{8!}{3! (8 - 3)!} = \frac{8 \cdot 7 \cdot 6 \cdot 5!}{(3 \cdot 2) 5!} = 56$

The fair spinner has four equal sections. What is the probability of the arrow landing on blue twice in two spins?

$\frac{1}{16}$
$\frac{1}{8}$
$\frac{1}{4}$
$\frac{1}{2}$

$P (blue, blue) = \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16}$

Note

This option is not possible with the given information.
This option is the probability of landing on blue once.
This option is the result when the probabilities are added.

Ms. Malone plans to survey her statistics class. She puts all of her students’ names in a hat and draws 30 random names. Name the sampling method she used.

Cluster
Stratified
Simple random
Systematic

Simple random: Any individual in the population has an equal chance of being selected at random from the population.

Note

The scenario does not reflect the other types of sampling.

In a normal distribution, what is the approximate percentage of data that lies within one standard deviation of the mean?

50%
68%
95%
99.7%

Note

This option is the percentage of data below the mean.
This option is the percentage of data within two standard deviations of the mean.
This option is the percentage of data within three standard deviations of the mean.

A set contains the numbers 1 through 20.
Determine P(prime or even).

0
$\frac{1}{20}$
$\frac{17}{20}$
$\frac{9}{10}$

$Prime : \{2, 3, 5, 7, 11, 13, 17, 19\} Even : \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20\} P (prime or even) = \frac{8}{20} + \frac{10}{20} - \frac{1}{20} = \frac{17}{20}$

Note

This option is the probability of selecting a number not contained in the set.
This option is the probability of a number that is prime AND even.
This option is the probability when an even prime number is counted twice.

A bag contains seven red and five yellow marbles. What is the probability of selecting three yellow marbles in a row without replacement?

$\frac{125}{1 728}$
$\frac{25}{264}$
$\frac{5}{144}$
$\frac{1}{22}$

$P (Y, Y, Y) = \frac{5}{12} \cdot \frac{4}{11} \cdot \frac{3}{10} = \frac{1}{22}$

Note

This option is the probability with replacement.
This option did not decrease the numerators by one.
This option did not decrease the denominators by one.

Of the 123 patients surveyed about their doctor’s office, 14.7% reported a ‘neutral’ experience. Determine the interval for neutral ratings when the margin of error is ±9.8%.

Between 14.7% and 24.5% of patients reported a ‘neutral’ experience.
Between 4.9% and 24.5% of patients reported a ‘neutral’ experience.
Between 113.2% and 132.8% of patients reported a ‘neutral’ experience.
Cannot be determined.

$14.7 - 9.8 = 4.9 14.7 + 9.8 = 24.5$

Note

This option is the upper half of the confidence interval.
This option used the number of patients rather than the neutral percentage.

A research organization conducted an experiment with 500 participants to reduce distracted driving. The participants were divided randomly into an experiment group and a control group.

Group Mean Score

Control,

$n = 250$
86

Treatment,

$n = 250$
80

A simulation with 10,000 trials was conducted with a standard deviation of 2.5. Calculate the z-score at the 95% confidence level for the observed difference.

–2.0
1.96
2.40
3.06

$Observed difference : 86 - 80 = 6 z = \frac{6}{2.5} = 2.4$

Note

This z-score is positive for the observed difference.
This option is the z-score at the 95% confidence level.
This option is the result when 1.96 is used instead of 2.5.