1 INTERMEDIATE ALGEBRA READINESS DIAGNOSTIC TEST PRACTICE Directions: Study the examples, work the problems, then check your answers at the end of each topic. If you don’t get the answer given, check your work and look for mistakes. If you have trouble, ask a math teacher or someone else who understands this topic. TOPIC 1: ELEMENTARY OPERATIONS A. Algebraic operations, grouping, evaluation : To evaluate an expression, first calculate the powers, then multiply and divide in order from left to right, and finally add and subtract in order from left to right. Parentheses have preference. example: 14 - 3 2 = 14 - 9 = 5 example: 2 • 4 + 3 • 5 = 8 + 15 = 23 example: 10 - 2 • 3 2 = 10 - 2 • 9 = 10 - 18 = -8 example: (10 - 2) • 3 2 = 8 • 9 = 72 Problems 1-7: Find the value: 1. 2 3 = 5. 0 4 = 2. -2 4 = 6. (-2) 4 = 3. 4 + 2 • 5 = 7. 1 5 = 4. 3 2 - 2 • 3 + 1 = Problems 8-13: Find the value if a = -3 , b = 2 , c = 0, d = 1 , and e = -3 : 8. a - e = 11. e d + b a - 2d e = 9. e 2 + (d - ab)c = 12. b e = 10. a - (bc - d) + e = 13. d c = Combine like terms when possible: example: 3x + y 2 - ( x + 2 y 2 ) = 3x - x + y 2 - 2 y 2 = 2 x - y 2 example: a - a 2 + a = 2a - a 2 Problems 14-20: Simplify: 14. 6 x + 3 - x - 7 = 18. 3a - 2 4( a - 2b) - 3a ( = 15. 2(3 - t ) = 19. 3 ( a + b) - 2( a - b) = 16. 10r - 5(2r - 3y) = 20. 1 + x - 2 x 3x - 4 x = 17. x 2 - ( x - x 2 ) = B. Simplifying fractional expressions : example: 27 36 = 9•3 9•4 = 9 9 • 3 4 = 1• 3 4 = 3 4 (note that you must be able to find a common factor - in this case 9 - in both the top and bottom in order to reduce a fraction.) example: 3a 12ab = 3a •1 3a •4 b = 3a 3a • 1 4 b = 1 • 1 4 b = 1 4 b (common factor: 3a) Problems 21-32: Reduce: 21. 13 52 = 22. 26 65 = 23. 3 6 3 + 9 = 28. x - 4 4 - x = 24. 6axy 15by = 29. 2( x 4)( x - 5) ( x - 5)( x - 4) = 25. 19a 2 95a = 30. x 2 - 9x x- 9 = 26. 14 x- 7 y 7 y = 31. 8( x -1) 2 6( x 2 -1) = 27. 5a b 5a+ c = 32. 2x 2 - x-1 x 2 - 2x 1 = example: 3 x • y 15 • 10x y 2 = 3 •10•x •y 15•x •y 2 = 3 3 • 5 5 • 2 1 • x x • y y • 1 y = 1 • 1• 2 • 1• 1 • 1 y = 2 y Problems 33-34: Simplify: 33. 4 x 6 • xy y 2 • 3y 2 = 34. x 2 - 3x x - 4 • xx - 4 ( 2x- 6 = C. Laws of integer exponents : I. a b • a c = a b c II. a b a c = a b - c III. (a b ) c = a bc IV. ab ( c = a c • b c V. a b ( c = a c b c VI. a 0 = 1 (if a ≠ 0 ) VII. a -b = 1 a b Problems 35-44: Find x: 35. 2 3 • 2 4 = 2 x 40. 8 = 2 x 36. 2 3 2 4 = 2 x 41. a x = a 3 • a 37. 3 -4 = 1 3 x 42. b 10 b 5 = b x 38. 5 2 5 2 = 5 x 43. 1 c -4 = c x 39. 2 4 ( 3 = 2 x 44. a 3y-2 a 2y-3 = a x Problems 45-59: Simplify: 45. 8 x 0 = 50. -3 ( 3 - 3 3 = 46. 3 -4 = 51. 2 x • 4 x -1 = 47. 2 3 • 2 4 = 52. 2 c+ 3 2 c-3 = 48. 0 5 = 53. 2 c 3 • 2 c - 3 = 49. 5 0 = 54. 8 x 2 x-1 =
18
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1 INTERMEDIATE ALGEBRA READINESS DIAGNOSTIC TEST PRACTICE
Directions: Study the examples, work the problems, then check your answers at the end of each topic. If
you don’t get the answer given, check your work and look for mistakes. If you have trouble, ask a math
teacher or someone else who understands this topic.
TOPIC 1: ELEMENTARY OPERATIONS
A. Algebraic operations, grouping, evaluation:
To evaluate an expression, first calculate the
powers, then multiply and divide in order from
left to right, and finally add and subtract in order
2. In what quadrant does the point a, b( ) lie, if a > 0 and b < 0?
Problems 3-6: For each given point, which of its
coordinates, x or y, is larger?
B. Distance between points: The distance between the points P1 x1, y1( ) and P2 x2, y2( ) is found by using the Pythagorean Theorem, which gives
P1P2 = x2 − x1( )2 + y2 − y1( )2 .
example: A(3, -1), B(-2, 4)
AB = 4 − −1( )( )2 + −2 − 3( )2 =
52 + −5( )2 = 50 = 25 2 = 5 2
Problems 7-10: Find the length of the segment
joining the given points:
7. (4, 0), (0, -3) 9. (2, -4), (0, 1)
8. (-1, 2), (-1, 5) 10. − 3, − 5( ), 3 3, − 6( ) C. Linear equations in two variables, slope,
intercepts, and graphing: The line joining the points P1 x1, y1( ) and P2 x2, y2( ) has slope y2−y1
x2−x1.
example: A(3,-1), B(-2,4) slope of
AB____
= 4− −1( )−2−3 = 5
−5 = −1
Problems 11-15: Find the slope of the line
joining the given points:
11. (-3, 1), and (-1, -4) 13. (3, -1), and (5, -1)
12. (0, 2), and (-3, 5)
6
3
4
5
-2 30
-2 0
0 1
12 14. 15.
To find the x-intercept (x-axis crossing) of an
equation, let y be zero and solve for x. For the y-
intercept, let x be zero and solve for y.
example: 3y − 4x = 12 if x = 0 , y = 4 so
y-intercept is 4. If y = 0 , x = −3 so x-intercept is –3.
The graph of y = mx + b is a line with slope m and
y-intercept b. To draw the graph, find one point on
it (such as (0, b)) and then use the slope to find
another point. Draw the line joining the two.
example: y = −32x + 5 has slope − 3
2 and y-
intercept 5. To graph the line, locate (0, 5).
From that point, go down 3 (top of slope
fraction), and over (right) 2 (bottom of
fraction) to find a second point. Join.
Problems 16-20: Find the slope and y-intercept,
and sketch the graph:
16. y = x + 4 19. x − y = −1 17. y = − 1
2x − 3 20. x = −3y + 2
18. 2y = 4x − 8
To find an equation of a non-vertical line, it is
necessary to know its slope and one of its points.
Write the slope of the line through x, y( ) and the known point, then write an equation which
says that this slope equals the known slope.
example: Find an equation of the line through
(-4, 1) and (-2, 0).
Slope = 1−0−4+2 = 1−2
Using (-2, 0) and x, y( ), Slope =
y−0x+2 = 1
−2 ; cross multiply, get
−2y = x + 2, or y = − 12x −1
Problems 21-25: Find an equation of the line:
21. Through (-3, 1) and (-1, -4)
22. Through (0, -2) and (-3, -5)
23. Through (3, -1) and (5, -1)
24. Through (8, 0), with slope –1
25. Through (0, -5), with slope 2 3
A vertical line has no slope, and its equation can
be written so it looks like x = k (where k is a number). A horizontal line has zero slope, and
its equation looks like y = k .
example: Graph on the same graph:
x + 3 = −1 and 1+ y = −3. The first equation is x = −4 ; the second is y = −4 .
Problems 26-27: Graph and write equation for:
26. The line through (-1, 4) and (-1, 2)
27. Horizontal line through (4, -1)
D. Linear inequalities in two variables:
example: Two variable graph: graph solution on
a number plane: x − y > 3 (This is an abbreviation for { x, y( ): x − y > 3}. Subtract x, multiply by –1, get y < x − 3 . Graph y = x − 3 , but draw a dotted
line, and shade the
side where y < x − 3 :
28. y < 3 31. x < y +1 29. y > x 32. x + y < 3 30. y ≥ 2
3x + 2 33. 2x − y >1
E. Graphing quadratic equations:
The graph of y = ax2 + bx + c is a parabola, opening
upward (if a > 0) or downward (if a < 0), and with line of symmetry. x = −b
2a, also called axis of
symmetry. To find the vertex V(h, k) of the parabola,
h = −b2a (since V is on the axis of symmetry), and k is
the value of y when h is substituted for x.
example: y = x2 −6x
a =1, b = −6, c = 0
Axis:
x = −b2a
= 62
= 3h = 3, k = 32 −18 = −9
Thus, vertex is (3, − 9)
13
example: y = 3 − x2 V(0, 3),
Axis: x = 0 , y-intercept: if x = 0 , y = 3 −02 = 3 x-intercept: if y = 0 , 0 = 3− x
2,
so 3 = x2, and x = ± 3
Problems 34-40: Sketch the graph:
34. y = x2 38. y = x +1( )2
35. y = −x2 39. y = x − 2( )2 −1 36. y = x
2 +1 40. y = x + 2( ) x −1( ) 37. y = x
2 − 3
Answers:
1.
2. IV
3. x 4. y
5. y
6. x 7. 5
8. 3
9. 29 10. 7
11. − 52
12. -1
13. 0
14. − 35
15. 3 4
16. m =1, b = 4
17. m = − 12 , b = −3
18. m = 2, b = −4
19. m =1, b =1
20. m = − 13 , b = 2
3
21. y = − 52 x −13
2
22. y = x − 2 23. y = −1 24. y = −x + 8 25. y = 2
3 x −5 26. x = −1
27. y = −1
28.
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
40.
TOPIC 7: LOGARITHMS and FUNCTIONS
A. Functions:
The area A of a square depends on its side
length s, and we say A is a function of s, and
write ‘A= f s( )’; for short, we read this as ‘A= f of s.’ There are many functions of s.
The one here is s2. We write this f s( ) = s
2 and
can translate: ‘the function of s we’re talking
about is s2’. Sometimes we write A s( ) = s
2.
This says the area A is a function of s, and
specifically, it is s2.
B. Function values and substitution:
If A s( ) = s2, A(3), read ‘A of 3’, means replace
every s in A s( ) = s2with 3, and find A when s is 3.
When we do this, we find A 3( )= 32 = 9 .
example: g x( ) is given: y = g x( )= πx2
4
-4
1
-1
-1
1
-1
(2,-1)
-2 1
-3
2
3
14
example: g(3) = π • 32 = 9π example: g 7( )= π • 72 = 49π example: g a( )= πa2 example: g x + h( )= π x + h( )2 = πx 2 +2πxh + πh 2
1. Given y = f x( )= 3x − 2 . Complete these ordered pairs: (3,___), (0,___),
( 12 ,___), (___, 10), (___,-1) ( x −1,___)
Problems 2-10: Given f x( ) = x2 − 4x + 2 . Find:
2. f 0( )=3. f 1( )=4. f −1( )=5. f −x( )=6. − f x( )=
7. f x( )− 2 =8. f x − 2( )=9. 2 f x( )=10. f 2x( )=
Problems 11-15: Given f x( )= xx+1 . Find:
11. f 1( )=12. f −2( )=13. f 0( )=
14. f −1( )=15. f x −1( )=
example: If k x( ) = x2 − 4x , for what x is
k x( ) = 0? If k x( ) = 0 , then x 2 − 4x = 0 and since x2 − 4x = x x − 4( ) = 0 , x can be either 0 or 4.
(These values of x: 0 and 4, are called ‘zeros of the
function’, because each makes the function zero.)
Problems 16-19: Find all real zeros of:
16. x x +1( )17. 2x2 − x − 3
18. x 2 −16x + 6419. x 2 + x + 2
Problems 20-23: Given f x( ) = x2 − 4x + 2 ,
find real x so that:
20. f x( )= −221. f x( )= 2
22. f x( )= −323. x is a zero of f x( )
Since y = f x( ), the values of y are the values of the function which correspond to specific values of x. The heights of the graph above (or below) the x-axis are the values of y and so also of the function. Thus for this graph, f 3( ) is the height (value) of the function at x = 3 and the value is 2: At x = −3 , the value (height) of f x( ) is zero; in other words, f −3( ) = 0 . Note that f 3( ) > f −3( ) , since 2 > 0 , and that f 0( ) < f −1( ), since f −1( )=1 and f 0( ) <1.
Problems 24-28: For this
graph, tell whether the
statement is true or false:
24. g −1( )= g 0( )25. g 0( )= g 3( )26. g 1( )> g −1( )
27. g −2( )> g 1( )28. g 2( )< g 0( )< g 4( )
C. Logarithms and exponents:
Exponential form: 23 = 8
Logarithmic form: log28 = 3
Both of the equations above say the same thing. ‘ log28 = 3’ is read ‘log base two of eight equals three’ and translates ‘the power of 2 which gives 8 is 3’.
Problems 29-32: Write the following information
in both exponential and logarithmic forms:
29. The power of 3 which gives 9 is 2.
30. The power of x which gives x3 is 3
31. 10 to the power –2 is 1100
.
32. 12 is the power of 169 which gives 13.
Problems 33-38: Write in logarithmic form:
33. 43 = 64 36. 1
10=10−1
34. 30 =1 37. a
b = c
35. 25 = 52 38. y = 3x
Problems 39-44: Write in exponential form:
39. log39 = 240. log31= 041. 5 = log2 32
42. 1= log4 443. y = loga x44. logb a = 2
Problems 45-50: Find the value:
45. 210 = 48. 610
310=
46. log4 410 = 49. log49 7 =
47. log6 6 = 50. log7 49 =
D. Logarithm and exponent rules:
exponent rules: log rules:
all quantities real) (base any positive real
number except 1)
ab • ac = ab+c log ab = log a + log b
ab
a c = ab−c log ab
= log a− log b
ab( )c = abc log ab = b log a
ab( )c = acbc loga ab = b
ab( )c = a c
b c aloga b( ) = b
a0 =1 (if a ≠ 0) loga b = logc b
logc a
a−b = 1
a b
ap
r = a pr = ar( )p (think of
p
r as
power
root) (base change rule)
15 Problems 51-52: Given log21024 =10 , find:
51. log210245 = 52. log2 1024 =
Problems 53-63: Solve for x in terms of y and z:
53. 3x = 3y • 3z 55. x 3 = y
54. 9y = 3z
3x 56. 3x = y
57. log x 2 = 3log y 58. log x = 2log y − log z 59. 3log x = log y 60. log x = log y + log z 61. log x + log y3 = log z 2 62. log7 3 = y; log72 = z; x = log32 63. y = loga 9; x = loga 3