Reciprocal Identities:

Quotient Identities:

Pythagorean Identities:

Double Angle Identities:

Logarithms:

Product property:

Quotient property:

Power property:

Property of equality:

Change of base formula:

Formula Sheet

1csc x = -.-

SlUX

1cotx=--tan x

1secx=--

cosx

SlUXtanx=--

cosxcosx

cotx=-.-SlUX

sin" x+ cos" x = I tan ' x + I= see" x -I+ cor' x = csc' X

sin2x = 2sinxcosx2 tan x

tan2x = 2I-tan x

cos2x = cos" x - sin" x= 1- Zsirr' x= Zcos" x-I

Y = log, x is equivalent to rx=a-

mlog, - = log, m - log, n

n

If log, m = log, n , then m = n

1log, n

og n=--a log, a

Slope-intercept form: y = mx + bPoint-slope form: y - Yl = m(x - Xl)Standard form: Ax + By + C = a

2. Solve lc:Jx= 2

3. Rewrite using a single logarithmic value for 08';35 + 1Qiglli

4. Rewrite using a single logarithmic value for lng:35 - In /j

Change the base to the common loqarithm for

5. ·'aBf = 6. ~aa;J.'f =

Complex Fractions

When simplifying complex fractions, multiply by a fraction equal to 1 which has a numerator and

denominator composed of the common denominator of all the denominators in the complex fraction.

Example:

_7 __ 6_x+l =5x+l

-2 3x-+--x x-4 =15---x-4

_7 __ 6_x+l x+l----,=--'----'--=- g--5 .r+Ix+l

-7x-7-65

=-7x-135

-2 3x-+--x x-4

15---x-4

-2x+ 8 + 3x2

5x2 - 20x-x =3x2 -2x+ 85x2 - 2lx

x(x- 4)g --'----'--x(x- 4)

-2(x-4)+3x(x)5(x)(x - 4) -lex)

= =

Simplify each of the following.25--a

7. --"a,,--_5+a

8.

42--x+2

5+~x+2

x 1 1-~----10. x+l x 11. 3x-4

x 1 32--+- x+--x+l x 3x-4

9.4-~2x-3155+--2x-3

Functions

To evaluate a function for a given value, simply plug the value into the function for x.

Recall: (f og JX) = f(g(x)) OR f[g(x)] read "fof gof x' Means to plug the inside function (in this

case g(x) ) in for x in the outside function (in this case, f(x».

Example: Given f(x) =2X2 + 1 and g(x) = x - 4 find f(g(x)).

f(g(x)) = f(x - 4)

= 2(x - 4i + 1= 2(x2 - 8x + 16) + 1=2x2-16x+32+1

f(g(x)) = 2X2 -16x + 33

Let f(x) = 2x + 1 and g(x) = 2X2 -l. Find each.

12. f(2) = _ 13. g(-3)= _ 14. f(t+l)= _

15. f[g(-2)J= __ -f(x + h) - f(x)17.

h16. g[f(m+2)J= _

Let f(x) = sinx Find each exactly.

19. f(23ll")= _

Let f(x) = X2, g(x) = 2x +5, and hex) = X2 -1. Find each.

20. h[f(-2)]= 21. f[g(X-1)]= __

Find f(x + h) - f(x) for the given function f.h

23. f(x)=9x+3 24. f(x)=5-2x

Intercepts and Points of Intersection

To find the x-intercepts, let y = 0 in your equation and solve.

To find the y-intercepts, let x = 0 in your equation and solve.

y- int. (Let x = 0)y = 02 - 2(0) - 3

y=-3

y - intercept (0, - 3)

Example: y = X2 - 2x - 3x - int. (Let y = 0)0= X2 -2x- 3

0= (x - 3)(x + 1)x=-lorx=3

x - intercepts (-1,0) and (3,0)

Find the x and y intercepts for each.

25. y= 2x-5

27.

26. Y =x2 +x-2

28. i =x3 -4x

Use substitution or elimination method to solve the system of equations.

Example:X2+Y -16x + 39 = 02 2 9 0x -y - =

(1st equation solved for y)

Plug what l is equalto into second equation.

(The rest is the same as

previous example)

Elimination Method

2x2-16x+30=0

X2 - 8x+ 15 = 0(x - 3)(x - 5) = 0x = 3 and x = 5Plug x = 3 and x = 5 into one original32 - y2 - 9 = 0 52 - i -9 = 0-i = 0 16 = y2

y=O y=±4

Points of Intersection (5,4), (5,-4) and (3,0)

Substitution Method

Solve one equation for one variable.

y2 =_X2 +16x-39

X2 _(_x2 + 16x- 39)- 9 = 0

2X2 -16x + 30 = 0X2 - 8x + 15 = 0(x - 3)(x - 5) = 0x = 3 or x - 5

Find the point(s) of intersection of the graphs for the given equations.x +y = 8 X2 +Y = 6 X2 - 4l - 20x - 64Y - 172= 0

29. 30. 31.4x - y = 7 x + y = 4 16x2 + 4l- 320x + 64y + 1600 = 0

Interval Notation

32. Complete the table with the appropriate notation or graph.

Solution Interval Notation Graph

-2 < x ~ 4

[-1,7)

~J-.8

Solve each equation. State your answer in BOTH interval notation and graphically.

33. 2x -120 34. -4 ~ 2x- 3 < 4 35. x x--->52 3

Domain and Range

Find the domain and range of each function. Write your answer in INTERVAL notation.

36. f(x) = X2 - 5 37. f(x) = -Jx + 3 38. f(x) = 3sinx 39. f(x) = _2_x-I

Inverses

To find the inverse of a function, simply switch the x and the y and solve for the new "y" value.

Example:

f(x) = ~x + 1 Rewrite f(x) as y

y = ~x + 1 Switch x and y

x = ~y+ 1 Solve for your new y

(x y = (~y+1) Cube both sides

x3 = y + 1 Simplify

y = x3 - 1 Solve for yf-I(X) = x3 -1 Rewrite in inverse notation

Find the inverse for each function.

40. f(x)=2x+l2

41. f(x)=~3

Also, recall that to PROVE one function is an inverse of another function, you need to show

that:

f(g(x)) = g(.f(x)) = x

Example:

x-9If: f(x) = -- and g(x) = 4x + 9 show f(x) and g(x) are inverses of each other.

4

(X-9)f(g(x)) = 4 -4- + 9 g(f(x))= (4X+:)-9

4x+ 9- 94

=x-9+9

=x4x4

=xf(g(x)) = g(.f(x)) = x therefore they are inverses

of each other.

Prove f and g are inverses of each other.

42. f(x) = x3

g(x) = ifb2

43. f(x)=9-x2,x?,O g(x)=.J9-x

Equation of a line

Point-slope form: y-y, = m(x-x,) Horizontal line: y = c (slope is 0)

Slope intercept form: y = mx + b Vertical line: x = c (slope is undefined)

44. Use slope-intercept form to find the equation of the line having a slope of 3 and a y-intercept of 5.

45. Determine the equation of a line passing through the point (5, -3) with an undefined slope.

46. Determine the equation of a line passing through the point (-4,2) with a slope of 0.

47. Use point-slope form to find the equation of the line passing through the point (0, 5) with a slope

of 2/3.

48. Find the equation of a line passing through the point (2, 8) and parallel to the line y = ~ x-I.6

49. Find the equation of a line perpendicular to the y- axis passing through the point (4, 7).

50. Find the equation of a line passing through the points (-3, 6) and (1,2).

51. Find the equation of a line with an x-intercept (2, 0) and a y-intercept (0, 3).

Radian and Degree Measure

U 180° t id f dl dse 0 get n 0 ra ians annradians

nradians .Use to get rid of degrees and

180°

convert to radians.convert to degrees.

52. Convert to degrees: a.5Jr6

4Jrb. c. 2.63

5

radians

53. Convert to radians: a. 45° b. -lr c.237"

Angles in Standard Position

54. Sketch the angle in standard position.l l zr6

SJrc.

3d. 1.8 radiansa.

Reference Triangles

55. Sketch the angle in standard position. Draw the reference triangle and label the sides, if

possible.

2a. -Jr b. 2250

3

c. d. 30'4

Unit Circle

You can determine the sine or cosine of a quadrantal angle by using the unit circle. The x-coordinate

of the circle is the cosine and the y-coordinate is the sine of the ;:mnll=~

(0,1)

Example: sin90° = 1 n V '\cos- = 0 (1,0)2 -1,0)

!" ./(0,-1)

56. a.) sin180° b.) cos270°

(0,1)

/' "'\-1,0) (1,0)-, /

(0,-1)

d.) sin zr

e.) cos360o f.) cos(-n)

Graphing Trig Functions

f(x) :'-sin (x)--- - -- 2

f(x) =-COS (x)

y = sin x and y = cos x have a period of 2 Jr and an amplitude of 1. Use the parent graphs above to

help you sketch a graph of the functions below. For f(x) = A sin(Bx + C) +K , A = amplitude, 2Jr =B

period,

C = phase shift (positive CIS shift left, negative CIS shift right) and K = vertical shift.B

Graph two complete periods of the function.

57. f(x) = 5sinx 58. f(x) = sin2x

59. f(x) = -cos(x - :) 60. f(x) = cosx - 3

Trigonometric Equations:

Solve each of the equations for 0::; x < 2:rr. Isolate the variable, sketch a reference triangle, find all

the solutions within the given domain, 0::; x < 2:rr. Remember to double the domain when solving for

a double angle. Use trig identities, if needed, to rewrite the trig functions. (See formula sheet at the

end of the packet.)

61.. 1smx=--

262. 2 cosx = 13

163. cos 2x = J2 64. . 7 1

Sllf" x =-2

65. sin2x = _ 132

66. 2cos2x-l-cosx=O

67. 4 cos ' x - 3 = 0 68. sin" x + cos2x - cosx = 0

Inverse Trigonometric Functions:

Recall: Inverse Trig Functions can be written in one of ways:

arcsin (x) sin " (x)

Inverse trig functions are defined only in the quadrants as indicated below due to their restricted

domains.

costx < 0

sini x c O

sin-1x > 0

cos' X > 0

tan' x > 0

tarr ' x< 0

Example:

Express the value of "y" in radians.-1

y = arctan J3 Draw a reference triangle.

~-1

This means the reference angle is 30° or Jr.6

-Jr Jr-<y<-2 2

So, y = - Jr so that it falls in the interval from6

stAnswer: y = - -

6

For each of the following, express the value for "y" in radians.

-13 ( )69. y = arcsin-2- 70. y = arccos \-1 71. y = arctan(-1)

Example: Find the value without a calculator.

cos( arctan~)

Draw the reference triangle in the correct quadrant first.

Find the missing side using Pythagorean Thm.

Find the ratio of the cosine of the reference triangle.

6cosB= --J6l

6

For each of the following give the value without a calculator.( 21 i . 121

72. tanl arccos}) 73. secl sm-I13)

. ( 12174. sml arctan5)

Vertical Asymptotes

Determine the vertical asymptotes for the function. Set the denominator equal to zero to find

the x-value for which the function is undefined. That will be the vertical asymptote.

176. f(x) = -2

X

2

77. f(x) = --;-x- -4

78. f(x)= ,2+xx-(l- x)

Horizontal Asymptotes

Determine the horizontal asymptotes using the three cases below.

Case I. Degree of the numerator is less than the degree of the denominator. The asymptote is

y = o.

Case II. Degree of the numerator is the same as the degree of the denominator. The asymptote is

the ratio of the lead coefficients.

Case III. Degree of the numerator is greater than the degree of the denominator. There is no

horizontal asymptote. The function increases without bound. (If the degree of the numerator is

exactly 1 more than the degree of the denominator, then there exists a slant asymptote, which is

determined by long division.)

Determine all Horizontal Asymptotes.

5x3 - 2X2 + 880. f(x) = ,

4x- 3x>+ 54x5

81. f(x)=-,-x--7

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